Class 12 - Maths - Matrices

Exercise 3.1

Question 1:

In the matrix A = [2       5      19      −7]

[35    -2       5/2    12]

[√3      1      −5       17]

write:

(i) The order of the matrix                                       (ii) The number of elements,

(iii) Write the elements a13, a21, a33, a24, a23

Answer:

(i) In the given matrix, the number of rows is 3 and the number of columns is 4.

Therefore, the order of the matrix is 3 * 4.

(ii) Since the order of the matrix is 3 * 4, there are 3 * 4 = 12 elements in it.

(iii) a13 = 19,a21 = 35, a33 = -5, a24 = 12, a23 = 5/2

Question 2:

If a matrix has 24 elements, what are the possible order it can have? What, if it has 13 elements?

Answer:

We know that if a matrix is of the order m * n, it has mn elements. Thus, to find all the

possible orders of a matrix having 24 elements, we have to find all the ordered pairs of natural

numbers whose product is 24.

The ordered pairs are: (1, 24), (24, 1), (2, 12), (12, 2), (3, 8), (8, 3), (4, 6), and (6, 4).

Hence, the possible orders of a matrix having 24 elements are:

1 * 24, 24 * 1, 2 * 12, 12 * 2, 3 * 8, 8 * 3, 4 * 6, and 6 * 4

(1, 13) and (13, 1) are the ordered pairs of natural numbers whose product is 13.

Hence, the possible orders of a matrix having 13 elements are 1 * 13 and 13 * 1.

Question 3:

If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?

Answer:

We know that if a matrix is of the order m × n, it has mn elements. Thus, to find all the

possible orders of a matrix having 18 elements, we have to find all the ordered pairs of natural

numbers whose product is 18.

The ordered pairs are: (1, 18), (18, 1), (2, 9), (9, 2), (3, 6,), and (6, 3).

Hence, the possible orders of a matrix having 18 elements are:

1 * 18, 18 * 1, 2 * 9, 9 * 2, 3 * 6, and 6 * 3

(1, 5) and (5, 1) are the ordered pairs of natural numbers whose product is 5.

Hence, the possible orders of a matrix having 5 elements are 1 * 5 and 5 * 1.

Question 4:

Construct a 2 * 2 matrix, A = [aij], whose elements are given by:

(i) aij = (i + j)2/2                          (ii) aij = i/j                                (iii) aij = (i + 2j)2/2

Answer:

(i) Given that aij = (i + j)2/2

In general a 2 * 2 matrix is given by A = [a11    a12]

[a21    a22]

Now,

a11 = (1 + 1)2/2 = 4/2 = 2

a12 = (1 + 2)2/2 = 9/2

a21 = (2 + 1)2/2 = 9/2

a22 = (2 + 2)2/2 = 16/2 = 8

Therefore, the required matrix is A = [2    9/2]

[9/2   8]

(ii) Given that aij = i/j

In general a 2 * 2 matrix is given by A = [a11    a12]

[a21    a22]

Now,

a11 = 1/1 = 1

a12 = 1/2

a21 = 2/1 = 2

a22 = 2/2 = 1

Therefore, the required matrix is A = [1    1/2]

[2      1]

(iii) Given that aij = (i + 2j)2/2

In general a 2 * 2 matrix is given by A = [a11    a12]

[a21    a22]

Now,

a11 = (1 + 2 * 1)2/2 = 9/2

a12 = (1 + 2 * 2)2/2 = 25/2

a21 = (2 + 2 * 1)2/2 = 16/2 = 8

a22 = (2 + 2 * 2)2/2 = 36/2 = 18

Therefore, the required matrix is A = [9/2    25/2]

[8           18]

Question 5:

Construct a 3 * 4 matrix, whose elements are given by

(i) aij = (1/2)|-3i + j|                            (ii) aij = 2i − j

Answer:

(i) Given, aij = (1/2)|-3i + j|

In general a 3 * 4 matrix is given by A = [a11    a12    a13    a14]

[a21    a22   a23    a24]

[a31    a32   a33    a34]

Now,

a11 = (1/2)|−3 * 1 + 1| = 1

a12 = (1/2)|−3 * 1 + 2| = 1/2

a13 = (1/2)|−3 * 1 + 3| = 0

a14 = (1/2)|−3 * 1 + 4| = 1/2

a21 = (1/2)|−3 * 2 + 1| = 5/2

a22 = (1/2)|−3 * 2 + 2| = 2

a23 = (1/2)|−3 * 2 + 3| = 3/2

a24 = (1/2)|−3 * 2 + 4| = 1

a31 = (1/2)|−3 * 3 + 1| = 4

a32 = (1/2)|−3 * 3 + 2| = 7/2

a33 = (1/2)|−3 * 3 + 3| = 3

a34 = (1/2)|−3 * 3 + 4| = 5/2

Therefore, the required matrix is

A = [1    1/2    0    1/2]

[5/2   2   3/2    1]

[4    7/2   3    5/2]

(ii) Given, aij = 2i - j

In general a 3 × 4 matrix is given by A = [a11    a12    a13    a14]

[a21    a22   a23    a24]

[a31    a32   a33    a34]

Now,

a11 = 2 * 1 - 1 = 1

a12 = 2 * 1 - 2 = 0

a13 = 2 * 1 - 3 = -1

a14 = 2 * 1 - 4 = -2

a21 = 2 * 2 - 1 = 3

a22 = 2 * 2 - 2 = 2

a23 = 2 * 2 - 3 = 1

a24 = 2 * 2 - 4 = 0

a31 = 2 * 3 - 1 = 5

a32 = 2 * 3 - 2 = 4

a33 = 2 * 3 - 3 = 3

a34 = 2 * 3 - 4 = 2

Therefore, the required matrix is

A = [1    0    -1    2]

[3   2     1     0]

[5    4    3      2]

Question 6:

Find the value of x, y, and z from the following equation:

(i) [4        3] = [y      z]                        (ii) [x + y        2] = [6      2]

[x         5]    [1      5]                             [5 + z      xy]     [5      8]

(iii) [x + y + z] =                             (iv) [4        3] = [y      z]

[x + z]                                           [x        5]     [1      5]

[y + z]        

Answer:

(i) [4        3] = [y      z]

[x         5]    [1      5]

As the given matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements, we get:

x = 1, y = 4, and z = 3

(ii) [x + y        2] = [6      2]

[5 + z      xy]     [5      8]

As the given matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements, we get:

x + y = 6, xy = 8, 5 + z = 5

Now,

5 + z = 5

=> z = 0

we know that:

(x − y)2 = (x + y)2 − 4xy

=> (x − y)2 = 36 − 32

=> (x − y)2 = 4

=> x − y = ±2

Now, when x − y = 2 and x + y = 6, we get x = 4 and y = 2

When x − y = − 2 and x + y = 6, we get x = 2 and y = 4

So, x = 4, y = 2, and z = 0 or x = 2, y = 4, and z = 0

(iii) [x + y + z] = 

[x + z]        

[y + z]        

As the two matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements, we get:

x + y + z = 9 ... (1)

x + z = 5 ........ (2)

y + z = 7 ........ (3)

From equation (1) and (2), we have:

y + 5 = 9

=> y = 4

Then, from equation (3), we have:

4 + z = 7

z = 3

Now, x + z = 5

=> x = 2

Hence, x = 2, y = 4 and z = 3.

(iv) [4        3] = [y      z]

[x        5]     [1      5]

As the two matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements, we get:

y = 4, z = 3 and x = 1

Question 7:

Find the value of a, b, c, and d from the equation:

[a - b        2a + c] = [-1      5]

[2a - b      3c + d]    [0      13]

Answer:

Given,

[a - b        2a + c] = [-1      5]

[2a - b      3c + d]    [0      13]

As the two matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements, we get:

a − b = −1 ...... (1)

2a − b = 0 ...... (2)

2a + c = 5 ....... (3)

3c + d = 13 ..... (4)

From equation (2), we have:

b = 2a

Then, from equation (1), we have:

a − 2a = −1

=> a = 1

and b = 2

Now, from equation (3), we have:

2 *1 + c = 5

=> c = 3

From equation (4) we have:

3 * 3 + d = 13

=> 9 + d = 13

=> d = 4

So, a = 1, b = 2, c = 3 and d = 4

Question 8:

A = [aij]m*n is a square matrix, if

(A) m < n                                (B) m > n                                  (C) m = n                     (D) None of these

Answer:

It is known that a given matrix is said to be a square matrix if the number of rows is equal to

the number of columns.

Therefore, A = [aij]m*n is a square matrix, if m = n

Hence, the correct answer is option C.

Question 9:

Which of the given values of x and y make the following pair of matrices equal

[3x + 7        5      ] = [0      y - 2]

[y + 1       2 − 3x]     [8        4   ]

(A) x = −1/3, y = 7                                        (B) Not possible to find

(C) y = 7, x = −2/3                                        (D) x = −1/3, y = −2/3

Answer:

It is given that

[3x + 7        5      ] = [0      y - 2]

[y + 1       2 − 3x]     [8        4   ]

Equating the corresponding elements, we get:

3x + 7 = 0

=> x = -7/3

5 = y – 2

=> y = 7

y + 1 = 8

=> y = 7

and 2 − 3x = 4

=> x = −2/3

We find that on comparing the corresponding elements of the two matrices, we get two

different values of x, which is not possible.

So, it is not possible to find the values of x and y for which the given matrices are equal.

Hence, the correct answer is option B.

Question 10:

The number of all possible matrices of order 3 * 3 with each entry 0 or 1 is:

(A) 27                            (B) 18                              (C) 81                                   (D) 512

Answer:

The given matrix of the order 3 × 3 has 9 elements and each of these elements can be either 0

or 1.

Now, each of the 9 elements can be filled in two possible ways.

