Class 12 - Maths - Determinants

Exercise 4.1

Evaluate the determinants in Exercises 1 and 2.

Question 1:

2           4

-5         -1

Given,   2         4

-5        -1

= 2 * (−1) – 4 * (−5)

= -2 + 20

= 18

Question 2:

(i)     cos θ    -sin θ                                   (ii)            x2 − x + 1          x - 1

sin θ      cos θ                                                      x + 1              x + 1

(i) Given,   cos θ       -sin θ

sin θ        cos θ

= cos θ * cos θ – (-sin θ) * sin θ

= cos2 θ + sin2 θ

= 1

(ii) Given   x2 − x + 1       x - 1

x + 1             x + 1

= (x2 − x + 1)(x + 1) − (x − 1)(x + 1)

= x3 − x2 + x + x2 − x + 1 − (x2 − 1)

= x3 + 1 − x2 + 1

= x3 − x2 + 2

Question 3:

If A =    1       2

4       2    , then show that |2A| = 4|A|

The given matrix is

A =    1       2

4       2

So, 2A = 2   1        2    =   2         4

4        2         8         4

LHS:

|2A| =    2        4

8        4

= 2 * 4 – 4 * 8

= 8 – 32

= -24

Now, |A| =    1        2

4        2

= 1 * 2 – 2 * 4

= 2 – 8

= -6

RHS:

4|A| = 4 * (-8) = -32

So, LHS = RHS

Question 4:

If           1      0      1

A =    0      1      2

0      0      4   , then show that |3A| = 27|A|

The give matrix is:

1      0      1

A =    0      1      2

0      0      4

It can be observed that in the first column, two entries are zero.

Thus, we expand along the first column (C1) for easier calculation.

Now, |A| =  1  1        2  - 0   0        1   + 0   0        1

0       4          0        4            1       2

=> |A| = 1(4 - 0) – 0 + 0

=> |A| = 4

So, 27|A| = 27 * 4 = 108   …………..1

1      0      1           3      0        3

3A =3    0      1      2    =    0       3        6

0      0      4           0       0       12

So, 3|A| = 3  3        6  - 0   0       3     + 0   0        3

0      12         0      12            3        6

=> 3|A| = 3(36 - 0) – 0 + 0

=> 3|A| = 108   ……….2

From equations 1 and 2, we get

|3A| = 27|A|

Hence, the given result is proved.

Question 5:

Evaluate the determinants

(i)        3        -1         -2         (ii)    3          -4         5

0         0         -1                  1           1        -2

3        -5          0                  2           3         1

(iii)      0          1           2        (iv)   2          -1        -2

-1         0         -3                 0           2        -1

-2         3          0                 3          -5         0

(i) Let

3        -1         -2

A =    0         0         -1

3        -5          0

It can be observed that in the second row, two entries are zero.

Thus, we expand along the second row for easier calculation.

|A| =  0  -1     -2  + 0  3      -2   - (-1)  3        -1

-5      0          3        0             3        -5

=> |A| = 0(0 - 10) + 0(0 + 6) + 1(-15 + 3)

=> |A| = -12

(ii) Let

3        -4          5

A =    1         1         -2

2         3          1

By expanding along the first row, we have:

|A| =  3  1      -2   + 4 1       -2    +   5   1       1

3        1          2       1               2      3

=> |A| = 3(1 + 6) + 4(1 + 4) + 5(3 - 2)

=> |A| = 21 + 20 + 5

=> |A| = 46

(iii) Let

0         1           2

A =    -1        0         -3

-2        3           0

By expanding along the first row, we have:

|A| =  0  0     -3    - 1  -1     -3    +   2   -1      0

3       0           -2      0              -2      3

=> |A| = 0(0 + 9) - 1(0 - 6) + 2(-3 + 0)

=> |A| = 6 - 6

=> |A| = 0

(iv) Let

2         -1         -2

A =    0          2         -1

3         -5           0

By expanding along the first column, we have:

|A| =  2  2       -1  - 0  -1     -2    +   3   -1     -2

-5       0         -5      0               2       -1

=> |A| = 2(0 - 5) - 0 + 3(1 + 4)

=> |A| = -10 + 15

=> |A| = 5

Question 6:

If           1      1       -2

A =    2      1       -3

5      4       -9     , then find |A|.

The given matrix is:

1      1       -2

A =    2      1       -3

5      4       -9

By expanding along the first row, we have:

|A| =  1  1      -3  - 1  2       -3    -2   2       1

4      -9         5       -9          5       4

=> |A| = 1(-9 + 12) – 1(-18 + 15) - 2(8 - 5)

=> |A| = 3 + 3 - 6

=> |A| = 6 – 6

=> |A| = 0

Question 7:

Find the value of x, if

(i)      2          4    =   2x          4            (ii)    2            3    =    x            3

2          1          6           x                     4            5          2x          5

(i) Given,

2          4    =   2x          4

2          1          6           x

=> 2 * 1 – 5 * 4 = 2x * x – 6 * 4

=> 2 – 20 = 2x2 - 24

=> -18 = 2x2 - 24

=> 2x2 = 24 – 18

=> 2x2 = 6

=> x2 = 3

=> x = ±√3

So, the value of x is ±√3

(ii) Given,

2          3    =    x           3

4          5         2x          5

=> 2 * 5 – 3 * 4 = x * 5 – 3 * 2x

=> 10 – 12 = 5x – 6x

=> -2 = -x

=> x = 2

So, the value of x is 2

Question 8:

If     x            2    =   6            2

18          x         18          6 , then x is equal to

(A) 6                                    (B) ±6                                (C) −6                                 (D) 0

Given,

x           2    =  6            2

18         x         18          6

=> x * x – 2 * 18 = 6 * 6 – 18 * 2

=> x2  = 36

=> x = ±6

Hence, the correct answer is B.

Exercise 4.2

Question 1:

Using the property of determinants and without expanding, prove that:

x             a             x + a

y             b            y + b

z             c             z + c

Given,

x             a             x + a              x             a             x            x             a              a

y             b            y + b     =         y             b            y     +     y             b              b

z             c             z + c               z              c            z            z              c              c

= 0 + 0    [Since two columns of the determinant are identical]

= 0

Question 2:

Using the property of determinants and without expanding, prove that:

a - b       b - c        c - a

b - c       c-  a        a - b   = 0

c - a       a - b        b - c

Let

a - b       b - c        c - a

Δ =    b - c       c-  a        a - b

c - a       a - b        b - c

Applying R1 -> R1 + R2, we get

a - c              b - a           c - a

Δ =    b - c              c-  a            a - b

-(a – c)      -(b – a)        -(c – b)

a - c              b - a           c - a

Δ =-    b - c              c-  a            a - b

a – c              b – a           c – b

Here, the two rows R1 and R3 are identical.

So, ∆ = 0

Question 3:

Using the property of determinants and without expanding, prove that:

2        7          65

3        8          75    = 0

5        9          86

Given,

2        7          65          2          7          63 + 2

3        8          75    =    3          8          72 + 3

5        9          86          5           9          81 + 5

=     2          7           63             2         7            2

3          8           72     +      3         8            3

5          9           81             5         9            5

=     2          7           63

3          8           72   + 0           [Since two columns are identical]

5          9           81

=     2          7         9 * 7

3          8         9 * 8

5          9         9 * 9

= 9    2          7         7

3          8         8                 [Since two columns are identical]

5          9         9

= 9 * 0

= 0

Question 4:

Using the property of determinants and without expanding, prove that:

1        bc          a(b + c)

1        ca          b(c + a)    = 0

1        ab          c(a + b)

Given,

1        bc          a(b + c)

1        ca          b(c + a)

1        ab          c(a + b)

Applying C3 -> C3 + C2, we get

=   1        bc          ab + bc + ca

1        ca           ab + bc + ca

1        ab           ab + bc + ca

= (ab + bc + ca)  1            bc             1

1            ca             1

1            ab            1

= (ab + bc + ca) * 0                          [Since two columns are identical]

= 0

Question 5:

Using the property of determinants and without expanding, prove that:

b + c       q + r        y + z           a          p           x

c + a       r + p        z + x   = 2   b          q          y

a + b       p + q       x + y           c           r          z

Let

b + c       q + r        y + z

Δ =     c + a       r + p        z + x

a + b       p + q       x + y

b + c       q + r        y + z          b + c       q + r        y + z

Δ =     c + a       r + p        z + x    +     c + a       r + p        z + x

a              p              x                 b             q              x

Δ = Δ1 + Δ2     ………….1

Now,

b + c       q + r        y + z

Δ1 =   c + a       r + p        z + x

a              p              x

Applying R2 -> R2 + R3, we get

b + c       q + r        y + z

Δ1 =       c             r               z

a              p              x

Applying R1 -> R1 – R2, we get

b       q         y

Δ1 =    c        r         z

a        p        x

Applying R1 R3 and R2 R3, we get

a        p         x

Δ1 = (-1)2    b        q         y

c         r         z

a        p         x

Δ1 =     b        q         y

c         r         z         ………….2

Now,

b + c       q + r        y + z

Δ2 =   c + a       r + p        z + x

b              q              y

Applying R1 -> R1 - R3, we get

c             r              z

Δ2 =    c + a       r + p        z + x

a              p              x

Applying R2 -> R2 – R1, we get

c         r         z

Δ2 =    a        p         x

b        q         y

Applying R1 R2 and R2 R3, we get

a        p         x

Δ2 = (-1)2    b        q         y

c         r         z

a        p         x

Δ2 =     b        q         y

c         r         z         ………….3

From equations 1, 2 and 3, we get

a        p         x

Δ = 2   b        q         y

c         r         z

Hence, the given result is proved.

Question 6:

By using the property of determinants, show that:

0        a           -b

-a        0           -c    = 0

b        c             0

Given,

0       a        -b

Δ  =  -a       0        -c

b       c          0

Applying R1 -> cR1, we get

0         ac         -bc

Δ = (1/c)     -a          0           -c

b           c             0

Applying R1 -> R1 – bR2, we get

ab         ac            0

Δ = (1/c)     -a          0           -c

b           c             0

b            c            0

Δ = (a/c)     -a          0           -c

b           c             0

Δ = (a/c) * 0                             [Since two rows are identical]

Δ = 0

Question 7:

By using the property of determinants, show that:

-a2        ab        ac

ba        -b2       bc    = 4a2b2c2

ca         cb        -c2

Given,

-a2        ab        ac

Δ  =   ba        -b2       bc

ca         cb        -c2

-a          b             c

Δ = abc   a          -b             c

a            b           -c     [taking out factors a, b, c from R1, R2 and R3]

-1          1             1

Δ = a2b2c2   1          -1             1

1            1           -1     [taking out factors a, b, c from C1, C2 and C3]

Applying R2 -> R2 + R1 and R3 -> R3 + R1, we get

-1          1             1

Δ = a2b2c2   0           0             2

0           2             0

Δ = a2b2c2 *(-1)  0           2

• 0

Δ = -a2b2c2 (0 - 4)

Δ = 4a2b2c2

Hence, the given result is proved.

