Matrices — Exercise 9(A)

Write the order of each of the following matrices :

(i) $\begin{bmatrix} 3 & -5 & 7 \\ -6 & 9 & 4 \end{bmatrix}$

(ii) $\begin{bmatrix} 2 & 5 \\ -3 & 6 \\ 1 & -8 \end{bmatrix}$

(iii) $\begin{bmatrix} 6 \\ 4 \\ -2 \end{bmatrix}$

(iv) $\begin{bmatrix} 8 & -3 \end{bmatrix}$

(v) $\begin{bmatrix} 11 & -8 \\ 3 & 6 \end{bmatrix}$

(vi) $\begin{bmatrix} 10 \end{bmatrix}$

Answer

(i) The matrix has 2 rows and 3 columns.

Hence, the order of matrix is (2 × 3).

(ii) The matrix has 3 rows and 2 columns.

Hence, the order of matrix is (3 × 2).

(iii) The matrix has 3 rows and 1 column.

Hence, the order of matrix is (3 × 1).

(iv) The matrix has 1 row and 2 columns.

Hence, the order of matrix is (1 × 2).

(v) The matrix has 2 rows and 2 columns.

Hence, the order of matrix is (2 × 2).

(vi) The matrix has 1 row and 1 column.

Hence, the order of matrix is (1 × 1).

Classify the following matrices :

(i) $\begin{bmatrix} 9 \\ 3 \\ -4 \end{bmatrix}$

(ii) $\begin{bmatrix} 7 & -8 \\ 9 & 4 \end{bmatrix}$

(iii) $\begin{bmatrix} 8 & -7 & 2 \end{bmatrix}$

(iv) $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

(v) $\begin{bmatrix} 6 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -2 \end{bmatrix}$

(vi) $\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

Answer

(i) Since, there is only once column and three rows in the matrix.

Hence, it is column matrix of order 3 x 1.

(ii) Since, no. of rows = no. of columns in the matrix.

Hence, it is a square matrix of order 2.

(iii) Since, there is only one row and 3 columns in the matrix.

Hence, it is a row matrix of order 1 x 3.

(iv) Since, there are 2 rows and 2 columns in the matrix.

Also, only the main diagonal element is 1 and the rest elements are zero.

Hence, it is a unit matrix of order 2.

(v) Since, there are 3 rows and 3 columns in the matrix.

Also, all the elements apart from main diagonal is zero.

Hence , it is a diagonal matrix of order 3.

(vi) Since, there are 2 rows and 3 columns in the matrix and all elements are zero.

Hence , it is a zero matrix of order 2 x 3.

Construct a (2 × 3) matrix, whose elements a_ij are given by a_ij = (3i − j).

Answer

Given,

a_ij = (3i − j).

∴ a₁₁ = [3(1) - 1] = 2, a₁₂ = [3(1) - 2] = 1, a₁₃ = [3(1) - 3] = 0.

a₂₁ = [3(2) - 1] = 5, a₂₂ = [3(2) - 2]= 4, a₂₃ = [3(2) - 3] = 3.

Hence, required matrix = $\begin{bmatrix} 2 & 1 & 0 \\ 5 & 4 & 3 \end{bmatrix}$.

Construct a (3 × 2) matrix [a_ij]_3×2 for which a_ij = (i × j).

Answer

Given,

a_ij = (i × j)

⇒ a₁₁ = 1 × 1 = 1, a₁₂ = 1 × 2 = 2.

⇒ a₂₁ = 2 × 1 = 2, a₂₂ = 2 × 2 = 4.

⇒ a₃₁ = 3 × 1 = 3, a₃₂ = 3 × 2 = 6.

Hence, required matrix = $\begin{bmatrix} 1 & 2 \\ 2 & 4 \\ 3 & 6 \end{bmatrix}$.

If a matrix has 4 elements, what are possible orders it can have?

Answer

If a matrix has 4 elements the possible orders are, 1 x 4, 2 x 2, 4 x 1.

Hence, possible orders are 1 x 4, 2 x 2, 4 x 1.

If a matrix has 6 elements, what are the possible orders it can have ?

Answer

If a matrix has 6 elements the possible orders are, 1 × 6, 6 × 1, 2 × 3, 3 × 2.

Hence, possible orders are 1 × 6, 6 × 1, 2 × 3, 3 × 2.

Find the values of a, b, c and d when $\begin{bmatrix} a + 3 & 4 \\ -2 & b - 6 \end{bmatrix}= \begin{bmatrix} -1 & c - 3 \\ d + 1 & 2 \end{bmatrix}$

Answer

Given,

$\begin{bmatrix} a + 3 & 4 \\ -2 & b - 6 \end{bmatrix} =\begin{bmatrix} -1 & c - 3 \\ d + 1 & 2 \end{bmatrix}$

∴ a + 3 = -1

⇒ a = -1 - 3 = -4.