Therefore, by the multiplication principle, the required number of possible matrices is

29 = 512

Hence, the correct answer is option D.

Exercise 3.2

Question 1:

Let A = [2       4], B = [1       3], C = [-2        5]

[3        2]         [-2     5]        [3          4]

Find each of the following

(i) A + B                  (ii) A – B                   (iii) 3A – C               (iv) AB                      (v)   BA

Answer:

(i) A + B = [2       4] + [1       3] = [2 + 1        4 + 3] = [3        7]

[3       2]    [-2      5]    [3 - 2         2 + 5]     [1       7]

(ii) A + B = [2       4] - [1       3] = [2 - 1               4 - 3] = [1        1]

[3       2]    [-2      5]    [3 – (-2)         2 - 5]     [5       -3]

(iii) 3A - C = 3[2       4] - [-2       5] = [3 * 2        3 * 4] = [-2        5]

[3       2]   [3        4]    [3 * 3         3 * 2]    [3         4]

[6        12] – [-2       5] = [6 + 2    12 - 5] = [8         7]

[9         6]      [3        4]    [9 – 3      6 - 4]    [6          2]

(iv) Matrix A has 2 columns. This number is equal to the number of rows in matrix B.

Therefore, AB is defined as:

AB = [2       4] * [1       3] = [2 * 1 + 4 * (-2)        2 * 3 + 4 * 5] = [2 - 8      6 + 20] = [-6      26]

[3       2]    [-2       5]   [3 * 1 + 2 * (-2)        3 * 3 + 2 * 5]    [3 - 4       9 + 10]    [-1      19]

(v) Matrix B has 2 columns. This number is equal to the number of rows in matrix A.

Therefore, BA is defined as:

BA = [1       3] * [2       4] = [1 * 2 + 3 * 3         1 * 4 + 3 * 2] = [2 + 9        4 + 6] = [11      10]

[-2     5]    [3       2]     [-2 * 2 + 5 * 3        -2 * 4 + 5 * 2]  [-4 + 15   -8 + 10]   [11       2]

Question 2:

Compute the following:

(i) [a       b] + [a      b]                               (ii) [a2 + b2       b2 + c2] + [2ab      2bc]

[-b      a]    [b      a]                                     [a2 + c2      a2 + b2]    [-2ac      -2ab]

(iii) [-1      4       -6]     [12     7      6]       (iv) [cos2 x     sin2 x] + [sin2 x      cos2 x]

[8        5       16] + [8       0      5]              [sin2 x      cos2 x]    [cos2 x     sin2 x]

[2        8         5]     [3      2       4]

Answer:

(i) [a       b] + [a      b] = [a + a       b + b] = [2a      2b]

[-b      a]    [b      a]    [-b + b     a + b]    [0         2a]

(ii) [a2 + b2       b2 + c2] + [2ab      2bc] = [a2 + b2 + 2ab      b2 + c2 + 2bc] = [(a + b)2   (b + c)2]

[a2 + c2      a2 + b2]    [-2ac      -2ab]    [a2 + c2 – 2ac      a2 + b2 – 2ab]     [(a - c)2    (a - b)2]

(iii) [-1      4       -6]     [12     7      6] = [-1 + 12    4 + 7     -6 + 6] = [11         11          0]

[8        5       16] + [8       0      5]     [8 + 8        5 + 0     16 + 5]    [16         5           21]

[2        8         5]     [3      2       4]    [2 + 3        8 + 2       5 + 4]     [5          10           9]

(iv) [cos2 x     sin2 x] + [sin2 x      cos2 x] = [cos2 x + sin2 x       sin2 x + cos2 x] = [1     1]

[sin2 x      cos2 x]     [cos2 x     sin2 x]     [sin2 x + cos2 x       cos2 x + sin2 x]    [1     1]

{Since sin2 x + cos2 x = 1}

Question 3:

Compute the indicated products

(i) [a     b][a      -b]                              (ii) 

[-b   a][b        a]                                    [2     3      4]



(iii) [1     -2][1     2      3]

[2       3][2     3     1]

(iv) [2     3     4][1     -3      5]

[3     4     5][0     2       4]

[4     5     6][3     0       5]

(v)  [2     1][1       0      1]

[3     2][-1     2       1]

[-1    1]

(vi) [3      -1        3][2         -3]

[-1     0         2][1          0]

[3           1]

Answer:

(i) [a     b][a      -b] = [a * a + b * b       a * (-b) + b * a]

[-b   a][b        a]    [-b * a + a * b     -b * (-b) + a * a]

= [a2 + b2       -ab + ab]

[-ab + ab       b2 + a2]

= [a2 + b2       0     ]

[0            a2 + b2]

(ii)   1                               1 * 2   1 * 3    1 * 4       =     2       3      4

2   [2    3     4] =      2 * 2   2 * 3    2 * 4              4       6       8

3                              3 * 2    3 * 3    3 * 4              6       9     12

(iii)    1     -2      1     2      3    =     1 * 1 – 2 * 2      1 * 2 – 2 * 3       1 * 3 – 2 * 1

2       3     2     3      1            2 * 1 + 3 * 2      2 * 2 + 3 * 3       2 * 3 + 3 * 1

=    1 – 4     2 – 6      3 – 2   =    -3           -4               1

2 + 6     4 + 9      6 + 3         8             13             9

(iv)    2      3        4       1        -3          5

3       4        5      0          2          4

4        5       6      3          0          5

=    2 * 1 + 3 * 0 + 4 * 3          2 * (-3) + 3 * 2 + 4 * 0              2 * 5 + 3 * 4 + 4 * 5

3 * 1 + 4 * 0 + 5 * 3          3 * (-3) + 4 * 2 + 5 * 0              3 * 5 + 4 * 4 + 5 * 5

4 * 1 + 5 * 0 + 6 * 3          4 * (-3) + 5 * 2 + 6 * 0              4 * 5 + 5 * 4 + 6 * 5

=    2 + 0 + 12        -6 + 6 + 0          10 + 12 + 20           14       0        42

3 + 0 + 15        -9 + 8 + 0          15 + 16 + 25   =      18       -1       56

4 + 0 + 18       -12 + 10 + 0      20 + 20 + 30           22       -2        70

(v)    2     1      1       0     1

3     2]    -1     2       1

-1     1

=    2 * 1 + 1 * (-1)         2 * 0 + 1 * 2           2 * 1 + 1 * 1

3 * 1 + 2 * (-1)         3 * 0 + 2 * 2           3 * 1 + 2 * 1

-1 * 1 + 1 * (-1)       -1 * 0 + 1 * 2         -1 * 1 + 1 * 1

=    2 - 1    0 + 2       2 + 1           1       2        3

3 - 2    0 + 4       3 + 2   =      1       4        5

-1 - 1    0 + 2    -1 + 1           -2      2        0

(vi)     3      -1       3        2         -3

-1     0         2        1          0

3          1

=    3 * 2 - 1 * 1 + 3 * 3         3 * (-3) - 1 * 0 + 3 * 1

-1 * 2 + 0 * 1 + 2 * 3       -1 * (-3) + 0 * 0 + 2 * 1

=    6 – 1 + 9    -9 – 0 + 3    =     14           -6

6 – 1 + 9     3 + 0 + 2             4             5

Question 4:

If A =    1         2          -3    , B =     3       -1        2  ,  C =    4           1              2

5          0            2                4        2         5              0           3               2

1         -1           1                -2       0         3              1           -2             3

Then compute A + B and B – C. Also verify that A + (B - C) = (A + B) – C

Answer:

A + B =    1         2          -3    +     3       -1        2

5          0            2          4        2         5

1         -1           1          -2       0         3

=    1 + 3    2 - 1       -3 + 2          4       1        -1

5 + 4    0 + 2       2 + 5   =     9       2         7

1 + 2   -1 + 0      1 + 3           3      -1        4

B - C =     3         -1          2    -     4        1         2

4          2           5          0        3         2

2         0            3          1       -2         3

=    3 - 4    -1 - 1      2 - 2           -1       -2        0

4 - 0    2 - 3       5 – 2    =     4       -1         3

2 - 1   0 – (-2)   3 - 3            1        2         0

A + (B - C) =    1         2          -3    +     -1       -2        0

5          0          2           4        -1         0

1         -1         1          -1         2         0

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

5 + 4        0 + (-1)         2 + 3   =    9       -1         5

1 + 1        -1 + 2            1 + 0         2        1         1

(A + B) - C =    4          1          -1    -    4         1         2

9          2           7          0        3          2

3         -1           4          1       -2         3

=    4 – 4          1 - 1          -1 - 2          0       0          -3

9 - 0          2 - 3             7 - 2   =    9       -1          5

3 - 1        -1 – (-2)        4 - 3         2         1          1

Hence, we have verified that A + (B - C) = (A + B) – C

Question 5:

If A =    2/3     1         5/3   , B =     2/5       3/5         1

1/3     2/3       4/3               1/5       2/5       4/5

7/3       2         2/3               7/5       6/5       2/5

Then compute 3A - 5B

Answer:

3A – 5B =3   2/3         1        5/3  - 5    2/5       3/5        1

1/3        2/3     4/3           1/5       2/5       4/5

7/3         2       2/3          7/5        6/5       2/5

=    2         3          5    +     2       3        5    =    0           0             0

1         2          4           1       2         4         0           0             0

7         6          2           7       6         2         0           0             0

Question 6:

Simplify cos θ    cos θ         sin θ     + sin θ    sin θ         -cos θ

-sin θ         cos θ                    cos θ         sin θ

Answer:

Given,    cos θ    cos θ         sin θ     + sin θ    sin θ         -cos θ

-sin θ         cos θ                    cos θ         sin θ

=    cos2 θ                      cos θ * sin θ    +      sin2 θ             -sin θ * cos θ

-sin θ * cos θ              cos2 θ                   sin θ * cos θ         sin2 θ