Question 8:

By using the property of determinants, show that:

(i)    1          a        a2

1          b       b2    = (a - b)(b - c)(c - a)

1          c        c2

(ii)   1          1         1

a          b        c    = (a - b)(b - c)(c - a)(a + b + c)

a3        b3        c3

(i) Given,

1          a        a2

Δ =    1          b       b2

1          c        c2

Applying R1 -> R1 – R3 and R2 -> R2 – R3, we get

0        a - c        a2 -  c2

Δ =   0        b - c       b2 – c2

1          c              c2

0        -1        -(a + c)

Δ = (b - c)(c - a)  0         1          b + c

1          c           c2

Applying R1 -> R1 + R2, we get

0         0        -a + b

Δ = (b - c)(c - a)  0         1          b + c

1         c           c2

0         0             -1

Δ = (b - c)(c - a)(c - a)  0         1          b + c

1         c           c2

Expanding along C1, we get

Δ = (b - c)(c - a)(c - a)  0              -1

1           b + c

Δ = (b - c)(c - a)(c - a)(0 + 1)

Δ = (b - c)(c - a)(c - a)

Hence, the given result is proved.

(ii) Given,

1          1         1

Δ =   a           b         c

a2        b2        c2

Applying C1 -> C1 – C3 and C2 -> C2 – C3, we get

0                  0            1

Δ =   a - c           b - c         c

a3 –c3       b3 –c3       c3

0                                                        0                                  1

Δ =   a - c                                                 b - c                               c

(a - c)(a2 + b2 + ab)       (b - c)(b2 + c2 + bc)                      c3

0                                            0                           1

Δ = (c - a)(b - c)   -1                                           1                          c

-(a2 + b2 + ab)                (b2 + c2 + bc)             c3

Applying C1 -> C1 + C2, we get

0                                            0                           1

Δ = (c - a)(b - c)   0                                           1                           c

(b2 - a2) + (bc - ac)        (b2 + c2 + bc)             c3

0                                            0                           1

Δ = (c - a)(b - c)(a - b)   0                                           1                           c

-(a + b + c)                     (b2 + c2 + bc)             c3

0                        0                           1

Δ = (c - a)(b - c)(a - b)(a + b + c)   0                        1                           c

-1                 (b2 + c2 + bc)             c3

Expanding along C1, we get

Δ = (c - a)(b - c)(a - b)(a + b + c)(-1)   0           1

1            c

Δ = (c - a)(b - c)(a - b)(a + b + c)(-1)(0 - 1)

Δ = (c - a)(b - c)(a - b)(a + b + c)

Hence, the given result is proved.

Question 9:

By using the property of determinants, show that:

x          x2        yz

y          y2       zx   = (x - y)(y - z)(z - x)(xy + yz + zx)

z          z2        xy

Let

x          x2        yz

Δ =   y          y2        zx

z          z2        xy

Applying R2 -> R2 – R1 and R3 -> R3 – R1, we get

x                x2            yz

Δ =   y - x       y2 – x2    zx - yz

z - x       z2 – x2    xy - yz

x                     x2                   yz

Δ =   y - x       -(x - y)(x + y)    z(x – y)

z - x        (z - x)(z + x)    -y(z – x)

x                x2           yz

Δ = (x - y)(z - x)  -1            -x - y           z

1             z + x          -y

Applying R3 -> R3 + R2, we get

x                x2           yz

Δ = (x - y)(z - x)  -1            -x - y           z

0             z - y         z - y

x                x2            yz

Δ = (x - y)(z - x)(z - y)  -1            -x - y            z

0               1               1

Expanding along R3, we get

Δ = [(x - y)(z - x)(z - y)]   (-1)   x         yz   + 1  x            x2

-1        z            -1       -x - y

Δ = [(x - y)(z - x)(z - y)][-xz – yz + (-x2 – xy + x2)]

Δ = -(x - y)(z - x)(z - y)(xy + yz + zx)

Δ = (x - y)(y - z)(z - x)(xy + yz + zx)

Hence, the given result is proved.

Question 10:

By using properties of determinants, show that:

(i)      x + 4         2x             2x

2x            x + 4         2x   = (5x + 4)(4 - x)2

2x             2x          x + 4

(ii)     y + k          y               y

y            y + k             y   = k2(3y + k)

y               y          y + k

(i) Let

x + 4         2x             2x

Δ =   2x            x + 4         2x

2x             2x          x + 4

Applying R1 → R1 + R2 + R3, we get

5x + 4      5x + 4      5x + 4

Δ =   2x            x + 4           2x

2x             2x             x + 4

1                1                1

Δ = (5x + 4)  2x            x + 4           2x

2x             2x             x + 4

Applying C2 → C2 − C1 and C3 → C3 − C1, we get

1                0                  0

Δ = (5x + 4)  2x          -x + 4              0

2x              0             -x + 4

1             0             0

Δ = (5x + 4)(4 - x)(4 - x)  2x           1             0

2x            0            1

Expanding along C3, we get

Δ = (5x + 4)(4 - x)(4 - x)  1             0

2x           1

Δ = (5x + 4)(4 - x)(4 - x)(1 - 0)

Δ = (5x + 4)(4 - x)2

Hence, the given result is proved.

(ii) Let

y + k         y             y

Δ =   y            y + k         y

y               y          y + k

Applying R1 → R1 + R2 + R3, we get

3y + k      3y + k      3y + k

Δ =   y                y + k            y

y                  y             y + k

Applying C2 → C2 − C1 and C3 → C3 − C1, we get

1           0             0

Δ = (3y + k)  y            k            0

y            0            k

1           0             0

Δ = k2(3y + k)   y            1            0

y            0            1

Expanding along C3, we get

Δ = k2(3y + k)   1             0

y              1

Δ = k2(3y + k)(1 - 0)

Δ = k2(3y + k)

Hence, the given result is proved.

Question 11:

By using properties of determinants, show that:

(i)      a – b - c         2a                  2a

2b               b – c - a            2b     = (a + b + c)2

2c                   2c           c – a - b

(ii)     x + y + 2z              x                              y

z                      y + z + 2x                      y   = 2(x + y + z)2

z                            x                z + x + 2y

(i) Let

a – b - c         2a                      2a

Δ =   2b               b – c - a                2b

2c                   2c             c – a - b

Applying R1 → R1 + R2 + R3, we get

a + b + c        a + b + c              a + b + c

Δ =   2b                  b – c - a                      2b

2c                       2c                     c – a - b

1                         1                              1

Δ = (a + b + c)  2b                  b – c - a                      2b

2c                       2c                     c – a - b

Applying C2 → C2 − C1 and C3 → C3 − C1, we get

1                            0                               0

Δ = (a + b + c)  2b                  -(a + b + c)                       0

2c                           0               -(a + b + c)

1               0                0

Δ = (a + b + c)3  2b            -1                0

2c             0               -1

Expanding along C3, we get

Δ = (a + b + c)3[-1 * (-1) - 0]

Δ = (a + b + c)3

Hence, the given result is proved.

(ii) Let

x + y + 2z              x                              y

Δ =    z                      y + z + 2x                      y

z                            x                z + x + 2y

Applying C1 → C1 + C2 + C3, we get

2(x + y + z)                x                              y

Δ =    2(x + y + z)           y + z + 2x                      y

2(x + y + z)                x                z + x + 2y

1                x                              y

Δ = 2(x + y + z)    1          y + z + 2x                      y

1                x                   z + x + 2y

Applying R2 → R2 − R1 and R3 → R3 − R1, we get

1                x                           y

Δ = 2(x + y + z)    0          x + y + z                     0

0                0                   x + y + z

1           x            y

Δ = 2(x + y + z)3   0           1           0

0           0           1

Expanding along R3, we get

Δ = 2(x + y + z)3(1 - 0)

Δ = 2(x + y + z)3

Hence, the given result is proved.

Question 12:

By using the property of determinants, show that:

1               x            x2

x2              1            x    = (1 – x3)2

x                x2          1

Let

1                x            x2

Δ =   x2              1            x

x                x2          1

Applying R1 → R1 + R2 + R3, we get

1 + x + x2     1 + x + x       1 + x + x2

Δ =   x2                       1                        x

x                        x2                       1

1                 1                1

Δ = (1 + x + x2)  x2                1                x

x                  x2               1

Applying C2 → C2 − C1 and C3 → C3 − C1, we get

1             0                0

Δ = (1 + x + x2)  x2          1 - x2          x – x2

x            x2 - x          1 - x

1             0             0

Δ = (1 + x + x2)(1 - x)(1 - x)  x2          1 + x        x

x             - x          1

1             0             0

Δ = (1 – x3)(1 - x)  x2          1 + x        x

x             - x          1

Expanding along R1, we get

Δ = (1 – x3)(1 - x)[1(1 + x) – x * (-x)]

Δ = (1 – x3)(1 - x)[1 + x + x2]

Δ = (1 – x3)(1 – x3)

Δ = (1 – x3)2

Hence, the given result is proved.

Question 13:

By using properties of determinants, show that:

1 + a2 – b2          2ab                      -2b

2ab                 1 - a2 + b2                 2a      = (1 + a2 + b2)3

2b                      -2a               1 - a2 – b2

Let

1 + a2 – b2         2ab                      -2b

Δ =     2ab                1 - a2 + b2                 2a

2b                       -2a               1 - a2 – b2

Applying R1 → R1 + bR3 and R2 → R2 - aR3, we get

1 + a2 + b2           0                 -b(1 + a2 + b2)

Δ =    0                     1 + a2 + b2        a(1 + a2 + b2)

2b                      -2a                     1 - a2 – b2

1               0                      -b

Δ = (1 + a2 + b2)2   0               1                       a

2b          -2a         1 - a2 – b2

Expanding along R1, we get

Δ = (1 + a2 + b2)2    1   1                      a        - b   0             1

-2a         1 - a2 – b2          2b         -2a

Δ = (1 + a2 + b2)2 [(1 - a2 – b2 + 2a2) – b(0 – 2b)]

Δ = (1 + a2 + b2)2 (1 + a2 + b2)

Δ = (1 + a2 + b2)3

Question 14:

By using the property of determinants, show that:

a2 + 1       ab        ac

ab         b2 + 1     bc       = 1 + a2 + b2 + c2

ca            cb       c2 + 1

Let

a2 + 1       ab          ac

Δ =   ab         b2 + 1        bc

ca            cb       c2 + 1

Taking out common factors a, b, and c from R1, R2, and R3 respectively, we have:

a + 1/a           b               c

Δ = abc  a                b + 1/b          c

a                     c           c + 1/c

Applying R2 -> R2 – R1 and R3 -> R3 – R1, we get

a + 1/a           b             c

Δ = abc  -1/a               1/b          c

-1/a                 0           1/c

Applying C1 -> aC1, C2 -> bC2 and C3 -> cC3, we get

a2 + 1        b2          c2

Δ = abc * 1/abc  -1               1            0

-1              0            1

a2 + 1       b2         c2

Δ =   -1             1           0

-1             0           1

Expanding along R3, we get

Δ = (-1)[ b2 * 0 – 1 * c2] + 0 + 1[a2 + 1 + b2]

Δ = c2 + a2 + 1 + b2

Δ = 1 + a2 + b2 + c2

Hence, the given result is proved.

Question 15:

Let A be a square matrix of order 3 * 3, then |kA| is equal to

1. k|A| B. k2|A| C. k3|A|                               D. 3k|A|

A is a square matrix of order 3 * 3.

Let

a1        a2      a3

A =   b1        b2      b3

c1        c2      c3

Then

ka1        ka2      ka3

kA =   kb1        kb2     kb3

kc1        kc2      kc3

Now,

ka1        ka2      ka3

|kA| =   kb1        kb2     kb3

kc1        kc2      kc3

a1        a2      a3

|kA| = (k * k * k)     b1        b2     b3

c1        c2      c3           [Taking out common factor k from each row]

a1        a2      a3

|kA| = k3  b1        b2     b3

c1        c2      c3

=> kA| = k3|A|

Hence, the correct answer is C.

