Matrices — Test Yourself

If a matrix A = $\begin{bmatrix*}[r] 0 & 1 \\ 2 & -1 \end{bmatrix*}$ and matrix B = $\begin{bmatrix*}[r] 3 \\ 1 \end{bmatrix*}$, then which of the following is possible :

1. A + B

2. A - B

3. AB

4. BA

Answer

Since, addition and subtraction of matrices requires each matrix to be of same order.

∴ A + B and A - B are not possible.

AB is possible because no. of columns in A (2) is equal to no. of rows in B (2).

BA is not possible because no. of columns in B (1) is not equal to no. of rows in A (2).

Hence, Option 3 is the correct option.

If M × $\begin{bmatrix*}[r] 3 & 2 \\ -1 & 0 \end{bmatrix*} = \begin{bmatrix*}[r] 3 & -1 \end{bmatrix*}$, the order of matrix M is :

1. 2 × 2

2. 2 × 1

3. 1 × 2

4. 1 × 3

Answer

We know that,

For matrix multiplication :

No. of columns in 1st matrix must be equal to the number of rows in 2nd matrix.

Resultant matrix order = No. of rows in 1st matrix × No. of columns in 2nd matrix.

Let order of matrix M be a × b.

⇒ M_{a × b} × $\begin{bmatrix*}[r] 3 & 2 \\ -1 & 0 \end{bmatrix*}_{2 \times 2} = \begin{bmatrix*}[r] 3 & -1 \end{bmatrix*}_{1 \times 2}$

b = 2 and a = 1.

Order of matrix M = 1 × 2.

Hence, Option 3 is the correct option.

If $\begin{bmatrix*}[r] 2x - y \\ x + y \end{bmatrix*} = \begin{bmatrix*}[r] 9 \\ 9 \end{bmatrix*}$, the value of x and y are :

1. x = 3 and y = 3

2. x = 3 and y = 9

3. x = 3 and y = 6

4. x = 6 and y = 3

Answer

Given,

$\begin{bmatrix*}[r] 2x - y \\ x + y \end{bmatrix*} = \begin{bmatrix*}[r] 9 \\ 9 \end{bmatrix*}$

⇒ 2x - y = 9 .......(1)

⇒ x + y = 9 ........(2)

Adding equation (1) and (2), we get :

⇒ 2x - y + x + y = 9 + 9

⇒ 3x = 18

⇒ x = $\dfrac{18}{3}$ = 6.

Substituting value of x in equation (2), we get :

⇒ 6 + y = 9

⇒ y = 9 - 6 = 3.

Hence, Option 4 is the correct option.

If matrix A = $\begin{bmatrix*}[r] x - y & x + y \\ y - x & y + x \end{bmatrix*}$ and matrix B = $\begin{bmatrix*}[r] x + y & y - x \\ x - y & y + x \end{bmatrix*}$, then A + B is :

1. $\begin{bmatrix*}[r] 2y & 2x \\ 0 & 2(x + y) \end{bmatrix*}$

2. $\begin{bmatrix*}[r] 2x & 2(x + y) \\ 0 & 0 \end{bmatrix*}$

3. $\begin{bmatrix*}[r] 2x & 2y \\ 0 & 2(x + y) \end{bmatrix*}$

4. $\begin{bmatrix*}[r] 2x - 2y & 2y \\ 0 & 0 \end{bmatrix*}$

Answer

Substituting values of A and B in A + B, we get :

$\Rightarrow A + B = \begin{bmatrix*}[r] x - y & x + y \\ y - x & y + x \end{bmatrix*} + \begin{bmatrix*}[r] x + y & y - x \\ x - y & y + x \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] x - y + x + y & x + y + y - x \\ y - x + x - y & y + x + y + x \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2x & 2y \\ 0 & 2(x + y) \end{bmatrix*}.$

Hence, Option 3 is the correct option.

Event A : Order of matrix A is 3 × 5.

Event B : Order of matrix B is 5 × 3.

Event C : Order of matrix C is 3 × 3.

Product of which two matrices gives a square matrix.

1. AB and AC

2. AB and BC

3. BA and BC

4. AB and BA

Answer

We know that,

Resultant matrix order = No. of rows in 1st matrix × No. of columns in 2nd matrix.

Resultant matrix (P) on multiplication of AB has order = 3 × 3.

Resultant matrix (Q) on multiplication of BA has order = 5 × 5.