=    cos2 θ + sin2 θ                                 cos θ * sin θ - sin θ * cos θ

-sin θ * cos θ + sin θ * cos θ               cos2 θ + sin2 θ

=     1          0

0          1               [Since cos2 θ + sin2 θ = 1]

Question 7:

Find X and Y, if

(i) X + Y =    7            0             X – Y =   3              0

2            5   and                   0             3

(ii) 2X + 3Y =    2            3             3X + 2Y =   2             -2

4            0   and                       -1            5

Answer:

(i) Given,

X + Y =    7            0

2            5

X - Y =    3            0

0            3       ………..2

Adding equation 1 and 2, we get

2X     =    7            0   +     3          0    +    7 + 3       0 + 0   =    10          0

2            5          0           3          2 + 0        5 + 3          2          8

=> X = (1/2)   10           0  =   5            0

2           8         1           4

Now, X + Y =    7            0

2            5

=>          5            0    + Y =     7          0

1            4                 2          5

=> Y =    7            0   -     5            0

2            5          1           4

=> Y =    7 - 5           0 - 0   =   2           0

2 - 1         5 - 4         1           1

(ii) Given,

2X + 3Y =    2             3

3            0     …………3

3X + 2Y =       2           -2

-1          5       ………..4

Multiply equation 3 by 2, we get

2(2X + 3Y) =2   2        3

3        0

4X + 6Y =       4            6

8           0       ………..5

Multiply equation 4 by 3, we get

3(3X + 2Y) =3   2        3

3        0

9X + 6Y =        6          -6

-3         15       ………..6

From equation 5 and 6, we get

=> (4X + 6Y) – (9X + 6Y) =    4           6     =   6           -6

8           0           -3        15

=> -5X =    -2         12

11       -15

=> X = -(1/5)   -2         12

11      -15

=> X =   2/5         -12/5

-11/5          3

Now,  2X + 3Y =    2             3

3             0

=>     2    2/5      -12/5     + 3y =    2            3

-11/5            3                  4            0

=>            4/5      -24/5     + 3y =    2            3

-22/5          6                     4            0

=> 3Y =    2             3   -     4/5        -24/5

4             0         -22/5       6

=> 3Y =    2 – 4/5       3 + 24/5

4 + 22/5       0 - 6

=> Y = (1/3)     6/5       39/5

42/5      -6

=> Y =   2/5       13/5

14/5      -2

Question 8:

Find X if

Y =    3            2             2X + Y =   1            0

1            4   and                   -3           2

Answer:

Given,

2X + Y =    1            0

-3           2

=> 2X +   3             2   =    1            0

1           4           -3         2

=> 2X =    1             0   -      3            2

-3           2          1           4

=> 2X =    1 - 3          0 - 2

-3 - 1        2 - 4

=> X = (1/2)   -2           -2

-4            -2

=> X =    -1           -1

-2           -1

Question 9:

Find x and y if  2   1        3   +   y          0  =   5          6

0         x        1          2        1          8

Answer:

Given,

2    1        3    +   y          0 =   5         6

0         x        1          2        1          8

=>                          2        6    +   y          0   =   5         6

0        2x       1          2       1          8

=>                          2 + y        6     =   5         6

1        2x + 2       1          8

Comparing the corresponding elements of these two matrices, we get:

2 + y = 5

=> y = 3

And 2x + 2 = 8

=> 2x = 6

=> x = 3

So, x = 3 and y = 3

Question 10:

Solve the equation for x, y, z and t if

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

y         t            0          2         4          6

Answer:

Given,

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

y         t            0          2         4          6

=>                           2x         2z +    3       -3   =      9        15

2y         2t       0          6          12      18

=>                          2x + 3     2z - 3     =    9        15

2y          2t + 6           12       18

Comparing the corresponding elements of these two matrices, we get:

2x + 3 = 9

=> 2x = 6

=> x = 3

2y = 12

=> y = 6

2z – 3 = 15

=> 2z = 18

=> z = 9

2t + 6 = 18

=> 2t = 12

=> t = 6

So, x = 3, y = 6, z = 9 and t = 6

Question 11:

If x     2   + y   -1   =     10

3            1            5     , find the values of x and y.

Answer:

Given,

x     2   + y   -1   =     10

3            1            5

=>     2x     +   -y   =     10

3x           y            5

=>     2x - y     =     10

3x + y            5

Comparing the corresponding elements of these two matrices, we get:

2x – y = 10 and 3x + y = 5

Adding these two equations, we get

5x = 15

=> x = 5

Now, 3x + y = 5

=> y = 5 − 3x

=> y = 5 − 9

=> y = −4

So, x = 3 and y = −4

Question 12:

Given, 3    x       y   =   x       6    +   4          x + y

z      w        -1    2w       z + w      3      , find the values of x, y, z and w.

Answer:

Given, 3    x       y   =   x       6    +   4          x + y

z      w        -1    2w       z + w      3

=>             3x     3y  =   x + 4                6 + x + y

3z    3w        -1 + z + w         2w + 3

Comparing the corresponding elements of these two matrices, we get:

3x = x + 4

=> 2x = 4

=> x = 2

3y = 6 + x + y

=> 2y = 6 + x

=> 2y = 6 + 2

=> 2y = 8

=> y = 4

3w = 2w + 3

=> w = 3

3z = -1 + z + w

=> 2z = -1 + w

=> 2z = -1 + 3

=> 2z = 2

=> z = 1

So, x = 2, y = 4, z = 1 and w = 3

Question 13:

If           cos x     -sin x      0

F(x) =    sin x      cos x      0

0            0          1

Show that F(x)F(y) = f(x + y)

Answer:

Given,

cos x     -sin x      0

F(x) =    sin x      cos x      0

0            0          1

Now,

cos y     -sin y      0

F(y) =    sin y      cos y      0

0            0          1

cos (x + y)     -sin (x + y)          0

F(x + y) =    sin (x + y)       cos (x + y)         0

0                            0                1

Now, F(x)F(y) =    cos x      -sin x       0      cos y      -sin y       0

sin x       cos x       0      sin y        cos y       0

cos x      -sin x       1      cos y      -sin y       1

=     cos x * cos y – sin x * sin y + 0          -cos x * sin y – sin x * cos y + 0            0

cos x * cos y – sin x * sin y + 0          -cos x * sin y – sin x * cos y + 0            0

0                                                              0                                          1

cos (x + y)     -sin (x + y)          0

=    sin (x + y)       cos (x + y)         0

0                          0                1

= F(x + y)

So, F(x)F(y) = f(x + y)

Question 14:

Show that

(i)     5          -1     2           1  ≠   2             1   5            -1

6           7      3          4        3             4    6            7

(ii)    1        2        3     -1       1        0            -1       1        0     1        2         3

0        1       0      0       -1         1    ≠     0        -1       1    0         1         0

1        1       0      1        1         0           2         3       4      1        1        0

Answer:

(i)     5          -1     2           1  =   5 * 2 – 1 * 3            5 * 1 – 1 * 4

6           7      3          4        6 * 2 + 7 * 3            6 * 1 + 7 * 4

=                10 - 3        5 - 4       =    7        1

12 + 21    6 + 28            33    34

2           1     5           -1  =   2 * 5 + 1 * 6            2 * (-1) + 1 * 7

3           4      6          7        3 * 5 + 4 * 6            3 * (-1) + 4 * 7

=          10 + 6       -2 + 7     =    16        5

15 + 24    -3 + 28          39      25

So    5          -1     2           1    ≠   2          1    5            -1

6           7      3          4         3            4    6            7

(ii)    1        2        3     -1       1        0

0        1       0      0       -1         1

1        1       0      1        1         0

=    1 * (-1) + 2 * 0 + 3 * 2               1 * 1 + 2 * (-1) + 3 * 3               1 * 0 + 2 * 1 + 3 * 4

0 * (-1) + 1 * 0 + 0 * 2               0 * 1 + 1 * (-1) + 0 * 3               0 * 0 + 1 * 1 + 0 * 4

1 * (-1) + 1 * 0 + 0 * 2               0 * 1 + 1 * (-1) + 0 * 3               1 * 0 + 1 * 1 + 0 * 4

=     5         8       14

0         -1          1

-1         0          1

-1        1       0     1       2        3

0        -1       1     0       1         0

2        3       4      1        1         0

=    -1 * 1 + 1 * 0 + 0 * 2                  -1 * 2 + 1 * 1 + 0 * 1               -1 * 3 + 1 * 0 + 0 * 0

0 * 1 + (-1) * 0 + 1 * 1                 0 * 2 + (-1) * 1 + 1 * 1          0 * 3 + (-1) * 0 + 1 * 0

2 * 1 + 3 * 0 + 4 * 1                    2 * 2 + 3 * 1 + 4 * 1                2 * 3 + 3 * 0 + 4 * 0

=     -1       -1       -3

1         0          0

6         11        6

So,   1        2        3     -1       1        0            -1       1        0     1        2         3

0        1       0      0       -1         1    ≠     0        -1       1    0         1         0

1        1       0      1        1         0           2         3       4      1        1        0

Question 15:

If A =    2        0       1

2         1        3

1        -1        0

Find A2 – 5A + 6I

Answer:

We have A2 = A * A

=> A2 =      2        0       1      2       0         1

2        1       3      2       1         3

1       -1       0     1       -1         0

=    2 * 2 + 0 * 2 + 1 * 1                   2 * 0 + 0 * 1 + 1 * (-1)            2 * 1 + 0 * 3 + 1 * 0

2 * 2 + 1 * 2 + 3 * 1                  2 * 1 + 1 * 1 + 3 * (-1)             2 * 1 + 1 * 3 + 3 * 0

1 * 2 + (-1) * 2 + 0 * 1              1 * 0 + (-1) * 1 + 0 * (-1)        1 * 1 + (-1) * 3 + 0 * 0