Question 16:

Which of the following is correct?

1. Determinant is a square matrix.
2. Determinant is a number associated to a matrix.
3. Determinant is a number associated to a square matrix.

We know that to every square matrix A = [aij] of order n. We can associate a number called the

determinant of square matrix A, where aij = (i, j)th element of A.

Thus, the determinant is a number associated to a square matrix.

Hence, the correct answer is option C.

Exercise 4.3

Question 1:

Find area of the triangle with vertices at the point given in each of the following:

(i) (1, 0), (6, 0), (4, 3)                       (ii) (2, 7), (1, 1), (10, 8)                 (iii) (−2, −3), (3, 2), (−1, −8)

(i) The area of the triangle with vertices (1, 0), (6, 0), (4, 3) is given by the relation,

1        0         1

Δ = (1/2)   6        0          1

4        3         1

=> Δ = (1/2)[1(0 - 3) – 0(6 - 4) + 1(18 - 0)]

=> Δ = (1/2)[-3 + 18]

=> Δ = 15/2

Hence, area of the triangle is 15/2 square units.

(ii) The area of the triangle with vertices (2, 7), (1, 1), (10, 8) is given by the relation,

2        7         1

Δ = (1/2)   1        1          1

10      8         1

=> Δ = (1/2)[2(1 - 8) – 7(1 - 10) + 1(8 - 10)]

=> Δ = (1/2)[-14 + 63 - 2]

=> Δ = (1/2)[-16 + 63]

=> Δ = 47/2

Hence, area of the triangle is 47/2 square units.

(iii)The area of the triangle with vertices (−2, −3), (3, 2), (−1, −8) is given by the relation,

-3       -3        1

Δ = (1/2)   3         2         1

-1      -8         1

=> Δ = (1/2)[-2(2 + 8) + 3(3 + 1) + 1(-24 + 2)]

=> Δ = (1/2)[-20 + 12 - 22]

=> Δ = (1/2)[-42 + 12]

=> Δ = -30/2

=> Δ = -15

Hence, area of the triangle is |-15| = 15 square units.

Question 2:

Show that points A(a, b + c), B(b, c + a), C(c, a + b) are collinear.

Area of ∆ABC is given by the relation,

a        b + c         1

Δ = (1/2)   b        c + a          1

c        a + b         1

Applying R2 -> R2 – R1 and R3 -> R3 – R1

a           b + c         1

Δ = (1/2)   b - a      a - b          1

c - a      a - c          1

a        b + c         1

Δ = (1/2) * (a - b) * (c - a)   -1          1            1

1          -1           1

Applying R3 -> R3 + R2

a        b + c         1

Δ = (1/2) * (a - b) * (c - a)   -1          1            0

0           0            0

Δ = (1/2) * (a - b) * (c - a) * 0              [Since all elements of R3 are 0]

Δ = 0

Since the area of the triangle formed by points A, B, and C is zero.

Hence, the points A, B, and C are collinear.

Question 3:

Find values of k if area of triangle is 4 square units and vertices are

(i) (k, 0), (4, 0), (0, 2)                                         (ii) (−2, 0), (0, 4), (0, k)

We know that the area of a triangle whose vertices are (x1, y1), (x2, y2), and (x3, y3) is the

absolute value of the determinant (∆), where

x1          y1          1

Δ = (1/2)   x2          y2          1

X3          y3          1

It is given that the area of triangle is 4 square units.

So, ∆ = ±4

(i) The area of the triangle with vertices (k, 0), (4, 0), (0, 2) is given by the relation,

k         0        1

Δ = (1/2)   4         0         1

0         2         1

=> ±4 = (1/2)[k(0 - 2) - 0(4 - 0) + 1(8 - 0)]

=> ±4 = (1/2)[-2k + 8]

=> ±4 = -k + 4

=> -k + 4 = ±4

When −k + 4 = − 4, then k = 8

When −k + 4 = 4, then k = 0

Hence, k = 0, 8

(ii) The area of the triangle with vertices (−2, 0), (0, 4), (0, k) is given by the relation,

-2         0       1

Δ = (1/2)   0         4         1

0         k         1

=> ±4 = (1/2)[-2(4 - k) - 0(0 - 0) + 1(0 - 0)]

=> ±4 = (1/2)[-8 +2k]

=> ±4 = k – 4

=> k – 4 = ±4

When k − 4 = − 4, then k = 0

When k − 4 = 4, then k = 8

Hence, k = 0, 8

Question 4:

(i) Find equation of line joining (1, 2) and (3, 6) using determinants.

(ii) Find equation of line joining (3, 1) and (9, 3) using determinants.

(i) Let P (x, y) be any point on the line joining points A (1, 2) and B (3, 6). Then, the points A, B,

and P are collinear. Therefore, the area of triangle ABP will be zero.

1         0         1

=>   (1/2)   3         6        1   = 0

x         y         1

=> (1/2)[1(6 - y) - 2(3 - x) + 1(3y – 6x)] = 0

=> 6 – y – 6 + 2x + 3y – 6x = 0

=> 2y – 4x = 0

=> y – 2x = 0

=> y = 2x

Hence, the equation of the line joining the given points is y = 2x.

(ii) Let P (x, y) be any point on the line joining points A (3, 1) and B (9, 3). Then, the points A, B,

and P are collinear. Therefore, the area of triangle ABP will be zero.

3         1         1

=>   (1/2)   9         3        1   = 0

x         y         1

=> (1/2)[3(3 - y) - 1(9 - x) + 1(9y – 3x)] = 0

=> 9 – 3y – 9 + x + 9y – 3x = 0

=> 6y – 2x = 0

=> x – 3y = 0

=> y = 2x

Hence, the equation of the line joining the given points is x − 3y = 0.

Question 5:

If the area of triangle is 35 square units with vertices (2, −6), (5, 4) and (k, 4). Then k is

1. 12 B. −2 C. −12, −2                             D. 12, −2

The area of the triangle with vertices (2, −6), (5, 4), and (k, 4) is given by the relation,

2       -6        1

Δ = (1/2)   5         4         1

k         4         1

=> Δ = (1/2)[2(4 - 4) + 6(5 - k) + 1(20 – 4k)]

=> Δ = (1/2)[30 – 6k + 20 – 4k]

=> Δ = (1/2)[50 – 10k]

=> Δ = 25 – 5k

It is given that the area of the triangle is ±35.

Therefore, we have:

=> 25 – 5k = ±35

=> 5(5 – k) = ±35

=> 5 – k = ±7

When 5 − k = −7, then k = 5 + 7 = 12

When 5 − k = 7, then k = 5 − 7 = −2

So, k = 12, −2

Hence, the correct answer is option D.

Exercise 4.4

Write Minors and Cofactors of the elements of following determinants:

Question 1:

(i)    2         -4                            (ii)    a            c

0           3                                     b           d

(i) The given determinant is

2         -4

0           3

Minor of element aij is Mij

M11 = minor of element a11 = 3

M12 = minor of element a12 = 0

M21 = minor of element a21 = −4

M22 = minor of element a22 = 2

Cofactor of aij is Aij = (−1)i + j Mij

A11 = (−1)1+1 M11 = (−1)2 * 3 = 3

A12 = (−1)1+2 M12 = (−1)3 * 0 = 0

A21 = (−1)2+1 M21 = (−1)3 * (−4) = 4

A22 = (−1)2+2 M22 = (−1)4 * (2) = 2

(ii) The given determinant is

a           c

b           d

Minor of element aij is Mij

M11 = minor of element a11 = d

M12 = minor of element a12 = b

M21 = minor of element a21 = c

M22 = minor of element a22 = a

Cofactor of aij is Aij = (−1)i + j Mij

A11 = (−1)1+1 M11 = (−1)2 * d = d

A12 = (−1)1+2 M12 = (−1)3 * b = -b

A21 = (−1)2+1 M21 = (−1)3 * c = -c

A22 = (−1)2+2 M22 = (−1)4 * a = a

Question 2:

(i)    1          0         0                   (ii)   1          0          4

0           1         0                          3           5        -1

0           0         1                          0           1          2

(i) The give matrix is:

1          0         0

0           1         0

0           0         1

By the definition of minors and cofactors, we have:

M11 = minor of a11 =   1       0   = 1

0       1

M12 = minor of a12 =   0       0   = 0

0       1

M13 = minor of a13 =   0       0   = 0

0       1

M21 = minor of a21 =   0       0   = 0

0       1

M22 = minor of a22 =   1       0   = 1

0       1

M23 = minor of a23 =   1       0   = 0

0       0

M31 = minor of a31 =   0       0   = 0

1       0

M32 = minor of a32 =   1       0   = 0

0       0

M33 = minor of a33 =   1       0   = 1

0       1

A11 = cofactor of a11 = (-1)1 + 1 M11 = 1

A12 = cofactor of a12 = (-1)1 + 2 M12 = 0

A13 = cofactor of a13 = (-1)1 + 3 M13 = 0

A21 = cofactor of a21 = (-1)2 + 1 M21 = 0

A22 = cofactor of a22 = (-1)2 + 2 M22 = 1

A23 = cofactor of a23 = (-1)2 + 3 M23 = 0

A31 = cofactor of a31 = (-1)3 + 1 M31 = 0

A32 = cofactor of a32 = (-1)3 + 2 M32 = 0

A33 = cofactor of a33 = (-1)3 + 3 M33 = 1

(ii) The give matrix is:

1          0         4

3           5        -1

0           1         2

By the definition of minors and cofactors, we have:

M11 = minor of a11 =   5      -1   = 10 + 1 = 11

1       2

M12 = minor of a12 =   3      -1   = 6 – 0 = 6

0       2

M13 = minor of a13 =   3       5   = 3 – 0 = 3

0       1

M21 = minor of a21 =   0       4   = 0 – 4 = -4

1       2

M22 = minor of a22 =   1       4   = 2 – 0 = 2

0       2

M23 = minor of a23 =   1       0   = 1 – 0 = 1

0       1

M31 = minor of a31 =   0       4   = 0 – 20 = -20

5      -1

M32 = minor of a32 =   1       4   = -1 – 12 = -13

3      -1

M33 = minor of a33 =   1       0   = 5 – 0 = 5

3       5

A11 = cofactor of a11 = (-1)1 + 1 M11 = 11

A12 = cofactor of a12 = (-1)1 + 2 M12 = -6

A13 = cofactor of a13 = (-1)1 + 3 M13 = 3

A21 = cofactor of a21 = (-1)2 + 1 M21 = 4

A22 = cofactor of a22 = (-1)2 + 2 M22 = 2

A23 = cofactor of a23 = (-1)2 + 3 M23 = -1

A31 = cofactor of a31 = (-1)3 + 1 M31 = -20

A32 = cofactor of a32 = (-1)3 + 2 M32 = 13

A33 = cofactor of a33 = (-1)3 + 3 M33 = 5

Question 3:

Using Cofactors of elements of second row, evaluate

5         2          8

Δ =     2         0          1

1          2         3

The given determinant is:

5         2          8

2         0          1

1          2         3

We have:

M21 = minor of a21 =   3       8   = 9 – 16 = -7

2       3

So, A21 = cofactor of a21 = (−1)2+1 M21 = 7

M21 = minor of a21 =   5       8   = 15 – 8 = 7

1       3

So, A22 = cofactor of a22 = (−1)2+2 M22 = 7

M23 = minor of a23 =   5       1   = 10 – 3 = 7

1       2

So, A22 = cofactor of a23 = (−1)2+3 M21 = -7

We know that ∆ is equal to the sum of the product of the elements of the second row with

their corresponding cofactors.