∴ b - 6 = 2

⇒ b = 6 + 2 = 8

∴ c - 3 = 4

⇒ c = 4 + 3 = 7

∴ d + 1 = -2

⇒ d = -2 - 1 = -3

Hence , a = -4, b = 8, c = 7, d = -3.

Find the values of x and y when $\begin{bmatrix} 5x + 3y \\ 2x - y \end{bmatrix} =\begin{bmatrix} 12 \\ 7 \end{bmatrix}$

Answer

Given,

$\begin{bmatrix} 5x + 3y \\ 2x - y \end{bmatrix} =\begin{bmatrix} 12 \\ 7 \end{bmatrix}$

∴ 5x + 3y = 12....(1)

∴ 2x - y = 7

y = 2x - 7....(2)

Substituting value of y from equation (2) in equation (1), we get :

⇒ 5x + 3(2x - 7) = 12

⇒ 5x + 6x - 21 = 12

⇒ 11x = 12 + 21

⇒ 11x = 33

⇒ x = $\dfrac{33}{11}$ = 3.

Substitute value of x in equation (2):

⇒ y = 2x - 7

⇒ y = 2(3) - 7

⇒ y = 6 - 7

⇒ y = -1.

Hence , x = 3 and y = -1.

Find the values of x, y, a and b when $\begin{bmatrix} x + y & a - b \\ a + b & 2x - 3y \end{bmatrix} =\begin{bmatrix} 5 & 3 \\ -1 & -5 \end{bmatrix}$

Answer

Given,

$\begin{bmatrix} x + y & a - b \\ a + b & 2x - 3y \end{bmatrix} =\begin{bmatrix} 5 & 3 \\ -1 & -5 \end{bmatrix}$

∴ x + y = 5

⇒ y = 5 - x ........(1)

∴ 2x - 3y = -5 ........(2)

Substituting value of y from equation(1) in (2), we get :

⇒ 2x - 3(5 - x) = -5

⇒ 2x - 15 + 3x = -5

⇒ 5x - 15 = -5

⇒ 5x = -5 + 15

⇒ 5x = 10

⇒ x = $\dfrac{10}{5}$ = 2.

Substitute value of x in equation 1 :

⇒ y = 5 - x

⇒ y = 5 - 2

⇒ y = 3.

∴ a - b = 3

⇒ a = 3 + b .........(3)

∴ a + b = -1 .........(4)

Substituting value of a from equation (3) in (4), we get :

⇒ 3 + b + b = -1

⇒ 2b = -1 - 3

⇒ 2b = -4

⇒ b = $\dfrac{-4}{2}$ = -2.

Substitute value of b in equation (3), we get :

⇒ a = -2 + 3

⇒ a = 1.

Hence, x = 2, y = 3, a = 1, b = -2.

Find the transpose of each of the matrices given below:

(i) $A = \begin{bmatrix} 2 & 3 \\ 5 & -4 \end{bmatrix}$

(ii) $B = \begin{bmatrix} 5 & 7 & -3 \end{bmatrix}$

(iii) $C = \begin{bmatrix} -2 \\ 6 \end{bmatrix}$

Answer

The matrix otained by interchanging the rows and column of a matrix is called the transpose of the matrix.

(i) Given,

$\Rightarrow A = \begin{bmatrix} 2 & 3 \\ 5 & -4 \end{bmatrix} \\[1em] \Rightarrow A^T = \begin{bmatrix} 2 & 5 \\ 3 & -4 \end{bmatrix}.$

Hence, the transpose of matrix A $= \begin{bmatrix} 2 & 5 \\ 3 & -4 \end{bmatrix}$.

(ii) Given,

$\Rightarrow B = \begin{bmatrix} 5 & 7 & -3 \end{bmatrix} \\[1em] \Rightarrow B^T = \begin{bmatrix} 5 \\ 7 \\ -3 \end{bmatrix}.$

Hence, the transpose of matrix B = $\begin{bmatrix} 5 \\ 7 \\ -3 \end{bmatrix}$

(iii) Given,

$\Rightarrow C = \begin{bmatrix} -2 \\ 6 \end{bmatrix} \\[1em] \Rightarrow C^T = \begin{bmatrix} -2 & 6 \\ \end{bmatrix}$

Hence, the transpose of matrix C $= \begin{bmatrix} -2 & 6 \\ \end{bmatrix}$.