=     5       -1          2

9        -2          5

0        -1        -2

So, A2 – 5A + 6I

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

9         -2          5               2          1          3             0          1          0

0          -1        -2               1         -1         0              0         0          1

=    5       -1          2       -      10          0         5     +      6         0          0

9         -2          5              10         5        15             0         6          0

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

=    5 – 10 + 6       -1 – 0 + 0        2 – 5 + 0

9 – 10 + 0       -2 – 5 + 6       5 – 15 + 0

0 – 5 + 0       -1 + 5 + 0        -2 – 0 + 0

=   1         -1         -3

-1        -1       -10

-5         4           4

Question 16:

If A =   1        0        2

0        2        1

2        0        3

Prove that A3 – 6A2 + 7A + 2I = 0

Answer:

We have A2 = A * A

=> A2 =      1        0       2      1       0         2

0        2       1      0       2         1

2        0       3     2        0         3

=     1 + 0 + 4                   0 + 0 + 0            2 + 0 + 6

0 + 0 + 2                   0 + 4 + 0            0 + 2 + 3

2 + 0 + 6                   0 + 0 + 0            4 + 0 + 9

=     5         0          8

2         4          5

8         0         13

Again A3 = A2 * A

=> A2 =      5        0       8      1       0         2

2        4       5      0       2         1

8        0      13     2        0         3

=     5 + 0 + 16                 0 + 0 + 0           10 + 0 + 24

2 + 0 + 10                 0 + 8 + 0            4 + 4 + 15

8 + 0 + 26                 0 + 0 + 0            15 + 0 + 39

=    21         0       34

12         8        23

34        0        55

Now, A3 – 6A2 + 7A + 2I

=     21        0       34    - 6    5         0          8      + 7     1         0          2       + 2    1         0            0

12        8       23             2         4          5                0         2           1               0          1            0

34        0       55             8         0         13               2         0           3               0          0            1

=     21        0       34    -       30       0         48      +       7         0         14       +      2         0            0

12        8       23            12       24        30              0         14         7               0          2            0

34        0       55             48         0       78              14         0        21               0          0           2

=    21 – 30 + 7 + 2           0 – 0 + 0 + 0           34 – 48 + 14 + 0

12 – 12 + 0 + 0           8 – 24 + 14 + 2       23 – 30 + 7 + 0

34 – 48 + 14 + 0         0 – 0 + 0 + 0           55 – 78 + 21 + 2

=    0         0         0

0         0         0

0         0         0

So, A3 – 6A2 + 7A + 2I = 0

Question 17:

If A =     3         -2               I =   1             0

4         -2      and          0             1  , find k so that A2 = kA – 2I

Answer:

We have A2 = A * A

=> A2 =    3           -2      3           -2

4           -2      4          -2

=     3 * 3 + (-2) * 4       3 * (-2) + (-2) * (-2)      =      1            -2

4 * 3 + (-2) * 4       4 * (-2) + (-2) * (-2)               4           -4

Now, A2 = kA – 2I

=>    1             -2    = k     3           -2     - 2     1            0

4            -4               4           -2              0             1

=>    1             -2    =       3k        -2k     -       2            0

4            -4               4k       -2k              0            2

=>    1             -2    =       3k - 2        -2k

4            -4               4k       -2k - 2

Comparing the corresponding elements, we have:

3k – 2 = 1

=> 3k = 3

=> k = 1

Thus, the value of k is 1.

Question 18:

If A =      0           -tan α/2

tan α/2        0        and I is the identity matrix of order 2, show that

I + A = (I - A)   cos α         -sin α

sin α           cos α

Answer:

LHS:

I + A =    1                       0        0               - tan α/2

0                       1        tan α/2          0

=    1           -tan α/2

tan α/2         1          ………………..1

RHS:

(I - A)   cos α         -sin α

sin α           cos α

=     1                       0        0               - tan α/2         cos α            -sin α

0                       1        tan α/2          0                sin α             cos α

=     1              tan α/2      cos α            -sin α

- tan α/2          1         sin α             cos α

=     cos α + sin α * tan α/2         -sin α + cos α * tan α/2

- cos α *  tan α/2 + sin α       sin α * tan α/2 + cos α      …………..2

=     1 – 2sin2 α/2 + 2 sin α/2 sin α/2 tan α/2        -2 sin α/2 cos α/2  + (2cos2 α/2 – 1) * tan α/2

- (2cos2 α/2 - 1) tan α/2 + 2sin α/2 cos α/2    2 sin α/2 cos α/2 tan α/2 + 1 – 2 sin2 α/2

=     1 – 2sin2 α/2 + 2sin2 α/2                                      -2 sin α/2 cos α/2  + 2 sin α/2 sin α/2 tan α/2

- 2sin α/2 cos α/2 + tan α/2 + 2sin α/2 cos α/2    2 sin2 α/2 + 1 – 2 sin2 α/2

=    1           -tan α/2

tan α/2         1

Thus, from equation 1 and 2, we get

LHS = RHS

So, I + A = (I - A)   cos α         -sin α

sin α          cos α

Question 19:

A trust fund has Rs 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of:

(a) Rs 1,800                                        (b) Rs 2,000

Answer:

(a) Let Rs x be invested in the first bond. Then, the sum of money invested in the second bond

will be Rs (30000 − x).

It is given that the first bond pays 5% interest per year and the second bond pays 7% interest

per year.

Therefore, in order to obtain an annual total interest of Rs 1800, we have:

[x         (30000 - x)][5/100] = 1800                  [SI for 1 year = (Principal * Rate)/100]

[7/100]

=> 5x/100 + 7(30000 - x)/100 = 1800

=> 5x + 210000 – 7x = 180000

=> 210000 – 2x = 180000

=> 2x = 210000 – 180000

=> 2x = 30000

=> x = 15000

Thus, in order to obtain an annual total interest of Rs 1800, the trust fund should invest Rs

15000 in the first bond and the remaining Rs 15000 in the second bond.

(b) Let Rs x be invested in the first bond. Then, the sum of money invested in the second bond

will be Rs (30000 − x).

Therefore, in order to obtain an annual total interest of Rs 2000, we have:

[x         (30000 - x)][5/100] = 2000                  [SI for 1 year = (Principal * Rate)/100]

[7/100]

=> 5x/100 + 7(30000 - x)/100 = 2000

=> 5x + 210000 – 7x = 200000

=> 210000 – 2x = 200000

=> 2x = 210000 – 200000

=> 2x = 10000

=> x = 5000

Thus, in order to obtain an annual total interest of Rs 2000, the trust fund should invest Rs

5000 in the first bond and the remaining Rs 25000 in the second bond.

Question 20:

The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are Rs 80, Rs 60 and Rs 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

Answer:

The bookshop has 10 dozen chemistry books, 8 dozen physics books, and 10 dozen economics

books.

The selling prices of a chemistry book, a physics book, and an economics book are respectively

given as Rs 80, Rs 60, and Rs 40.

The total amount of money that will be received from the sale of all these books can be

represented in the form of a matrix as:

12[10    8    10]





= 12[10 * 80 + 8 * 60 + 10 * 40]

= 12[800 + 480 + 400]

= 12 * 1680

= 20160

Thus, the bookshop will receive Rs 20160 from the sale of all these books.

Question 21:

Assume X, Y, Z, W and P are matrices of order 2 * n, 3 * k, 2 * p, n * 3 and p * k respectively. The restriction on n, k and p so that PY + WY will be defined are:

1. k = 3, p = n B. k is arbitrary, p = 2 C. p is arbitrary, k = 3         D. k = 2, p = 3

Answer:

Matrices P and Y are of the orders p * k and 3 * k respectively.

Therefore, matrix PY will be defined if k = 3.

Consequently, PY will be of the order p * k.

Matrices W and Y are of the orders n * 3 and 3 * k respectively.

Since the number of columns in W is equal to the number of rows in Y, matrix WY is well-

defined and is of the order n * k.

Matrices PY and WY can be added only when their orders are the same.

However, PY is of the order p * k and WY is of the order n * k.

Therefore, we must have p = n

Thus, k = 3 and p = n are the restrictions on n, k, and p so that will be defined.

Question 22:

Assume X, Y, Z, W and P are matrices of order 2 * n, 3 * k, 2 * p, n * 3 and p * k respectively.     If n = p, then the order of the matrix 7X – 5Z is

1. p * 2 B. 2 * n C. n * 3                               D. p * n

Answer:

Matrix X is of the order 2 * n.

Therefore, matrix 7X is also of the same order.

Matrix Z is of the order 2 * p, i.e., 2 * n        [Since n = p]

Therefore, matrix 5Z is also of the same order.

Now, both the matrices 7X and 5Z are of the order 2 * n.

Thus, matrix 7X − 5Z is well-defined and is of the order 2 * n.

Hence, the correct answer is option B.