∆ = a21A21 + a22A22 + a23A23

= 2 * 7 + 0 * 7 + 1 * (−7)

= 14 − 7

= 7

Question 4:

Using Cofactors of elements of third column, evaluate

5         2          8

Δ =     2         0          1

1          2         3

The given determinant is:

5         2          8

2         0          1

1          2         3

We have:

M13 = minor of a21 =   1       y   = z – y

1       z

So, A13 = cofactor of a13 = (−1)1+3 M13 = z - y

M23 = minor of a23 =   1       x   = z - x

1       z

So, A23 = cofactor of a23 = (−1)2+3 M23 = -(z - x) = x - z

M33 = minor of a33 =   1       x   = y - x

1       y

So, A33 = cofactor of a33 = (−1)3+3 M33 = y - x

We know that ∆ is equal to the sum of the product of the elements of the second row with

their corresponding cofactors.

∆ = a13A13 + a23A23 + a33A33

= yz(z - y) + zx(x - z) + xy(y - x)

= yz2 – y2z + x2z – xz2 + xy2 – x2y

= (x2z – y2z) + (yz2 – xz2) + (xy2 – x2y)

= z(x2 – y2) - z2 (x – y) - xy(x – y)

= z(x – y)(x + y) - z2(x – y) - xy(x – y)

= (x - y)[z(x + y) - z2 - xy]

= (x - y)[zx + zy - z2 - xy]

= (x - y)[zx - z2 + zy - xy]

= (x - y)[z(x – z) + y(z – x)]

= (x - y)[-z(z – x) + y(z – x)]

= (x - y)(y - z)(z - x)

So, Δ = (x - y)(y - z)(z - x)

Question 5:

For the matrices A and B, verify that (AB)′ = B’A’where

(i) A =    1

-4   , B = [-1       2       1]

3

(ii) A =   0

1   , B = [1       5       7]

2

(i) AB =     1                                   -1          2          1

-4   [-1       2       1] =    4          -8         -4

3                                    -3         6          3

So, (AB)’ =   -1         4          -3

2         -8           6

1         -4          3

Now, A’ = [1       -4         3], B’ =    -1

2

1

So, B’A’ =    -1                                   -1          4        -3

2   [1       -4       3] =     3          -8        6

1                                    1          -4         3

Hence, it is verified that (AB)’ = B’A’

(ii) AB =    0                                    0          0          0

1    [1       5       7] =     1           5          7

2                                   2          10       14

So, (AB)’ =   0          1          2

0         5          10

0         7          14

Now, A’ = [0        1         2], B’ =     1

5

7

So, B’A’ =     1                                   0          1           2

5   [0       1       2] =     0          5          10

7                                    0         7         14

Hence, it is verified that (AB)’ = B’A’

Exercise 4.5

Find adjoint of each of the matrices in Exercises 1 and 2.

Question 1:

1            2

3            4

Let

A =   1            2

3            4

Now,

A11 = 4, A12 = -3, A21 = -2, A22 = 1

So, adj(A) =   A11          A21   =    4           -2

A12          A22         -3           1

Question 2:

1           -1             2

2             3             5

-2            0             1

Let

A =     1           -1             2

2             3             5

-2            0             1

Now,

A11 =    3            5   = 3 – 0 = 3

0            1

A12 = -   2           5   = -(2 + 10) = -12

-2           1

A13 =    2            3   = 0 + 6 = 6

-2           0

A21 = -   -1          2   = -(-1 - 0) = 1

0           1

A22 =    1            2   = 1 + 4 = 5

-2           1

A23 = -   1          -1   = -(0 - 2) = 2

-2          0

A31 =   -1           2   = -5 – 6 = -11

3            5

A32 = -   1           2   = -(5 - 4) = -1

2           5

A33 =    1           -1   = 3 + 2 = 5

2            3

Hence, adj(A) =     A11         A21        A31              3            1          -11

A12         A22        A32     =     -12         5             -1

A13         A23        A33              6           2              5

Verify A * (adj A) = (adj A) * A = | A | * I in Exercises 3 and 4.

Question 3:

2            3

-4          -6

Let

A =   2            3

-4          -6

Now,

|A| = 2 * (-6) – (-4) * 3

= -12 + 12

= 0

So, |A|I = 0 *   1            0    =    0             0

0            1           0            0

Now, A11 = -6, A12 = 4, A21 = -3, A22 = 2

So, adj(A) =   A11          A21   =    -6           -3

A12          A22          4             2

Now, A(adj A) =   2          3     -6         -3

-4         -6     4           2

=   -12 + 12         -6 + 6

24 – 24       12 – 12

=   0          0

0          0

also, A(adj A) =    6          -3    2          3

4           2    -4        -6

=   -12 + 12     -18 + 18

8 – 8          12 – 12

=   0          0

0          0

Hence, A * (adj A) = (adj A) * A = | A | * I

Question 4:

1            -1             2

3             0            -2

1             0             3

Let

A =    1           -1             2

3            0            -2

1            0             3

Now,

|A| = 1(0 - 0) + 1(9 + 2) + 2(0 - 0) = 11

1            0             0          11          0             0

So, |A|I = 11 *   0            1             0    =     0           11           0

0            0             1           0            0           11

Now,

A11 = 0, A12 = -11, A13 = 0

A21 = 3, A22 = 1, A23 = -1

A31 = 2, A32 = 8, A33 = 3

So, adj (A) =   0           3            2

-11        1           8

0          -1           3

Now,

1           -1            2      0            3            2

A(adj A) =    3            0            -2    -11         1             8

1            0             3     0           -1            3

=    0 + 11 + 0            3 – 1 – 2             2 – 8 + 6

0 + 0 + 0               9 + 0 + 2             6 + 0 – 6

0 + 0 + 0               3 + 0 – 3             2 + 0 + 9

=    11           0           0

0           11          0

0            0          11

Also,

0            3            2    1          -1             2

(adj A)A =    -11         1             8    3           0             -2

0           -1            3     1           0              3

=    0 + 9 + 2            0 + 0 + 0             0 – 6 + 6

-11 + 3 + 8        11 + 0 + 0          -22 - 2 + 24

0 - 3 + 3             0 + 0 + 0             0 + 2 + 9

=    11           0           0

0           11          0

0            0          11

Hence, A * (adj A) = (adj A) * A = | A | * I

Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11.

Question 5:

2           -2

4            3

Let

A =   2            -2

4            3

|A| = 2 * 3 – (-2) * 4

= 6 + 8

= 14

Now,

A11 = 3, A12 = -4, A21 = 2, A22 = 2

So, adj(A) =   A11          A21   =    3            2

A12          A22         -4           2

= (1/14)    3            2

-4           2

Question 6:

-1          5

-3          2

Let

A =   -1           5

-3           2

|A| = (-1) * 2 – 5 * (-3)

= -2 + 15

= 13

Now,

A11 = 2, A12 = 3, A21 = -5, A22 = -1

So, adj(A) =   A11          A21   =    2           -5

A12          A22         3            -1

= (1/13)    2           -5

3           -1

Question 7:

1            2            2

0            2            4

0            0            5

Let

A =   1             2             3

0            2             4

0            0             5

|A| = 1(10 - 0) – 2(0 - 0) + 3(0 - 0) = 10

Now,

A11 = 10 – 0 = 10, A12 = -(0 - 0) = 0, A13 = 0 – 0 = 0

A21 = -(10 - 0) = -10, A22 = 5 – 0 = 5, A23 = -(0 - 0) = 0

A31 = 8 – 6 = 2, A32 = -(4 - 0) = -4, A33 = 2 – 0 = 2

Now, (adj A) =    10         -10          2

0             -5         -4

0              0           2

So, A-1 = (adj A)/|A| = (1/10)  10         -10           2

0            -5          -4

0              0           2

Question 8:

1            0            0

3            3            0

5            2          -1

Let

A =   1             0             0

3            3             0

5            3            -1

|A| = 1(-3 - 0) – 0 + 0 = -3

Now,

A11 = -3 – 0 = -3, A12 = -(-3 - 0) = 3, A13 = 6 – 15 = -9

A21 = -(0 - 0) = 0, A22 = -1 – 0 = -1, A23 = -(2 - 0) = -2

A31 = 0 – 0 = 0, A32 = -(0 - 0) = 0, A33 = 3 – 0 = 3

Now, (adj A) =    -3           0            0

3          -1            0

-9          -2            3

So, A-1 = (adj A)/|A| = (-1/3)  -3            0             0

3          -1             0

-9          -2             3

Question 9:

2            1            3

4           -1            0

-7           2           1

Let

A =   2             1              3

4            -1             0

-7            2             1

|A| = 2(-1 - 0) – 1(4 - 0) + 3(8 - 7)

= -2 – 4 + 3

= -3

Now,

A11 = -1 – 0 = -1, A12 = -(4 - 0) = -4, A13 = 8 – 7 = 1

A21 = -(1 - 6) = 5, A22 = 2 + 21 = 23, A23 = -(4 + 7) = -11

A31 = 0 + 3 = 3, A32 = -(0 - 12) = 12, A33 = -2 – 4 = -6

Now, (adj A) =    -1           5            3

-4         23          12

1         -11          -6

So, A-1 = (adj A)/|A| = (-1/3)   -1           5            3

-4         23           12

1         -11          -6

Question 10:

1           -1           2

0            2          -3

3           -2          4

Let

A =   1            -1             2

0            2             -3

3           -2             4

|A| = 1(8 - 6) - 0 + 3(3 - 4)

= 2 –3

= -1

Now,

A11 = 8 – 6 = 2, A12 = -(0 + 9) = -9, A13 = 0 – 6 = -6

A21 = -(-4 + 4) = 0, A22 = 4 - 6 = -2, A23 = -(-2 + 3) = -1

A31 = 3 - 4 = -1, A32 = -(-3 - 0) = 3, A33 = 2 – 0 = 2

Now, (adj A) =     2           0           -1

-9          -2            3

-6          -1            2

So, A-1 = (adj A)/|A| = (-1)   2           0            -1     =    -2           0            1

-9         -2             3            9           2            -3

-6         -1             2            6           1            -2

Question 11:

1              0                 0

0          cos α          sin α

0          sin α       - cos α

Let

A =   1               0                0

0          cos α          sin α

0          sin α          -cos α

|A| = 1(-cos2 α – sin2 α) - 0 + 0

= -(cos2 α + sin2 α)

= -1

Now,

A11 = -cos2 α – sin2 α = -1, A12 = 0, A13 = 0

A21 = 0, A22 = - cos α, A23 = - sin α

A31 = 0, A32 = - sin α, A33 = cos α

Now, (adj A) =    -1            0                0

0         -cos α        -sin α

0         -sin α         cos α

So, A-1 = (adj A)/|A| = (-1)   -1            0                0       =   1            0                0

0         -cos α        -sin α        0          cos α         sin α

0         -sin α         cos α        0           sin α        -cos α

Question 12:

Let A =     3           7            B =    6            8

2           5    and           7            9   , verify that (AB)-1 = B-1A-1

Given,

A =     3           7

2           5

Now, |A| = 3 * 5 – 7 * 2 = 15 – 14 = 1

A11 = 5, A12 = -2, A21 = -7, A22 = 3

So, adj A =    5           -7

-2           3

Now, A-1 = (adj A)/|A| =    5            -7

2             3

Again given,

B =     6           8

7           9

Now, |B| = 6 * 9 – 8 * 7 = 52 – 56 = -2

B11 = 9, B12 = -7, B21 = -8, B22 = 6

So, adj B =    9          -8

-7           6

Now, B-1 = (adj B)/|B| = (-1/2)   9            -8   =    -9/2           4

-7            6          7/2           -3

Now, B-1A-1 =    -9/2            4     5           -7

7/2             -3    -2           3

=   -4/2 – 8        63/2 + 12  =  -61/2          87/2

35/2 + 6       -49/2 – 9        47/2          -67/2        …………..1

Again, AB =    3              7    6             8

2              5   7              9

=     18 + 49           24 + 63

12 + 35            16 + 45

=    67         87

47         61

Now, |AB| = 67 * 61 – 87 * 47 = 4087 – 4089 = -2

So, adj (AB) =    61        -87

-47        67

Now, (AB)-1 = (adj AB)/|AB| = (-1/2)   61        -87   =   -61/2         87/2

-47        67         47/2         -67/2      …………..1

From equations 1 and 2, we get

(AB)-1 = B-1A-1

Hence, the give result is proved.