Exercise 3.3

Question 1:

Find the transpose of each of the following matrices:

(i)     5            (ii)      1          -1      (iii)     -1          5        6

1/2                   2            3                 √3        5         6

-1                                                         2         3         -1

Answer:

(i) Let A =     5

1/2   , then AT = [5         1/2          -1]

-1

(ii) Let A =   1            -1   , then AT =     1             2

2             3                          -1            3

(iii) Let A =    -1          5          6    , then AT =    -1        √3         2

√3        5          6                           5          5           3

2          3          -1                          6          6           -1

Question 2:

If A =   -1        2          3                              -4         1          -5

5          7         9                     B =    1          2           0

-2        1         1        and                1          3           1    then verify that

(i) (A + B)’ = A’ + B’

(ii) (A - B)’ = A’ - B’

Answer:

Given, A =   -1        2          3

5          7         9

-2         1          1

So, A’ =  -1         5         2

2          7         1

3          9          1

And

B =   -4         1         1

1          2         3

-5         0          1

So, B’ =  -4         1         1

1          2         3

-5         0          1

(i) A’ + B’ =    -1        5          -2              -4        1         1               -5         6          -1

2         7            1      +     1          2         3         =     3          9           4

3         9            1             -5         0         1                -2         9          2

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

5         7            9      +     1          2          0         =     6         9          9

-2        1            1              1         3           1               -1         4          2

So, (A + B)’ =   -5         6         -1

3          9           4

-2         9           2

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

(ii) A’ - B’ =   -1        5          -2              -4        1         1                3          4          -3

2         7            1      -      1          2         3         =    1          5           -2

3         9            1             -5         0         1               8          9            0

A - B =     -1         2            3              -4        1         -5              3          1          8

5         7            9      -      1          2          0      =      4          5          9

-2        1            1              1         3           1              -3         -2         0

So, (A - B)’ =    3          4         -3

1          5         -2

8          9           0

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

Question 3:

If A’ =   3              4             B =     -1         2           1

-1             2                         1         2           3

0               1    and

Then verify that

(i) (A + B)’ = A’ + B’

(ii) (A - B)’ = A’ - B’

Answer:

(i) We know that A = (A’)’

Therefore, we have

A =    3          -1         0

4           2          1

B’ =   -1             1

2              2

1              3

Now, A + B =    3          -1         0     +   -1          2          1     =     2          1           1

4           2         1            1          2          3             5         4           4

So, (A + B)’ =    2             5

• 4

1             4

A’ + B’ =   3             4              -1           1                  2             5

-1            2      +      2            2       =        1              4

0            1               1            3                 1              4

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

A + B =    3          -1         0     -    -1          2          1     =     4          -3          -1

4           2         1            1          2          3            3           0          -2

So, (A - B)’ =     4             3

-3             0

-1           -2

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

-1            2      -       2            2       =        -3            0

0            1               1            3                 -1           -2

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

Question 4:

If A’ =   -2            3           B =  -1           0

1           2   and           1           2   , the find (A + 2B)’

Answer:

We know that (A’)’ = A

So, A =    -2          1

3           2

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

3          2               1           2

=     -2        1        +     -2          0      =       -4          1

3          2               2           4               5            6

So, (A + 2B)’ =   -4          5

• 6

Question 6:

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

(i) A =    1    , B = [-1       2      1]

-4

3

(ii) A =   0    , B = [1       5      7]

1

2

Answer:

(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] =     2         -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’

Question 6:

(i) If A =    cos α      sin α

-sin α      cos α  , the verify that A’A = I

(ii) If A =    sin α      cos α

-cos α     sin α  , the verify that A’A = I

Answer:

(i)    A =    cos α      sin α

-sin α      cos α

So, A’ =    cos α      -sin α

sin α        cos α

Now A’A =   cos α      -sin α     cos α       sin α

sin α      cos α      -sin α       cos α

=    cos α * cos α + (-sin α) * (-sin α)               cos α * sin α + (-sin α) * cos α

sin α * cos α + cos α * (-sin α)                   sin α * sin α + cos α * cos α

=    cos2 α + sin2 α                                             cos α * sin α - sin α * cos α

sin α * cos α - cos α * sin α                              sin2 α + cos2 α

=    1              0

0              1

= I

Hence, it is verified that A’A = I

(ii)   A =    sin α        cos α

-cos α     sin α

So, A’ =    sin α      -cos α

cos α       sin α

Now A’A =   sin α      -cos α     sin α        cos α

cos α      sin α      -cos α      sin α

=    sin α * sin α + (-cos α) * (-cos α)               sin α * cos α + (-cos α) * sin α

cos α * sin α + sin α * (-cos α)                   cos α * cos α + sin α * sin α

=    sin2 α + cos2 α                                             sin α * cos α - sin α * cos α

sin α * cos α - sin α * cos α                              cos2 α + sin2 α

=    1              0

0              1

= I

Hence, it is verified that A’A = I

Question 7:

(i) Show that the matrix    1           -1            5

-1           2            1

5           1             3

is a symmetric matrix.

(ii) Show that the matrix   0           1            -1

-1          0             1

1          -1            0

is a skew-symmetric matrix.

Answer:

A’ =    1          -1           5

-1          2            1   = A

5           1            3

=> A’ = A

Hence, A is a symmetric matrix.

(ii) We have

A’ =    0          -1           1           0           1            -1

1           0          -1   = -   -1          0             1  = -A

-1         1            0            1         -1            0

=> A’ = -A

Hence, A is a skew-symmetric matrix.

Question 8:

For the matrix A =    1           5

6           7   , verify that

(i) (A + A’) is a symmetric matrix.                       (ii) (A + A’) is a skew-symmetric matrix.

Answer:

Given, A =    1           5

6           7

Now,  A’ =    1           6

5           7

Now, A + A’ =   1             5     +      1            6     =       2          11

6            7              5            7             11        14

So, (A + A’)’ =   2          11    = A + A’

11       14

Hence, A + A’ is a symmetric matrix.

(ii)      A - A’ =    1             5    -        1            6     =      0          -1

6            7              5            7             1          0

So, (A - A’)’  =   0            1    = -(A - A’)

-1           0

Hence, A + A’ is a skew-symmetric matrix.

Question 9:

Find (A + A’)/2 and (A – A’)/2, when A =   0          a            b

-a         0            c

-b        -c            0

Answer:

The given matrix is A =  0          a            b

-a         0            c

-b        -c            0

So, A’ =  0          -a         -b

a           0          -c

-b          c           0

Now, A + A’ =   0          a            b           0        -a          -b           0           0          0

-a          0           c     +    a          0           -c    =    0            0          0

-b        -c          0            b          c            0          0            0          0

So, (A + A’)/2 =       0         0           0

0         0           0

0         0           0

Now, A - A’ =    0          a            b           0        -a          -b           0          2a       2b

-a          0           c     -     a          0           -c    =    -2a        0        2c

-b        -c          0            b          c            0          -2b      -2c       0

So, (A - A’)/2 =       0          a            b

-a         0            c

-b        -c           0

Question 10:

Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

(i)    3                5

1               -1

(ii)    6           2          2

-2          3          -1

2         -1           3

(iii)     3          3          -1

-2        -2           1

-4        -5           2

(iv)   1                 5

-1               2

Answer:

(i) Let A =   3                 5

1                -1

Then A’ =   3                 5

• -1

Now, A + A’ =    3                 5     +    3                 1    =     6                 6

1                -1           5               -1            6               -2

Let P = (A + A’)/2 = (1/2)   6                6     =     3                3

6                -2           3               -1

Now, P’ =   3                 3   = P

3                -1

Thus, P = (A + A’)/2 is a symmetric matrix.

Now, A - A’ =     3                 5     -     3                 1    =     0                4

1                -1           5               -1            -4               0

Let Q = (A - A’)/2 =  (1/2)  0                 4     =    0                2

-4                0           -2               0

Now, Q’ =  0                 2   = -Q

-2                0

Thus, Q = (A - A’)/2 is a skew-symmetric matrix.

Representing A as the sum of P and Q:

P + Q =  3                 3     +    0                 2    =     3                 5   = A

3                -1          1                 0            1                -2

(ii) Let A =    6         -2          2

-2         3          -1

2          -1          3

Then, A’ =    6         -2          2

-2         3          -1

2          -1          3

Now, A + A’ =   6          -2           2            6          -2          2           12        -4          4

-2          3          -1    +     -2          3         -1           -4         6          -2

2          -1           3            2          -1          3           4         -2           6

Let P = (A + A’)/2 = (1/2)   12         -4          4            6          -2          2

-4          6          -2     =    -2          3         -1

4          -2           6            2          -1          3

Then, P’ =    6         -2          2     = P

-2         3          -1

2          -1          3

Thus, P = (A + A’)/2 is a symmetric matrix.

Now, A - A’ =    6          -2           2            6          -2          2           0          0           0

-2          3          -1    -      -2          3         -1           0          0           0

2          -1           3            2          -1          3           0          0           0

Let Q = (A - A’)/2 =    0           0          0

0           0          0

0           0          0

Now, Q’ =    0           0          0     = -Q

0           0           0

0           0           0

Thus, Q = (A - A’)/2 is a skew-symmetric matrix.

Representing A as the sum of P and Q:

P + Q =    6          -2           2            6          -2          2           0          0           0    = A

-2          3          -1    +     -2          3         -1     =    0          0           0

2          -1           3            2          -1          3           0          0           0

(iii) Let A =    3         3         -1

-2        -2          1

-4        -5          2

Then, A’ =    3         -2         -4

3         -2         -5

-1          1          2

Now, A + A’ =   3           3          -1           3          -2         -4           6         1            -5

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

-4         -5          2            -1         1           2           -5        -4            4

Let P = (A + A’)/2 = (1/2)   6          1           -5            3            1/2          -5/2

1          -4          -4     =    1/2         -2              -2

-5         -4           4           -5/2       -2                2

Then, P’ =    3           1/2         -5/2     = P

1/2         -2            -2

-5/2      -2              2

Thus, P = (A + A’)/2 is a symmetric matrix.

Now, A - A’ =    3           3          -1            3          -2         -4           0          5           3

-2         -2          1    -       3           -2        -5    =     -5         0           6

-4         -5          2            -1           1         2           -3         -6           0

Let Q = (A - A’)/2 =    0           5          3            0           5/2      3/2

-5          0          6     =     -5/2       0          3

-3         -6          0            -3/2      -3         0

Now, Q’ =    0            -5/2          -3/2     = -Q

5/2           0               -3

3/2           3                0

Thus, Q = (A - A’)/2 is a skew-symmetric matrix.

Representing A as the sum of P and Q:

P + Q =    3          1/2         -5/2            0          5/2       3/2           3          3         -1    = A

1/2        -2            -2        +   -5/2       0           3        =   -2        -2          1

-5/2      -2              2             -3/2      -3          0               -4        -5         2

(iv) Let A =    1                 5

-1                2

Then A’ =   1                -1

• 2

Now, A + A’ =    1                 5     +    1                 1    =     2                 4

-1                2          5                  2           4                 4

Let P = (A + A’)/2 = (1/2)   2                 4     =    1                2

4                 4           2                2

Now, P’ =   1                 2   = P

2                 2

Thus, P = (A + A’)/2 is a symmetric matrix.

Now, A - A’ =     1                 5     -    1                -1    =      0                6

-1               2           5                 2            -6               0

Let Q = (A - A’)/2 =  (1/2)  0                 6     =    0                 3

-6                0           -3               0

Now, Q’ =  0                 3   = -Q

-3                0

Thus, Q = (A - A’)/2 is a skew-symmetric matrix.

Representing A as the sum of P and Q:

P + Q =   1                2     +    0                 3    =      1                5   = A

2                2           -3                 0           -1               2

Question 11:

If A, B are symmetric matrices of same order, then AB − BA is a

1. Skew symmetric matrix B. Symmetric matrix
2. Zero matrix D. Identity matrix

Answer:

A and B are symmetric matrices, therefore, we have:

A’ = A and B’ = B    ………1

Consider (AB - BA)’ = (AB)’ – (BA)’             [Since (A - B)’ = A’ – B’]

= B’A’ – A’B’                [Since (AB)’ = B’A’]

= BA – AB                    [From equation 1]

= -(AB - BA)

So, (AB - BA)’ = - (AB - BA)’

Thus (AB - BA)’ is a skew-symmetric matrix.

Hence, the correct answer is option A.

Question 12:

If A =    cos α     -sin α

sin α      cos α  , then A + A’ = I, then the value of α is

1. π/6 B. π/3 C. π                                 D. 3π/2

Answer:

Given, A =   cos α     -sin α

sin α      cos α

So, A’ =         cos α     sin α

-sin α      cos α

Now, A + A’ = I

=>    cos α      -sin α   +    cos α      sin α     =       1               0

sin α       cos α          -sin α     cos α              0               1

=>    2cos α    -sin α   =    1              0

0            2cos α          0             0

Comparing the corresponding elements of the two matrices, we have:

2cos α = 1

=> cos α = 1/2

=> cos α = π/3

=> α = π/3

Hence, the correct answer is option B.

Exercise 3.4

Using elementary transformations, find the inverse of each of the matrices, if it exists in Exercises 1 to 17.

Question 1:

• -1

2          3

Answer:

Let A =   1          -1

2           3

We know that A = IA

=>     1          -1   =   1          0   A

2         3          0           1

=>     1          -1   =   1          0   A                       [R2 -> R2 – 2R1]

0         5          -2         1

=>     1          -1   =   1             0   A                     [R2 -> R2/5]

0         1          -2/5    1/5

=>     1          0   =    3/5      1/5    A                   [R1 -> R1 + R2]

0         1          -2/5     1/5

So, A-1 =  3/5         1/5

-2/5         1/5

Question 2:

• -1

2          3

Answer:

Let A =   2           1

1           1

We know that A = IA

=>     2          1    =   1          0   A

1          1          0           1

=>     1          0    =   1          -1   A                       [R1 -> R1 – R2]

1         1          0          1

=>     1          0   =    1              -1    A                   [R2 -> R2 – R1]

0         1          -1            2

So, A-1 =  1                -1

-1                 2

Question 3:

• 3

2            7

Answer:

Let A =   1            3

2           7

We know that A = IA

=>     1           3   =   1          0   A

2         7          0           1

=>     1           3   =   1          0   A                       [R2 -> R2 – 2R1]

0          1          -2         1

=>     1          0   =    7              -3   A                   [R1 -> R1 - 3R2]

0         1          -2             1

So, A-1 =   7                 -3

-2                 1

Question 4:

• 3

5            7

Answer:

Let A =   2            3

5           7

We know that A = IA

=>     2           3   =   1          0   A

5         7          0           1

=>     1       3/2   =   1/2       0   A                       [R1 -> R1/2]

5         7          0          1

=>     1       3/2   =   1/2           0   A                   [R2 -> R2 - 5R1]

0     -1/2        -5/2         1

=>     1          0   =    -7             3   A                   [R1 -> R1 + 3R2]

0     -1/2          5/2        -1

=>     1           0   =   -7             3   A                   [R2 -> -2R1]

0          1         5             -2

So, A-1 =  -7                3

5                -2

Question 5:

2            1

7            4

Answer:

Let A =   2           1

7          4

We know that A = IA

=>     2           1   =   1          0   A

7          4         0           1

=>     1       1/2   =   1/2       0   A                       [R1 -> R1/2]

7          4          0          1

=>     1       1/2   =   1/2           0   A                   [R2 -> R2 - 7R1]

0      1/2        -7/2         1

=>     1          0   =    4             -1   A                   [R1 -> R1 - R2]

0      1/2         -7/2         1

=>     1          0   =    4             -1   A                   [R2 -> 2R2]

0         1         -7             2

So, A-1 =  4                 -1

-7                2

Question 6:

2            5

1            3

Answer:

Let A =   2           5

1          3

We know that A = IA

=>     2           5   =   1          0   A

1          3         0           1

=>     1       5/2   =   1/2       0   A                       [R1 -> R1/2]

1          3          0          1

=>     1       5/2   =   1/2           0   A                   [R2 -> R2 - R1]

0      1/2        -1/2         1

=>     1          0   =    3             -5   A                   [R1 -> R2 - 5R2]

0      1/2         -1/2         1

=>     1          0   =    -3            -5   A                   [R2 -> 2R2]

0         1         -1             2

So, A-1 =  3                -5

-1                2

Question 7:

3            1

5            2

Answer:

Let A =   3           1

5          2

We know that A = IA

=>     3           1   =   1          0   A

5           2         0           1

=>     1           1   =   1           0   A                       [C1 -> C1 – 2C2]

1          2         -2         3

=>     1           0   =   1               0   A                   [C1 -> C2 - C1]

1           1        -2              3

=>     1          0   =    2             -1   A                   [C1 -> C1 – C2]

0         1         -5              3

So, A-1 =  2                 -1

5                  3

Question 8:

4            5

3            4

Answer:

Let A =   4           5

3          4

We know that A = IA

=>     4           5   =   1          0   A

3          4         0           1

=>     1           1   =   1          -1   A                       [R1 -> R1 – R2]

3          4          0          1

=>     1           1   =   1              -1   A                   [R2 -> R2 - 3R1]

0          1        -3              4

=>     1          0   =    4             -5   A                   [R2 -> R1 - R2]

0         1         -3             4

So, A-1 =  4                 -5

-3                4

Question 9:

4            5

3            4

Answer:

Let A =   3          10

2          7

We know that A = IA

=>     3         10   =   1          0   A

2           7         0           1

=>     1           3   =   1          -1   A                       [R1 -> R1 – R2]

2          7          0          1

=>     1           3   =   1              -1   A                   [R2 -> R2 - 2R1]

0          1        -2              3

=>     1          0   =    7            -10   A                   [R2 -> R1 - 3R2]

0         1         -2             3

So, A-1 =  7               -10

-2                3

Question 10:

4            5

3            4

Answer:

Let A =   3          -1

-4         2

We know that A = IA

=>     3         -1   =   1          0    A

-4          2         0           1

=>     1          -1   =   1          0   A                       [C1 -> C1 + 2C2]

0          2         2          1

=>     1           0   =   1               1   A                   [C2 -> C1 + C2]

0          2         2              3

=>     1          0   =    1           1/2   A                   [C2 -> C2/2]

0         1          2           3/2

So, A-1 =  1              1/2

2              3/2

Question 11:

2           -6

1           -2

Answer:

Let A =   2          -6

1         -2

We know that A = IA

=>     2         -6   =   1          0    A

1          -2         0           1

=>     2          0   =    1          3   A                       [C1 -> C2 + 3C1]

1          1         0          1

=>     2           0   =   -2             3   A                   [C1 -> C1 - C2]

0          1         -1             1

=>     1          0   =    -1             3   A                   [C1 -> C1/2]

0         1          -1/2        1

So, A-1 =  -1                3

-1/2             1

Question 12:

6           -3

-2           1

Answer:

Let A =   6          -3

-2         1

We know that A = IA

=>     6         -3   =   1          0    A

-2          1         0           1

=>     1      -1/2   =   1/6       0   A                       [R1 -> R1/6]

-2         1         0            1

=>     1      -1/2   =   1/6           0   A                   [R2 -> R2 + 2R1]

0         0         1/3            1

Now, in the above equation, we can see all the zeros in the second row of the matrix on the

L.H.S. Therefore, A−1 does not exist.

Question 13:

2           -3

-1           2

Answer:

Let A =   2          -3

-1         2

We know that A = IA

=>     2         -3   =    1          0   A

-1          2        0           1

=>     1          -1   =   1           1   A                       [R1 -> R1 + R2]

-1          2          0          1

=>     1          -1   =   1              1   A                   [R2 -> R2 + R1]

0          1         1              2

=>     1          0   =    2               3  A                   [R2 -> R1 + R2]

0         1         1               2

So, A-1 =  2                  3

1                  2

Question 14:

2            1

4            2

Answer:

Let A =   2           1

4          2

We know that A = IA

=>     2           1   =   1          0    A

4           2         0           1

=>     0           0   =   1      -1/2   A                       [R1 -> R1 - R1/2]

4           2         0            1

Now, in the above equation, we can see all the zeros in the first row of the matrix on the

L.H.S. Therefore, A−1 does not exist.