Question 13:

If A =    3           7

-1          2    , show that A2 – 5A + 7I = 0. Hence find A-1

Given,

A =    3           7

-1          2

A2 = A * A =  3              1    3              1

-1            2    -1             2

=    9 – 1            3 + 2

-3 – 2          -1 + 4

=    8             5

-5            3

Now,

A2 – 5A + 7I =    8             5   - 5   3             1   + 7   1             0

-5            3          -1            2            0             1

=    8             5   -    15           5    +     7             0

-5            3         -5          10           0             7

=   8 – 15 + 7          5 – 5 + 0

-5 + 5 + 0          3 – 10 + 7

=   0             0

0             0

Hence, A2 – 5A + 7I = 0

=> A * A – 5A + 7I = 0

=> A * A – 5A = -7I

=> A * A(A-1) – 5A(A-1) = -7I(A-1)

=> A * (AA-1) – 5(AA-1) = -7I(A-1)

=> AI – 5I = -7A-1

=> A – 5I = -7A-1

=> A-1 = (-1/7)(A – 5I)

=> A-1 = (1/7)(5I - A)

=> A-1 = (1/7)     5             0    -    3             1

0             5          -1           2

=> A-1 = (1/7)   5 - 3           0 - 1

0 + 1         5 - 2

=> A-1 = (1/7)   2           -1

1            3

Question 14:

For the matrix A =    3           2

1           1    , find the numbers a and b such that A2 + aA + bI = O.

Given,

A =    3           2

1           1

A2 = A * A =  3             2     3             2

1             1     1             1

=    9 + 2            6 + 2

3 + 1            2 + 1

=    11           8

4            3

Now, A2 + aA + bI = 0

=> A * A + aA + bI = 0

=> A * A + aA = -bI

=> A * A(A-1) + aA(A-1) = -bI(A-1)

=> A * (AA-1) + a(AA-1) = -bI(A-1)

=> AI + aI = -bA-1

=> A + aI = -bA-1

=> A-1 = (-1/b)(A + aI)    ……………1

=> A-1 = (1/1)    1           -2   =   1             -2

-1           3         -1            3

From equation 1, we have

=>   1             -2   = (-1/b)     3             2    -    a             0

-1            3                       1             1         0             a

=>   1             -2   = (-1/b)     3 + a             2

-1            3                       1             1 + a

=>   1             -2   =   (-3 – a)/b           -2/b

-1            3         1/b                  (-1 – a)/2

Comparing the corresponding elements of the two matrices, we get

=> -1/b = -1

=> b = 1

And (-3 – a)/b = 1

=> -3 – a = b

=> -3 – a = 1

=> a = -3 – 1

=> a = -4

Hence, −4 and 1 are the required values of a and b respectively.

Question 15:

For the matrix A =    1           1           1

1           2          -3

2          -1           3  , show that A3 – 6A2 + 5A + 11I = O. Hence find A-1

Given,

1           1           1

A =    1           2          -3

2          -1           3

A2 = A * A

=   1           1            1      1           1            1

1          2            -3     1           2           -3

2         -1            3      2          -1           3

=    1 + 1 + 2            1 + 2 – 1            1 – 3 + 3

1 + 2 - 6             1 + 4 + 3            1 – 6 – 9

2 - 1 + 6             2 - 2 – 3             2 + 3 + 9

=     4            2            1

-3          8          -14

7          -3           14

A3 = A2 * A

=    4           2            1     4           2            1

-3         8         -14     -3          8         -14

7         -3          14      7         -3          14

=    4 + 2 + 2              4 + 4 – 1               4 – 6 + 3

-3 + 8 - 28        -3 + 16 + 14         -3 – 24 – 42

7 - 3 + 28           7 - 6 – 14               7 + 9 + 42

=     8            7            1

-23       27        -69

32       -13         58

So, A3 – 6A2 + 5A + 11I

=    8           7              1          4            2          1            1          1           1               1         0            0

-23       27          -69  - 6  -3           8         -14  + 5   1          2          -3   + 11   0          1            0

32       -13         58           7           -3         14           2         -1           3              0         0            1

=    8           7              1       24          12         6        5          5           5         11         0            0

-23       27          -69  -  -18          48      -84  +   5         10       -15   +   0          11            0

32       -13         58        42         -18       84       10       -5         15         0         0            11

=    8 – 24 + 5 + 11                7 – 12 + 5 + 0                        1 – 6 + 5 + 0

-23 + 18 + 5 + 0              27 – 48 + 5 + 11                 -69 + 84 - 15 + 0

32 – 42 + 10 + 0            -13 + 18 - 5 + 0                  58 – 84 + 15 + 11

=    0          0            0

0          0            0

0          0            0

Hence, A3 – 6A2 + 5A + 11I = 0

Now, A3 – 6A2 + 5A + 11I = 0

=> A * A * A – 6(A * A) + 5A + 11I = 0

=> (A * A * A) * A-1 – 6(A * A) * A-1 + 5A * A-1 + 11I * A-1 = 0

=> (A * A) * (A * A-1) – 6A * (A * A-1) + 5(A * A-1) = -11(I * A-1)

=> A2 – 6A + 5I = -11A-1

=> A-1 = (-1/11)(A2 – 6A + 5I)   ………………1

Now, A2 – 6A + 5I

=    4           2              1          1            1          1             1         0            0

-3           8          -14  - 6   1            2          -3   + 5   0          1            0

7          -3           14          2           -1           3           0         0            1

=    4           2              1         6            6           6            5          0            0

-3           8          -14    -    6           12       -18     +    0          5            0

7          -3           14          12        -6          18           0          0            5

=   4 - 6 + 5            2 - 6 + 0            1 - 6 + 0

-3 - 6 + 0             8 - 12 + 5         -14 + 18 + 0

7 - 12 + 0        -3 + 6 + 0           14 - 18 + 5

=    3            -4           -5

-9            1            4

-5            3            1

From equation 1, we get

A-1 = (-1/11)   3            -4           -5    = (1/11)  -3            4             5

-9            1            4                      9            -1           -4

-5            3            1                     5            -3           -1

Question 16:

For the matrix A =    2          -1          1

-1           2         -1

1         -1           2   , show that A3 – 6A2 + 9A - 4I = O. Hence find A-1

Given,

2           -1          1

A =   -1           2          -1

1          -1           2

A2 = A * A

=    2           -1           1    2           -1           1

-1          2           -1    -1           2          -1

1          -1            2    1           -1           2

=    4 + 1 + 1           -2 - 2 – 1            2 + 1 + 2

-2 - 2 - 1            1 + 4 + 1           -1 – 2 – 2

2 + 1 + 2           -1 - 2 – 2            1 + 1 + 4

=     6           -5            5

-5          6           -5

5          -5            6

A3 = A2 * A

=    6           -5           5    2          -1            1

-5          6           -5   -1           2           -1

5          -5            6     1         -1            2

=    12 + 5 + 5           -6 - 10 – 5             6 + 5 + 10

-10 - 6 - 5            5 + 11 + 5            -5 – 6 – 10

10 + 5 + 6           -5 - 10 – 6             5 + 5 + 12

=     22         -12        21

-21        22        -21

21        -21         22

So, A3 – 6A2 + 9A - 4I

=    22        -21         21          6           -5          5            2         -1           1             1         0           0

-21         22       -21   - 6  -5           6           -5  + 5   -1          2         -1    + 4    0         1           0

21        -21        22           5           -5           6           1          -1          2              0         0           1

=    22       -21          21       36         -30       30       18         -9         9         4         0            0

-21        22         -21  -  -30          36      -30  +   -9         18        -9   +   0          4            0

21       -21         22        30         -30       36        9          -9        18        0          0            4

=    22 – 36 + 18 + 4             -21 + 30 - 9 + 0                     21 – 30 + 9 + 0

-21 + 30 - 9 + 0               22 – 36 + 18 + 4                 -21 + 30 - 9 + 0

21 – 30 + 9 + 0               -21 + 30 - 9 + 0                    22 – 36 + 18 + 4

=    0          0            0

0          0            0

0          0            0

Hence, A3 – 6A2 + 9A - 4I = 0

Now, A3 – 6A2 + 9A - 4I = 0

=> A * A * A – 6(A * A) + 9A - 4I = 0

=> (A * A * A) * A-1 – 6(A * A) * A-1 + 9A * A-1 - 4I * A-1 = 0

=> (A * A) * (A * A-1) – 6A * (A * A-1) + 9(A * A-1) = 4(I * A-1)

=> A2 – 6A + 9I = 4A-1

=> A-1 = (1/4)(A2 – 6A + 9I)   ………………1

Now, A2 – 6A + 9I

=    6           -5            5          2            -1          1            1         0            0

-5           6            -5   - 6  -1            2         -1   + 9   0          1            0

5          -5            6           1           -1           2           0         0            1

=    6           -5            5          12          -6          6            9          0           0

-5           6            -5    -    -6           12         -6     +    0          9            0

5          -5             6           6          -6          12           0          0           9

=   6 - 12 + 9         -5 + 6 + 0            5 - 6 + 0

-5 + 6 + 0             6 - 12 + 9         -5 + 6 + 0

5 - 6 + 0          -5 + 6 + 0            6 - 12 + 9

=    3             1           -1

1             3            1

-1           -1           3

From equation 1, we get

A-1 = (1/4)   3             1           -1

1             3            1

-1           -1           3

Question 17:

Let A be a nonsingular square matrix of order 3 * 3. Then |adj A| is equal to

1. |A| B. |A|2 C. |A|3                                        D. 3|A|

We know that,

|A|         0           |A|

(adj A)A = |A|I =    0           |A|          0

0             0           |A|

|A|           0           |A|

=> |(adj A)A| =     0            |A|          0

0              0           |A|

1         0          0

=> |(adj A)||A| = |A|3 =    0        1          0   = |A|3 I

0        0          1

So, => |(adj A)| = |A|2

Hence, the correct answer is option B.

Question 18:

If A is an invertible matrix of order 2, then det (A-1) is equal to

1. det (A) B. 1/ det (A) C. 1                                  D. 0

Since A is an invertible matrix, then A-1 exists and A-1 = (adj A)/|A|

Given, the matrix A is of order 2,

Let A =   a          b

c          d

Then |A| = ad – bc

And adj A =   d         -b

-c         a

=   d/|A|        -b/|A|

-c/|A|        a/|A|

So, A-1 =  d/|A|        -b/|A|

-c/|A|        a/|A|

= (1/|A|2)  d           -b

-c           a

= (1/|A|2) * (ad - bc)

= (1/|A|2) * |A|

= |A|

So, det(A-1) = 1/det(A)

Hence, the correct answer is option B.