Question 15:

2         -3          3

2           2          3

3          -2          2

Answer:

Let A =   2         -3           3

2          2           3

3         -2           2

We know that A = IA

=>  2          -3           3          1          0           0   A

2          2            3    =    0          1            0

3         -2           2          0         0             1

=>  2          -3           3          1          1            0   A

0          5            0    =   -1         1           0                      [R2 -> R2 – R1]

3         -2           2          0          0            1

=>  2         -3            3          1          0          -1   A

0          5            0    =   -1/5     1/5        0                     [R2 -> R2/5]

3         -2           2          0          0             1

=>  -1         -1           1          1          0          -1   A

0          1            0    =   -1/5     1/5        0                     [R1 -> R1 - R3]

3         -2           2          0          0             1

=>  -1           0         1          4/5       1/5       -1  A

0          1           0    =   -1/5      1/5        0                      [R1 -> R1 + R2 and R3 -> R3 + 2R2]

3          0           2          -2/5      2/5       1

=>  -1           0         1          4/5       1/5       -1  A

0          1           0    =   -1/5      1/5        0                      [R3 -> R3 + 3R1]

0          0           5           2           1         -2

=>  -1           0         1          4/5       1/5       -1  A

0          1           0    =   -1/5      1/5        0                      [R3 -> R3/5]

0          0           1           2/5      1/5   -2/5

=>  -1           0         0          2/5        0       3/5  A

0          1           0    =   -1/5      1/5       0                      [R1 -> R1 – R3]

0          0           1           2/5      1/5   -2/5

=>  1           0           0          -2/5         0             3/5   A

0          1           0    =   -1/5        1/5             0             [R1 -> (-1)R1]

0          0            1          2/5        1/5          -2/5

So, A-1 =  -2/5          0               3/5

-1/5         1/5               0

2/5          1/5            -2/5

Question 16:

1          3          -1

-3         0          -5

2          5           0

Answer:

Let A =   1          3          -1

-3         0          -5

2          5           0

We know that A = IA

=>  1          -3          -2          1          0           0   A

-3         0          -5    =    0          1            0

2          5           0          0         0             1

=>  1           3           0          1           0           0   A

0          9         -11    =   3          1            0                      [R1 -> R1 + 3R3 and R2 -> R2 + 8R3]

0         -1           4          -2         0            1

=>  1           0           0          -5          0           3   A

0          1          21    =   -13       1            8                      [R1 -> R1 + 3R3 and R2 -> R2 + 8R3]

0         -1           4          -2         0             1

=>  1           0          10         -5          0           3  A

0          1          21    =   -13        1           8                      [R3 -> R3 + R2]

0          0          25          -15       1           9

=>  1           0           0          1          -2/5        -3/5      A

0          1           0    =   -2/5       4/25      11/25             [R1 -> R1 - 10R3 and R2 -> R2 - 21R3]

0          0            1          -3/5    1/25        9/25

So, A-1 =   1            -2/5          -3/5

-2/5        4/25       11/25

-3/5       1/25         9/25

Question 17:

2          5          -1

5          1           0

0          1           3

Answer:

Let A =   2          0          -1

5          1           0

0          1           3

We know that A = IA

=>  2          0          -1          1          0           0   A

5          1          0     =    0          1            0

0          1           3          0         0             1

=>  1           0       -1/2         1/2      0           0   A

5          1            0    =   0          1            0                      [R1 -> R1/2]

0          1           3           0         0            1

=>  1           0       -1/2         1/2       0           0   A

0          1        5/2    =   -5/2      1           0                      [R2 -> R2 - 5R!]

0          1           3           0          0            1

=>  1           0      -1/2          1/2       0           0  A

0          1        5/2    =   -5/2      1           0                      [R3 -> R3 - R2]

0          0        1/2          5/2      -1          1

=>  1           0      -1/2          1/2           0              0      A

0          1        5/2    =   -5/2          1              0             [R3 -> 2R3]

0          0            1          5             -2              2

=>  1           0           0           3             -1              1      A

0          1            0    =   -15           6              -5             [R1 -> R1 + R3/2 and R2 -> R2 - 5R3/2]

0          0            1          5             -2              2

So, A-1 =   3              -1                1

-15             6               -5

5              -2                 2

Question 18:

Matrices A and B will be inverse of each other only if

1. AB = BA B. AB = BA = 0 C. AB = 0, BA = I                  D. AB = BA = I

Answer:

We know that if A is a square matrix of order m, and if there exists another square matrix

B of the same order m, such that AB = BA = I, then B is said to be the inverse of A. In

this case, it is clear that A is the inverse of B.

Thus, matrices A and B will be inverses of each other only if AB = BA = I.

Hence, the correct answer is option D.

Miscellaneous Exercise on Chapter 3

Question 1:

Let A =  0          0    , show that (aI + bA)n = anI + nan-1bA, where I is the identity matrix of order 3

0          1

and n є N.

Answer:

Given, A =   0          0

0          1

We shall prove the result by using the principle of mathematical induction.

For n = 1, we have:

P(1): (aI + bA)1 = a1I + 1 * a1-1bA = aI + bA, n є N

Therefore, the result is true for n = 1.

Let the result be true for n = k.

That is,

P(k): (aI + bA)k = akI + kak-1bA

Now, we prove that the result is true for n = k + 1.

Consider (aI + bA)k+1 = (aI + bA)k * (aI + bA)

= (akI + kak-1bA) * (aI + bA)

= ak-1I + kakbAI + akbIA + kak-1b2A2

= ak-1I + (k + 1)akbA + kak-1b2A2   ………….1

Now,

A2 =    0          0     0          1    =   0            0  = O

0          1     0          1          0           0

From Equation 1, we have:

(aI + bA)k+1 = ak-1I + (k + 1)akbA + O

=> (aI + bA)k+1 = ak-1I + (k + 1)akbA

Therefore, the result is true for n = k + 1.

Thus, by the principle of mathematical induction, we have:

(aI + bA)n = anI + nan-1bA, where A =   0            1 , n є N

• 0

Question 2:

If A =   1           1             1                                       An =   3n-1       3n-1      3n-1

1           1             1                                                 3n-1       3n-1      3n-1  , n є N

1           1             1  , then prove that                 3n-1       3n-1      3n-1

Answer:

Given, A =   31-1       31-1      31-1  =    30          30        30    =   1           1             1

31-1       31-1      31-1         30          30        30         1           1             1

31-1       31-1      31-1         30          30        30         1           1             1

We shall prove the result by using the principle of mathematical induction.

For n = 1, we have:

P(1):   31-1       31-1      31-1  =    30          30        30    =   1           1             1

31-1       31-1      31-1         30          30        30         1           1             1  = A

31-1       31-1      31-1         30          30        30         1           1             1

Therefore, the result is true for n = 1.

Let the result be true for n = k.

That is

P(k):   3k-1       3k-1      3k-1

3k-1       3k-1      3k-1

3k-1       3k-1      3k-1

Now, we prove that the result is true for n = k + 1.

Now, Ak+1 = A * Ak

=    1            1            1    3k-1       3k-1      3k-1

1            1            1    3k-1       3k-1      3k-1

1            1            1    3k-1       3k-1      3k-1

=    3.3k-1       3.3k-1      3.3k-1

3.3k-1       3.3k-1      3.3k-1

3.3k-1       3.3k-1      3.3k-1

=    3(k - 1) + 1       3(k - 1) + 1      3(k - 1) + 1

3(k - 1) + 1       3(k - 1) + 1      3(k - 1) + 1

3(k - 1) + 1       3(k - 1) + 1      3(k - 1) + 1

Therefore, the result is true for n = k + 1.

Thus by the principle of mathematical induction, we have:

An =  3n-1       3n-1      3n-1

3n-1       3n-1      3n-1

3n-1       3n-1      3n-1

Question 3:

If A =   3        -4                                   An =  1 + 2n         -4n

1         1   , then prove that           n              1 – 2n     where n is any positive integer.

Answer:

Given, A =   3              -4

1               -1

We shall prove the result by using the principle of mathematical induction.

For n = 1, we have:

P(1):   1 + 2      -4     =   3              -4  = A

1           1 - 2        1             -1

Therefore, the result is true for n = 1.

Let the result be true for n = k.

That is,

P(k):   1 + 2k         -4k

1             1 – 2k

Now, we prove that the result is true for n = k + 1.

Consider, Ak+1 = Ak . A

=  1 + 2k          -4k     3            -4

1              1 – 2k     1            -1

=   3(1 + 2k) – 4k          -4(1 + 2k) + 4k

3k + 1 – 2k               -4k – 1(1 – 2k)

=   3 + 6k – 4k            -4 - 8k + 4k

3k + 1 – 2k            -4k – 1 + 2k

=   3 + 2k           -4 - 4k

1 + k             -1 - 2k

=   1 + 2(k + 1)          4(k + 1)

1 + k                1 – 2(k + 1)

Therefore, the result is true for n = k + 1.

Thus, by the principle of mathematical induction, we have:

An =  1 + 2n            -4n

1                  1 – 2n  , n є N

Question 4:

If A and B are symmetric matrices, prove that AB − BA is a skew symmetric matrix.

Answer:

It is given that A and B are symmetric matrices. Therefore, we have:

A’ = A and B’ = B   ……..1

Now, (AB - BA)’ = (AB)’ – (BA)’                [Since (A - B)’ = A’ – B’]

= B’A’ – A’B’                  [Since (AB)’ = B’A’]

= BA – AB                      [From equation 1]

= -(AB − BA)

So, (AB − BA)’ = -(AB − BA)

Thus, (AB − BA) is a skew-symmetric matrix.

Question 5:

Show that the matrix B’AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.

Answer:

We suppose that A is a symmetric matrix,

then   A’ = A    .............1

Consider (B’AB)’ = {B’(AB)}’

= (AB)’(B’)’               [Since (AB)’ = B’A’]

= B’A’(B)                   [Since (B’)’ = B]

= B’(A’B)

= B’AB                     [From equation 1]

So, (B’AB)’ = B’AB

Thus, if A is a symmetric matrix, then B’AB is a symmetric matrix.