Exercise 4.6

Examine the consistency of the system of equations in Exercises 1 to 6.

Question 1:

x + 2y = 2

2x + 3y = 3

The given system of equations is:

x + 2y = 2

2x + 3y = 3

The given system of equations can be written in the form of AX = B, where

A =   1           2     X =     x    and B =    2

2          3   ,            y                      3

Now,

|A| = 1 * 3 – 2 * 2 = 3 – 4 = -1 ≠ 0

So, A is non-singular.

Therefore, A−1 exists.

Hence, the given system of equations is consistent.

Question 2:

2x - y = 5

x + y = 4

The given system of equations is:

2x - y = 5

x + y = 4

The given system of equations can be written in the form of AX = B, where

A =   2          -1     X =     x    and B =    5

1          1   ,            y                      4

Now,

|A| = 2 * 1 – 1 * (-1) = 2 + 1 = 3 ≠ 0

So, A is non-singular.

Therefore, A−1 exists.

Hence, the given system of equations is consistent.

Question 3:

x + 3y = 5

2x + 6y = 8

The given system of equations is:

x + 3y = 5

2x + 6y = 8

The given system of equations can be written in the form of AX = B, where

A =   1           3     X =     x    and B =    5

2          6   ,            y                      5

Now,

|A| = 1 * 6 – 2 * 3 = 6 – 6 = 0

So, A is a singular matrix.

Now,

-2          1

adj(A)B =   6          -3       5     =     30 – 24       =      6    ≠ O

-2          1        8            -10 + 8               -2

Thus, the solution of the given system of equations does not exist.

Hence, the system of equations is inconsistent.

Question 4:

Examine the consistency of the system of equations.

x + y + z = 1                       2x + 3y + 2z = 2                         ax + ay + 2az = 4

The given system of equations is:

x + y + z = 1

2x + 3y + 2z = 2

ax + ay + 2az = 4

This system of equations can be written in the form AX = B, where

1             1             1                  x                       1

A =   2             3             2     , X =     y     and B =     2

a              a            2a                 z                       4

Now,

|A| = 1(6a – 2a) – 1(4a – 2a) + 1(2a – 3a)

= 4a – 2a – a

= a ≠ 0

So, A is non-singular.

Therefore, A−1 exists.

Hence, the given system of equations is consistent.

Question 5:

Examine the consistency of the system of equations.

3x - y - 2z = 2                                2y - z = -1                         3x - 5y = 3

The given system of equations is:

3x - y - 2z = 2

2y - z = -1

3x - 5y = 3

This system of equations can be written in the form AX = B, where

3            -1            -2                  x                       2

A =   0             2             -1     , X =     y     and B =    -1

3             -5              0                 z                        3

Now,

|A| = 3(0 – 5) -0 + 3(1 + 4) = -15 + 15 = 0

So, A is a singular matrix.

Now,

adj(A) =   -5            10             5       2             -10 – 10 + 15              -5

-3             6              3       -1    =        -6 – 6 + 9          =      -3    ≠ O

-6           12             6       3              -1 – 12 + 18               -6

Thus, the solution of the given system of equations does not exist.

Hence, the system of equations is inconsistent.

Question 6:

Examine the consistency of the system of equations.

5x − y + 4z = 5                                      2x + 3y + 5z = 2                         5x − 2y + 6z = −1

The given system of equations is:

5x − y + 4z = 5

2x + 3y + 5z = 2

5x − 2y + 6z = −1

This system of equations can be written in the form of AX = B, where

5            -1              4                x                       5

A =   2             3              5    , X =     y     and B =     2

5             -2              6                 z                       -1

Now,

|A| = 5(18 + 10) + 1(12 – 25) + 4(-4 – 15)

= 5 * 28 + 1 * (-13) + 4 * (-19)

= 140 – 13 – 76

= 51 ≠ 0

So, A is non-singular.

Therefore, A−1 exists.

Hence, the given system of equations is consistent.

Solve system of linear equations, using matrix method, in Exercises 7 to 14.

Question 7:

5x + 2y = 4

7x + 3y = 5

The given system of equations is:

5x + 2y = 4

7x + 3y = 5

The given system of equations can be written in the form of AX = B, where

A =   5           2     X =     x    and B =    4

7           3   ,           y                      5

Now,

|A| = 5 * 3 – 7 * 2 = 15 – 14 = 1 ≠ 0

So, A is non-singular matrix.

Therefore, its inverse exists.

So, A-1 =    3           -2

-7            5

Now,

X = A-1B =   3            -2      4

-7            5      5

x    =      12 - 10    =     2

y           -28 + 25         -3

So, x = 2 and y = -3

Question 8:

2x - y = -2

3x + 4y = 3

The given system of equations is:

2x - y = -2

3x + 4y = 3

The given system of equations can be written in the form of AX = B, where

A =   2          -1     X =     x    and B =   -2

3           4   ,           y                      3

Now,

|A| = 2 * 4 – 3 * (-1) = 8 + 3 = 11 ≠ 0

So, A is non-singular matrix.

Therefore, its inverse exists.

So, A-1 = (1/11)    4           1

-3           2

Now,

X = A-1B = (1/11)  4             1     -2

-3            2      3

x    =  (1/11)      -8 + 3    = (1/11)   -5    =     -5/11

y                           6 + 6                     12          12/11

So, x = -5/11 and y = 12/11

Question 9:

4x - 3y = 3

3x - 5y = 7

The given system of equations is:

4x - 3y = 3

3x - 5y = 7

The given system of equations can be written in the form of AX = B, where

A =   4          -3     X =     x    and B =    3

3          -5   ,           y                      7

Now,

|A| = (-5) * 4 – 3 * (-3) = -20 + 9 = -11 ≠ 0

So, A is non-singular matrix.

Therefore, its inverse exists.

So, A-1 = (-1/11)   -5          3   = (1/11)   5          -3

-3           4                     3         -4

Now,

X = A-1B =  (1/11)  5           -3      3

3           -4      7

x    =   (1/11)    15 - 21   =  (1/11)   -6     =     -6/11

y                         9 - 28                       -19         -19/11

So, x = -6/11 and y = -19/11

Question 10:

5x + 2y = 3

3x + 2y = 5

The given system of equations is:

5x + 2y = 3

3x + 2y = 5

The given system of equations can be written in the form of AX = B, where

A =   5           2     X =     x    and B =    3

3            2   ,           y                      5

Now,

|A| = 5 * 2 – 3 * 2 = 10 - 6 = 4 ≠ 0

So, A is non-singular matrix.

Therefore, its inverse exists.

Question 11:

2x + y + z = 1                           x – 2y – z = 3/2                          3y – 5z = 9

The given system of equations can be written in the form of AX = B, where

2          1            1                x                        1

A =     1          -2          -1    , X =     y     and B =   3/2

0           3           -5                z                        9

Now,

|A| = 2(10 + 3) – 1(-5 - 3) + 0

= 2 * 13 – 1 * (-8)

= 26 + 8

= 34 ≠ 0

Thus, A is non-singular. Therefore, its inverse exists.

Now,

A11 = 13, A12 = 5, A13 = 3

A21 = 8, A22 = -10, A23 = -6

A31 = 1, A32 = 3, A33 = -5

So, A-1 =  adj(A) /|A| = (1/34)   13         8           1

5         -10          3

3          -6          -5

So, X = A-1B = (1/34)   13           8          1       1

5           -10        3     3/2

3            -6        -5       9

x     = (1/34)    13 + 12 + 9

y                       5 – 15 + 27

z                        3 – 9 - 45

x     = (1/34)     34      =       1

y                        17              1/2

z                        -51            -3/2

So, x = 1, y = 1/2 and z = -3/2

Question 12:

x − y + z = 4                                   2x + y − 3z = 0                                            x + y + z = 2

The given system of equations can be written in the form of AX = B, where

1         -1            1                x                        4

A =     2           1           -3    , X =     y     and B =     0

1           1            1                 z                        2

Now,

|A| = 1(1 + 3) + 1(2 + 3) + 1(2 - 1)

= 4 + 5 + 1

= 10 ≠ 0

Thus, A is non-singular. Therefore, its inverse exists.

Now,

A11 = 4, A12 = -5, A13 = 1

A21 = 2, A22 = 0, A23 = -2

A31 = 2, A32 = 5, A33 = 3

So, A-1 =  adj(A) /|A| = (1/10)   4           2           2

-5           0           5

1          -2           3

So, X = A-1B = (1/10)   4            2          2       4

-5           0          5       0

1           -2         3       2

x     = (1/10)    16 + 0 + 4

y                       -20 + 0 + 10

z                        4 + 0 + 6

x     = (1/10)     20      =       2

y                        -10             -1

z                         10               1

So, x = 2, y = -1 and z = 1

Question 13:

2x + 3y + 3z = 5                             x − 2y + z = −4                                 3x − y − 2z = 3

The given system of equations can be written in the form of AX = B, where

2           3            3               x                        5

A =     1           -2            1   , X =     y     and B =    -4

3           -1           -2               z                        3

Now,

|A| = 2(4 + 1) - 3(-2 - 3) + 3(-1 + 6)

= 2 * 5 – 3 * (-5) + 3 * 5

= 10 + 15 + 15

= 40 ≠ 0

Thus, A is non-singular. Therefore, its inverse exists.

Now,

A11 = 5, A12 = 5, A13 = 5

A21 = 3, A22 = -13, A23 = 11

A31 = 9, A32 = 1, A33 = -7

So, A-1 =  adj(A) /|A| = (1/40)   5           3            9

5         -13           1

5          11          -7

So, X = A-1B = (1/40)   5            3          9       5

5          -13         1     -4

5           11        -7       3

x     = (1/40)    25 - 12 + 27

y                       25 + 52 + 3

z                        25 - 44 - 21

x     = (1/40)     40      =       1

y                        80               2

z                        -40             -1

So, x = 1, y = 2 and z = -1

Question 14:

x − y + 2z = 7                             3x + 4y − 5z = −5                            2x − y + 3z = 12

The given system of equations can be written in the form of AX = B, where

1          -1            2               x                        7

A =     3            4           -5   , X =    y     and B =     -5

2           -1            3               z                       12

So, |A| = 1(12 - 5) + 1(9 + 10) + 2(-3 - 8)

= 7 + 19 + 2 * (-11)

= 7 + 19 - 22

= 4 ≠ 0

Thus, A is non-singular. Therefore, its inverse exists.

Now,

A11 = 7, A12 = -19, A13 = -11

A21 = 1, A22 = -1, A23 = -1

A31 = -3, A32 = 11, A33 = 7

So, A-1 =  adj(A) /|A| =   (1/4)    7           1           -3

-19       -1           11

-11       -1             7

So, X = A-1B =   (1/4)     7          1          3       7

-19        -1         11     -5

-11       -1          7      12

x     =  (1/4)      49 - 5 - 36

y                      -133 + 5 + 132

z                        -77 + 5 + 84

x     =   (1/4)     8      =       2

y                        4               1

z                        12             3

So, x = 2, y = 1 and z = 3

Question 15:

If A =   2      -3         5

3       2        -4

1        1        -2    , find A–1. Using A–1 solve the system of equations

2x – 3y + 5z = 11

3x + 2y – 4z = – 5

x + y – 2z = – 3

Given,

2      -3          5

A =    3       2         -4

1        1        -2

So, |A| = 2(-4 + 4) + 3(-6 + 4) + 5(3 - 2)

= 0 - 6 + 5

= -1 ≠ 0

Now,

A11 = 0, A12 = 2, A13 = 1

A21 = -1, A22 = -9, A23 = -5

A31 = 2, A32 = 23, A33 = 13

So, A-1 =  adj(A) /|A| = (-1)   0       -1           2             0         1            -2

2       -9          23     =    -2        9          -23

1       -5          13            -1        5         -13     ……………1

The given system of equations can be written in the form of AX = B, where

2         -3           5               x                         11

A =      3          2           -4   , X =    y     and B =      -5

1          1           -2               z                        -3

The solution of the system of equations is given by

X = A-1B

So, X = A-1B =   0          1          -2       11

-2         9         -23      -5                              [From equation 1]

-1        5         -13      -3

x     =        0 – 5 + 6

y           -22 - 45 + 69

z           -11 - 25 + 39

x     =     1

y            2

z            3

So, x = 1, y = 2 and z = 3

Question 16:

The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs 70. Find cost of each item per kg by matrix method.

Let the cost of onions, wheat, and rice per kg be Rs x, Rs y,and Rs z respectively.

Then, the given situation can be represented by a system of equations as:

4x + 3y + 2z = 60

2x + 4y + 6z = 90

6x + 2y + 3z = 70

This system of equations can be written in the form of AX = B, where

4          3            2                x                       60

A =      2          4            6    , X =    y     and B =     90

6          2            3                z                       70

|A| = 4(12 - 12) - 3(6 - 36) + 2(4 - 24)

= 0 – 3 * (-30) + 2 * (-20)

= 90 - 40

= 50 ≠ 0

Now,

A11 = 0, A12 = 30, A13 = -20

A21 = -5, A22 = 0, A23 = 10

A31 = 10, A32 = -20, A33 = 10

So, A-1 =  adj(A) /|A| = (1/50)   0          -5           10

30         0          -20

-20       10          10

So, X = A-1B = (1/50)   0          -5         10       60

30         0         -20      90

-20       10        10      70

x     = (1/50)       0 - 450 + 700

y                         1800 + 0 - 1400

z                       -1200 + 900 + 700

x     = (1/50)    250    =      5

y                       400            8

z                       400            8

So, x =5, y = 8 and z = 8

Hence, the cost of onions is Rs 5 per kg, the cost of wheat is Rs 8 per kg, and the cost of rice is

Rs 8 per kg.

Miscellaneous Exercise on Chapter 4

Question 1:

Prove that the determinant   x           sin θ          cos θ

-sin θ      -x                1

cos θ        1                x        is independent of θ.

Let

x           sin θ          cos θ

Δ =  -sin θ      -x                1

cos θ        1                x

Δ = x(x2 - 1) – sin θ(-x * sin θ – cos θ) + cos θ(-sin θ + x * cos θ)

Δ = x3 - x + x * sin2 θ – sin θ * cos θ - sin θ *cos θ + x * cos2 θ

Δ = x3 - x + x * (sin2 θ + cos2 θ)

Δ = x3 - x + x

Δ = x3            [Independent of θ]

Hence the given determinant is independent of θ.

Question 2:

Without expanding the determinant, prove that

a         a2         bc             1         a2         a3

b         b2        ca      =      1         a2         b3

c         c2         ab             1         a2          c3

LHS:

a         a2         bc

b         b2        ca

c         c2         ab

= (1/abc)   a2        a3        abc

b2        b3        abc

c2        c3        abc                 [Apply R1 -> aR1, R2 -> bR2, R3 -> cR3]

= (abc/abc)   a2        a3            1

b2        b3            1

c2        c3             1             [Taking out factor abc from C3]

=  a2        a3            1

b2        b3            1

c2        c3             1

=  1            a2        a3

1            b2        b3

1            c2        c3                  [Apply C1 C3 and C2 C3]

= RHS

Hence, the give result is proved.

Question 3:

Evaluate   cos α * cos β         cos α * cos β       -sin α

-sin β                           cos β                  0

sin α * cos β         sin α * sin β         cos α

Given,

cos α * cos β         cos α * cos β       -sin α

Δ =      -sin β                           cos β                  0

sin α * cos β         sin α * sin β         cos α

Expanding along C3, we get

Δ = -sin α(-sin α * sin2 β – cos2 β * sin α) + 0 + cos α(cos α * cos2 β + cos α * sin2 β)

Δ = sin2 α(sin2 β + cos2 β) + cos2 α(cos2 β + sin2 β)

Δ = sin2 α * 1 + cos2 α * 1

Δ = sin2 α + cos2 α

Δ = 1

Question 4:

If a, b and c are real numbers and

b + c        c + a       a + b

Δ =    c + a        a + b       b + c  = 0

a + b       b + c       c + a

Show that either a + b + c = 0 or a = b = c.

Given,

b + c        c + a       a + b

Δ =   c + a        a + b       b + c

a + b       b + c       c + a

Applying R1 -> R1 + R2 + R3, we get

2(a + b + c)        2(a + b + c)       2(a + b + c)

Δ =   c + a                          a + b                  b + c

a + b                         b + c                   c + a

1                1               1

Δ = 2(a + b + c)  c + a         a + b          b + c

a + b        b + c          c + a

Applying C2 -> C2 – C1 and C3 -> C3 – C1, we get

1                0                0

Δ = 2(a + b + c)  c + a         b - c          b - a

a + b        c - a          c - b

Expanding along R1, we get:

Δ = 2(a + b + c)[1 * {(b - c)(c - b) – (b - a)(c - a)}]

Δ = 2(a + b + c)[-b2 – c2 + 2bc – bc + ba + ca – a2]

Δ = 2(a + b + c)[ab + bc + ca + ba – a2 - b2 – c2]

It is given that Δ = 0

=> 2(a + b + c)[ab + bc + ca + ba – a2 - b2 – c2] = 0

Either a + b + c = 0 or ab + bc + ca + ba – a2 - b2 – c2 = 0

Now, 2ab + 2bc + 2ca + 2ba – 2a2 - 2b2 – 2c2 = 0

=> (a - b)2 + (b - c)2 + (c - a)2 = 0

=> (a - b)2 = (b - c)2 = (c - a)2 = 0        [Since (a - b)2, (b - c)2, (c - a)2 are non negative]

=> (a - b) = (b - c) = (c - a) = 0

=> a = b = c

Hence, if ∆ = 0 then either a + b + c = 0 or a = b = c.

Question 5:

Solve the equations   x + a           x             x

x              x + a         x       = 0, a ≠ 0

x                  x         x + a

Given,

x + a           x             x

x              x + a           x      = 0

x                  x         x + a

Applying R1 -> R1 + R2 + R3, we get

3x + a       3x + a        3x + a

x                x + a             x       = 0

x                  x              x + a

1                  1                  1

(3x + a)    x                x + a             x       = 0

x                  x              x + a

Applying C2 -> C2 – C1 and C3 -> C3 – C1, we get

1             0            0

(3x + a)    x             a            0    = 0

x             0            a

Expanding along R1, we get:

(3x + a)(1 * a2 – 0) = 0

=> a2(3x + a) = 0

=> 3x + a = 0            [Since a ≠ 0]

=> x = -a/3

Question 6:

Prove that   a2                 bc            ac + c2

a2 + ab         b2                 ac       = 4a2b2c2

ab             b2 + bc            c2

Given,

a2                 bc            ac + c2

Δ =  a2 + ab         b2                 ac

ab             b2 + bc            c2

Taking out common factors a, b and c from C1, C2 and C3, we get

a               c            a + c

Δ = abc  a + b         b               a

b             b + c          c

Applying R2 -> R2 – R1 and R3 -> R3 – R1, we get

a               c            a + c

Δ = abc  b            b - c           -c

b - a          b             -a

Applying R2 -> R2 + R1, we get

a               c            a + c

Δ = abc  a + b         b               a

b - a          b             -a

Applying R3 -> R3 + R2, we get

a               c            a + c

Δ = abc  a + b         b               a

2b            2b              0

a             c            a + c

Δ = 2ab2c  a + b         b               a

1             1              0

Applying C2 -> C2 – C1, we get

a          c - a         a + c

Δ = 2ab2c  a + b        -a                a

1             1                0

Expanding along R3, we get:

Δ = 2ab2c[a(c - a) + a(a + c)]

Δ = 2ab2c[ac – a2 + a2 + ac]

Δ = 2ab2c * 2ac

Δ = 4a2b2c2

Hence, the given result is proved.

Question 7:

If A-1 =   3            -1            1                    1             2            -2

-15          6            -5   and B =   -1            3              0

5            -2             2                    0            -2             1  , find (AB)-1

We know that (AB)-1 = B-1A-1

1           2          -2

B =      -1          3            0

0          -2           1

|B| = 1 * 3 – 2 * (-1) – 2 * 2

= 3 + 2 - 4

= 1

Now,

B11 = 3, B12 = 1, B13 = 2

B21 = 2, B22 = 1, B23 = 2

B31 = 6, B32 = 2, B33 = 5

So, B-1 =  adj(B) /|B| =  3              2             6

1             1             2

2             2             5

Now, (AB)-1 = B-1A-1

=    3               2            6     3             -1             -1

1               1             2    -15           6             -5

2               2             5    5              -2             2

=   9 – 30 + 30             -3 + 12 – 12             3 – 10 + 12

3 – 15 + 10              -1 + 6 – 4                   1 – 5 + 4

6 – 30 + 25             -2 + 12 – 10             2 – 10 + 10

=    9             -3             5

-2              1             0

1               0            2

Question 8:

Let A =   1            -2            1

-2             3            1

1             1             5   , verify that

Given,

1            -2            1

A =  -2             3            1

1             1             5

|A| = 1(15 - 1) + 2(-10 - 1) + 1 (-2 - 3)

= 14 – 22 - 5

= -13

Now,

A11 = 14, A12 = 11, A13 = -5

A21 = 11, A22 = 4, A23 = -3

A31 = -5, A32 = -3, A33 = -1

So,

14           11          -5

-5             -3           -1

So, A-1 =  adj(A) /|A| = (-1/13)   14           11              -5     = (1/13)   -14           -11             5

11             4               -3                      -11             -4             3

-5             -3               -1                       5                3              1

(i) |adj(A)| = 14(-4 - 9) – 11(-11 - 15) – 5(-33 + 20)

= 14 * (-13) – 11 * (-26) – 5 * (-13)

= -182 + 286 + 65

= 169

We have

-13          26           -13

-13          -13           -65

26          -39           -13

-13          -13           -65

= (1/13)    -1           2           -1

2          -3           -1

-1          -1           -5

Now, A-1 = (1/13)   -14           -11               5

-11           -4                3

5               3                1

Now, A-1 =  -14/13          -11/13              5/13

-11/13           -4/13               3/13

5/13               3/13                1/13

(-4/169 – 9/169)                -(-11/169 – 15/169)               (-33/169 + 20/169)

adj(A-1 ) =    -(-11/169 – 15/169)            (-14/169 – 25/169)               -(-42/169 + 55/169)

(-33/169 + 20/169)             -(-42/169 + 55/169)               (56/169 – 121/169)

= (1/169)   -13           26           -13

26           -39           -13

-13          -13          -65

= (1/13)    -1           2           -1

2          -3           -1

-1         -1           -5

(ii)  We have

A-1 =  (1/13)   -14         -11               5

-11          -4                3

5              3                1

And

2           -3             -1

-1          -1             -5

Now, | A-1| = (1/13)3[-14 * (-13) + 11 * (-26) + 5 * (-13)]

= (1/13)3(-169)

= -1/13

So, (A-1)-1 =  adj(A-1) /| A-1| = [1/(-13)] * (1/13)   -1             2              -1