Now, we suppose that A is a skew-symmetric matrix.

Consider (B’AB)’ = {B’(AB)}’

= (AB)’(B’)’               [Since (AB)’ = B’A’]

= B’A’(B)                   [Since (B’)’ = B]

= B’(A’B)

= B’(-A)B                     [From equation 1]

= - B’AB

So, (B’AB)’ = -B’AB

Thus, if A is a skew-symmetric matrix, then B’AB is a skew-symmetric matrix.

Hence, if A is a symmetric or skew-symmetric matrix, then B’AB is a symmetric or skew

symmetric matrix accordingly.

Question 6:

Solve system of linear equations, using matrix method.

2x – y = -2

3x + 4y = 3

Answer:

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

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

3           4  ,        y     and            3

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

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

Now, A-1 = adj A/|A|= (1/11)   4             1

-3          2

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

-3          2      3

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

y                                  -3          2      3

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

y                        6 + 6                      11         12/11

Hence, x = -5/11, y = 12/11

Question 7:

For what value of x,

1          2         0       0

[1        2          1]  2          0          1       2     = O?

1          0         2        x

Answer:

We have,

1          2         0       0

[1        2          1]  2          0          1       2     = O

1          0         2        x

0

=> [1 + 4 + 1        2 + 0         0 + 2 + 2]     2     = O

x

0

=> [6          2          4]    2     = O

x

=> [6 * 0 + 2 * 2 + 4 * x] = O

=> [4 + 4x] = 

=> 4 + 4x = 0

=> 4x = -4

=> x = -1

Thus, the required value of x is -1.

Question 8:

If A =   3          1

-1         2  , show that A2 - 5A + 7I = O

Answer:

Given,  A =   3          1

-1         2

Now,

A2 = A * A =   3          1    3            1

-1         2    -1          2

=  3 * 3 + 1 * (-1)          3 * 1 + 1 * 2

(-1) * 3 + 2 * (-1)     -1 * 1 + 2 * 2

=   9 – 1       3 + 2

-3 – 2    -1 + 4

=    8           5

-5          3

LHS: 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

= O

= RHS

Hence, A2 - 5A + 7I = O

Question 9:

Find x, if

1          0          2       x

[x        -5        -1]  0          2           1      4     = O?

2          0          3       1

Answer:

We have,

1          0          2       x

[x        -5        -1]  0          2           1      4     = O?

2          0          3       1

x

=> [x + 0 - 2      0 – 10 + 0     2x - 5 - 3]   4     = O

1

x

=> [x - 2    -10      2x - 8]    4     = O

1

=> [x(x – 2) - 40 + 2x - 8] = O

=> [x2 – 2x – 40 + 2x - 8] = 

=> [x2 – 48] = 

=> x2 – 48 = 0

=> x2 = 48

=> x = ±4√3

Thus, the required value of x is ±4√3.

Question 10:

A manufacturer produces three products x, y, z which he sells in two markets. Annual sales are indicated below:

 Market Product
 I 10000 2000 18000 II 6000 20000 8000

(a) If unit sale prices of x, y and z are Rs 2.50, Rs 1.50 and Rs 1.00, respectively, find the total revenue in each market with the help of matrix algebra.

(b) If the unit costs of the above three commodities are Rs 2.00, Rs 1.00 and 50 paise respectively. Find the gross profit.

Answer:

(a) The unit sale prices of x, y, and z are respectively given as Rs 2.50, Rs 1.50, and Rs 1.00.

Consequently, the total revenue in market I can be represented in the form of a matrix as:

2.50

[10000       2000         18000]   1.50

1.00

= 10000 * 2.50 + 2000 * 1.50 + 18000 * 1.0

= 25000 + 3000 + 18000

= 46000

The total revenue in market II can be represented in the form of a matrix as:

2.50

[6000       20000         8000]  1.50

1.00

= 6000 * 2.50 + 20000 * 1.50 + 8000 * 1.00

= 15000 + 30000 + 8000

= 53000

Therefore, the total revenue in market I is Rs 46000 and the same in market II is Rs 53000.

(b) The unit cost prices of x, y, and z are respectively given as Rs 2.00, Rs 1.00, and 50 paise.

Consequently, the total cost prices of all the products in market I can be represented in the

form of a matrix as:

2.00

[10000       2000         18000]   1.00

0.50

= 10000 * 2.00 + 2000 * 1.00 + 18000 * 0.50

= 20000 + 2000 + 9000

= 31000

Since the total revenue in market I isRs 46000, the gross profit in this market is (Rs 46000 − Rs

31000) Rs 15000.

The total cost prices of all the products in market II can be represented in the form of a matrix

as:

2.00

[6000       20000         8000]  1.00

0.50

= 6000 * 2.00 + 20000 * 1.00 + 8000 * 0.50

= 12000 + 20000 + 4000

= 36000

Since the total revenue in market II is Rs 53000, the gross profit in this market is (Rs 53000 − Rs

36000) Rs 17000.

Question 11:

Find the matrix X so that X   1         2         3     =    -7       -8         -9

4         5          6           2         4          6

Answer:

Given,

X   1         2         3     =    -7       -8         -9

4         5          6           2         4          6

The matrix given on the R.H.S. of the equation is a 2 * 3 matrix and the one given on the L.H.S.

of the equation is a 2 * 3 matrix.

Therefore, X has to be a 2 * 2 matrix.

Now, let X =  a             c

b            d

Therefore, we have:

a              c    1         2         3     =    -7       -8         -9

b             d    4         5          6           2         4          6

=>    a + 4c       2a + 5c         3a + 6c    =    -7       -8         -9

b + 4d       2a + 5d        3b + 6d    =    2         4          6

Equating the corresponding elements of the two matrices, we have:

a + 4c = -7, 2a + 5c = -8, 3a + 6c = -9

b + 4d = 2, 2a + 5d = 4, 3b + 6d = 6

Now, a + 4c = -7

=> a = -7 – 4c

So, 2a + 5c = -8

=> -14 – 8c + 5c = -8

=> -3c = 6

=> c = -2

So, a = -7 – 4 * (-2) = -7 + 7 = 1

Now, b + 4d = 2

=> b = 2 – 4d

So, 2b + 5d = 4

=> 4 – 8d + 5d = 5

=> 4 – 3d = 4

=> 3d = 0

=> d = 0

So, b = 2 – 4 * 0 = 2

Thus, a = 1, b = 2, c = −2, d = 0

Hence, the required matrix X is   1            -2

2             0

Question 12:

If A and B are square matrices of the same order such that AB = BA, then prove by induction that ABn = BnA. Further, prove that (AB)n = AnBn for all n є N.

Answer:

A and B are square matrices of the same order such that AB = BA.

For n = 1, we have:

P(1): AB = BA          [Given]

=> AB1 = B1A

Therefore, the result is true for n = 1.

Let the result be true for n = k.

P(k): ABk = BkA    ………….1

Now, we prove that the result is true for n = k + 1.

ABk+1 = ABk.B

= (BkA)B           [From equation 1]

= Bk.(AB)          [Associative rule]

= Bk.(BA)          [Since AB = BA]

= (BkB)A

= Bk+1A

Therefore, the result is true for n = k + 1.

Thus, by the principle of mathematical induction, we have ABn = BnA, n є N.

Now, we prove that (AB)n = AnBn for all n є N

For n = 1, we have:

(AB)1 = A1B1 = AB

Therefore, the result is true for n = 1.

Let the result be true for n = k.

(AB)k = AkBk    …………..2

Now, we prove that the result is true for n = k + 1.

(AB)k+1 = (AB)k .(AB)

= (AkBk).(AB)           [From equation 2]

= Ak(Bk A)B              [Associative rule]

= Ak(ABk)B               [(AB)n = AnBn for all n є N]

= (Ak A)( Bk B)         [Associative rule]

= Ak+1 Bk+1

Therefore, the result is true for n = k + 1.

Thus, by the principle of mathematical induction, we get (AB)n = An Bn, for all natural numbers.

Choose the correct answer in the following questions

Question 13:

If A =    α          β

γ         - α   is such that A2 = I, then

1. 1 + α2 + βγ = 0 B. 1 - α2 + βγ = 0 C. 1 - α2 - βγ = 0             D. 1 + α2 - βγ = 0

Answer:

Given,

A =    α          β

γ         - α

Now, A2 = A . A

=    α           β     α            β

γ         - α      γ         - α

=    α2 + βγ             αβ – αβ

αγ – αγ             βγ - α2

=    α2 + βγ                    0

0                      βγ - α2

Now, A2 = I

=>   α2 + βγ                    0    =  1              0

0                      βγ - α2        0             1

On comparing the corresponding elements, we have:

α2 + βγ = 1

=> α2 + βγ – 1 = 0

=> 1 - α2 – βγ = 0

Hence, the correct answer is option C.

Question 14:

If the matrix A is both symmetric and skew symmetric, then

1. A is a diagonal matrix B. A is a zero matrix
2. A is a square matrix D. None of these Answer

Answer:

If A is both symmetric and skew-symmetric matrix, then we should have

A’ = A and A’ = -A

=> A = -A

=> A + A = 0

=> 2A = 0

=> A = 0

Therefore, A is a zero matrix.

Hence, the correct answer is option B.

Question 15:

If A is square matrix such that A2 = A then (I + A)3 – 7A is equal to

1. A B. I − A                          C. I                                   D. 3A

Answer: C

(I + A)3 – 7A = I3 + A3 + 3I2A + 3IA2 – 7A

= I3 + A.A2 + 3I2A + 3IA2 – 7A

= I + A.A + 3A + 3A – 7A              [Since A2 = A and I3 = I]

= I + A2 + 6A – 7A

= I + A – A                                      [Since A2 = A]

= I

So, (I + A)3 – 7A = I

.