2            -3              -1

-1            -1              -5

=    1             -2                1

-2             3                1

1              1                5

= A

So, (A-1)-1 = A

Question 9:

Evaluate       x             y             x + y

y          x + y            x

x + y          x              y

Given,

x             y             x + y

Δ =    y            x + y             x

x + y         x                y

Applying R1 -> R1 + R2 + R3, we get

2(x + y)            2(x + y)            2(x + y)

Δ =    y                         x + y                    x

x + y                     x                       y

1              1              1

Δ = 2(x + y)   y            x + y           x

x + y          x              y

Applying C2 -> C2 – C1 and C3 -> C3 – C1, we get

1              0              0

Δ = 2(x + y)   y                x           x - y

x + y         -y              -x

Expanding along R1, we get:

Δ = 2(x + y)[-x2 + y(x - y)]

=> Δ = 2(x + y)[-x2 + xy – y2]

=> Δ = -2(x + y)[x2 + y2 – xy]

=> Δ = -2(x3 + y3)

Question 10:

Evaluate      1             x             y

1           x + y         y

1              x          x + y

Given,

1             x             y

Δ =    1          x + y          y

1             x          x + y

Applying R2 -> R2 – R1 and R3 -> R3 – R1, we get

1            x           y

Δ =   0            y           0

0            0           x

Expanding along C1, we get:

Δ = 1(xy - 0) = xy

Using properties of determinants in Exercises 11 to 15, prove that:

Question 11:

α           α2         β + γ

β           β2         γ + α    = (α - β)( β - γ)(γ - α)(α + β + γ)

γ            γ2         α + β

Given,

α           α2         β + γ

Δ =     β           β2         γ + α

γ            γ2         α + β

Applying R2 -> R2 – R1 and R3 -> R3 – R1, we get

α                 α2              β + γ

Δ =    β - α         β2 – α2        α - β

γ - α         γ2 – α2        α - γ

α            α2          β + γ

Δ = (β - α)( γ - α)   1           β + α         -1

1           γ + α         -1

Applying R3 -> R3 – R2, we get

α            α2          β + γ

Δ = (β - α)( γ - α)   1           β + α         -1

0           γ - β            0

Expanding along R3, we get:

Δ = (β - α)( γ - α)[-( γ - β)(-α – β - γ)]

Δ = (β - α)(γ - α)(γ - β)(α + β + γ)

Δ = (α - β)( β - γ)(γ - α)(α + β + γ)

Hence, the given result is proved.

Question 12:

x           x2         1 + px3

y           y2         1 + py3    = (1 + pxyz)(x - y)(y - z)(z - x)

z            z2         1 + pz3

Given,

x           x2         1 + px3

Δ =   y           y2         1 + py3

z            z2         1 + pz3

Applying R2 -> R2 – R1 and R3 -> R3 – R1, we get

x                x2             1 + px3

Δ =   y - x         y2 - x2        p(y3 - x3)

z - x         z2 - x2        p(z3 - x3)

x                x2                 1 + px3

Δ = (y – x)(z - x)    1              y + x        p(y2 + x2 + xy)

1              z + x        p(z2 + x2 + zx)

Applying R3 -> R3 – R2, we get

x                x2                        1 + px3

Δ = (y – x)(z - x)    1              y + x                   p(y2 + x2 + xy)

0              z - y                  p(z - y)(x + y + z)

x                x2                    1 + px3

Δ = (y – x)(z - x)(z - y)    1              y + x               p(y2 + x2 + xy)

0                 1                     p(x + y + z)

Expanding along R3, we get:

Δ = (y – x)(z - x)(z - y)[(-1) * p * (xy2 + x3 + x2y) + 1 + px3 + p(x + y + z)(xy)]

Δ = (y – x)(z - x)(z - y)[-pxy2 - px3 - px2y + 1 + px3 + px2y + pxy2 + pxyz]

Δ = (x – y)(y - z)(z - x)(1 + pxyz)

Hence, the given result is proved.

Question 13:

3a          -a + b       -a + c

-b + a        3b         -b + c    = 3(a + b + c)(ab + bc + ca)

-c + a      -c + b         3c

Given,

3a          -a + b       -a + c

Δ =    -b + a        3b         -b + c

-c + a      -c + b         3c

Applying C1 -> C1 + C2 + C3, we get

a + b + c       -a + b       -a + c

Δ =   a + b + c          3b         -b + c

a + b + c       -c + b         3c

1         -a + b       -a + c

Δ = (a + b + c)   1            3b         -b + c

1          -c + b         3c

Applying R2 -> R2 – R1 and R3 -> R3 – R1, we get

1          -a + b        -a + c

Δ = (a + b + c)   0          2b + a        a - b

0           a - c          2c + a

Expanding along C1, we get:

Δ = (a + b + c)[(2b + a)(2c + a) – (a - b)(a - c)]

Δ = (a + b + c)[4bc + 2ab + 2ac + a2 – a2 + ac + ab - bc]

Δ = (a + b + c)(3ab + 3bc + 3ca)

Δ = 3(a + b + c)(ab + bc + ca)

Hence, the given result is proved.

Question 14:

1          1 + p               1 + p + q

2          3 + 2p          4 + 3p + 2q    = 1

3          6 + 3p        10 + 6p + 3q

Given,

1          1 + p               1 + p + q

Δ =   2          3 + 2p          4 + 3p + 2q

3          6 + 3p        10 + 6p + 3q

Applying R2 -> R2 – 2R1 and R3 -> R3 – 3R1, we get

1         1 + p       1 + p + q

Δ =   0           1              2 + p

0           3             7 + 3p

Applying R3 -> R3 – 3R2, we get

1         1 + p       1 + p + q

Δ =   0           1              2 + p

0           0                 1

Expanding along C1, we get:

Δ = 0 + 0 + 1[1 * 1 – 0 *(2 + p)]

Δ = 1

Hence, the given result is proved.

Question 15:

sin α         cos α       cos(α + δ)

sin β         cos β       cos(β + δ)    = 0

sin γ         cos γ        cos(γ + δ)

Given,

sin α        cos α       cos(α + δ)

Δ =   sin β         cos β       cos(β + δ)

sin γ         cos γ        cos(γ + δ)

sin α * sin δ        cos α * cos δ       cos α * cos δ - sin α * sin δ

Δ = 1/(sin δ * cos δ)  sin β * sin δ        cos β * cos δ       cos β * cos δ - sin β * sin δ

sin γ * sin δ        cos γ * cos δ        cos γ * cos δ - sin γ * sin δ

Applying C1 -> C1 + C3, we get

cos α * cos δ        cos α * cos δ       cos α * cos δ - sin α * sin δ

Δ = 1/(sin δ * cos δ)  cos β * cos δ        cos β * cos δ       cos β * cos δ - sin β * sin δ

cos γ * cos δ         cos γ * cos δ        cos γ * cos δ - sin γ * sin δ

Δ = 0        [Since two columns C1 and C2 are identical]

Hence, the given result is proved.

Question 16:

Solve the system of the following equations

2/x + 3/y + 10/z = 4

4/x – 6/y + 5/z = 1

6/x + 9/y – 20/z = 2

Let 1/x = p, 1/y = q and 1/z = r

Then the given system of equations is as follows:

2p + 3q + 10r = 4

4p – 6q + 5r = 1

6p + 9q – 20r = 2

This system of equations can be written in the form of AX = B, where

2          3           10               p                        4

A =      4         -6            5    , X =    q     and B =      1

6          9          -20               r                        2

|A| = 2(120 - 45) - 3(-80 - 30) + 10(36 + 36)

= 150 + 330 + 720

= 1200

Now,

A11 = 75, A12 = 110, A13 = 72

A21 = 150, A22 = -100, A23 = 0

A31 = 75, A32 = 30, A33 = -24

So, A-1 =  adj(A) /|A| = (1/1200)   75         150           75

110      -100           30

72          0             -24

So, X = A-1B = (1/1200)   75         150         75     4

110      -100        30     1

72          0          -24     2

p     = (1/1200)     300 + 150 + 150

q                              400 - 100 + 60

r                                288 + 0 - 48

x     = (1/1200)    600    =      1/2

y                           400            1/3

z                            200            1/5

So, p =1/2, q = 1/3 and r = 1/5

Hence, x = 2, y = 3 and z = 5

Question 17:

If a, b, c, are in A.P., then the determinant

x + 2        x + 3       x + 2a

x + 3        x + 4       x + 2b

x + 4        x + 5       x + 2c

1. 0 B. 1 C. x                                          D. 2x

Given,

x + 2        x + 3       x + 2a

Δ =    x + 3        x + 4       x + 2b

x + 4        x + 5       x + 2c

x + 2            x + 3           x + 2a

Δ =    x + 3            x + 4          x + (a + c)

x + 4           x + 5            x + 2c                [If a, b, c are in AP then 2b = a + c]

Applying R1 -> R1 – R2 and R3 -> R3 – R2, we get

-1             -1                 a - c

Δ =    x + 3       x + 4         x + (a + c)

1                1                 c - a

Applying R1 -> R1 + R3, we get

0               0                    0

Δ =    x + 3       x + 4         x + (a + c)

1                1                 c - a

Δ = 0                  [Since all the elements of the first row (R1) are zero]

Hence, the correct answer is option A.

Question 18:

If x, y, z are nonzero real numbers, then the inverse of matrix

A =   x         0          0

0          y          0

0          0          z    is

(A)   x-1         0           0                          (B)                x-1         0           0

0          y-1           0                                    xyz    0           y-1          0

0           0          z-1                                              0           0           z-1

(C)                  x         0           0             (D)                      1          0           0

(1/xyz)   0          y           0                        (1/xyz)    0          1           0

0          0           z                                         0          0           1

Given,

A =   x         0          0

0          y          0

0          0          z

So, |A| = x(yz - 0) + 0 + 0

= xyz

Now,

A11 = yz, A12 = 0, A13 = 0

A21 = 0, A22 = xz, A23 = 0

A31 = 0, A32 = 0, A33 = xy

So, A-1 = adj(A) /|A| = (1/xyz)   yz         0           0

0           xz          0

0           0           xy

=  1/x           0             0

0             1/y            0

0              0            1/z

=  x-1             0           0

0             y-1           0

0             0            z-1

Hence, the correct answer is option A.

Question 19:

Let A =    1             sin θ           1

-sin θ       1           sin θ

-1         -sin θ            1   , where 0 ≤ θ ≤ 2π, then

A). Det (A) = 0                 B). Det (A) є (2, ∞)              C). Det (A) є (2, 4)            D). Det (A) є [2, 4]

Given,

1             sin θ           1

A =   -sin θ       1           sin θ

-1         -sin θ            1

Now, |A| = 1(1 + sin2 θ) – sin θ(-sin θ + sin θ) + 1(sin2 θ + 1)

=> |A| = 1 + sin2 θ + sin2 θ + 1

=> |A| = 2 + 2 sin2 θ

=> |A| = 2(1 + sin2 θ)

Now, 0 ≤ θ ≤ 2π

=> sin 0 ≤ sin θ ≤ sin 2π

=> 0 ≤ sin θ ≤ 1

=> 0 ≤ sin2 θ ≤ 1

=> 1 + 0 ≤ 1 + sin2 θ ≤ 1 + 1

=> 1 ≤ 1 + sin2 θ ≤ 2

=> 2 * 1 ≤ 2(1 + sin2 θ) ≤ 2 * 2

=> 2 ≤ 2(1 + sin2 θ) ≤ 4

So, Det (A) є [2, 4]

Hence, the correct answer is option D.