Both P and Q are square matrix of order 3 and 5 respectively.

Hence, Option 4 is the correct option.

Two matrices A and B each of order 2 x 2.

Assertion (A) : A X B = 0 $\nRightarrow$ A = 0 or B = 0.

Reason (R) : Let $A = \begin{bmatrix*}[r] 2 & 2 \\ 5 & 5 \end{bmatrix*}$ ≠ 0 and $B = \begin{bmatrix*}[r] -4 & 3 \\ 4 & -3 \end{bmatrix*}$ ≠ 0 but A x B = $\begin{bmatrix*}[r] 2 & 2 \\ 5 & 5 \end{bmatrix*}\begin{bmatrix*}[r] -4 & 3 \\ 4 & -3 \end{bmatrix*}$ = 0.

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true and R is correct reason for A.

4. Both A and R are true and R is incorrect reason for A.

Answer

It is not necessarily true in case of matrices that if A X B = 0

Then, either A = 0 or B = 0.

So, assertion (A) is true.

According to reason, $A = \begin{bmatrix*}[r] 2 & 2 \\ 5 & 5 \end{bmatrix*}$ and $B = \begin{bmatrix*}[r] -4 & 3 \\ 4 & -3 \end{bmatrix*}$

$\Rightarrow AB = \begin{bmatrix*}[r] 2 & 2 \\ 5 & 5 \end{bmatrix*}.\begin{bmatrix*}[r] -4 & 3 \\ 4 & -3 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 2 \times (-4) + 2 \times 4 & 2 \times 3 + 2 \times (-3)\ 5 \times (-4) + 5 \times 4 & 5 \times 3 + 5 \times (-3) \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] -8 + 8 & 6 - 6\ -20 + 20 & 15 - 15 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 0 & 0\ 0 & 0 \end{bmatrix*}\\[1em]$

So, reason (R) is true.

Thus, Both A and R are true and R is correct reason for A.

Hence, option 3 is the correct option.

Matrix A = $\begin{bmatrix*}[r] x & y \end{bmatrix*}$ and Matrix B = $\begin{bmatrix*}[r] a \\ b \end{bmatrix*}$.

Assertion (A) : Product BA is possible and order of resulting matrix is 2 x 2.

Reason (R) : The product BA of two matrices A and B is possible only if number of rows in matrix B Is same as number of columns in matrix A.

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true and R is correct reason for A.

4. Both A and R are true and R is incorrect reason for A.

Answer

∴ Order of matrix A = 1 x 2

∴ Order of matrix B = 2 x 1

Since product of matrix is possible, only when the number of columns in the first matrix is equal to number of rows in second.

Since,

Number of columns in B = 1

Number of rows in A = 1

∴ Product BA is possible.

We know that,

The no. of rows in the resulting matrix is equal to the no. of rows in first matrix and no. of columns equals to the no. of columns in second matrix.

∴ Order of matrix BA = 2 x 2

So, assertion is true. But reason says the product BA is possible only if number of rows in matrix B is same as number of columns in matrix A, which is false.

Hence, option 1 is the correct option.

A, B and C are three matrices each of order 2 x 2.

Statement 1 : If A x B = A x C ⇒ B = C

Statement 2 : Cancellation law is applicable in matrix multiplication.

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

We know that,

Matrix multiplication does not satisfy the cancellation law.

∴ Statement 2 is false.

Thus, if A x B = A x C, we cannot conclude that B = C.

∴ Statement 1 is false.

Hence, option 2 is the correct option.

Matrix A = $\begin{bmatrix*}[r] 2 & -2 \\ -2 & 2 \end{bmatrix*}$ and matrix B = $\begin{bmatrix*}[r] 5 & 5 \\ 5 & 5 \end{bmatrix*}$

Statement 1 : AB = 0

Statement 2 : AB = 0, even if A ≠ 0 and B ≠ 0.

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Given,

Matrix A = $\begin{bmatrix*}[r] 2 & -2 \\ -2 & 2 \end{bmatrix*}$ and matrix B = $\begin{bmatrix*}[r] 5 & 5 \\ 5 & 5 \end{bmatrix*}$

Solving,

$\Rightarrow AB = \begin{bmatrix*}[r] 2 & -2 \\ -2 & 2 \end{bmatrix*}. \begin{bmatrix*}[r] 5 & 5 \\ 5 & 5 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 2 \times 5 + (-2) \times 5 & 2 \times 5 + (-2) \times 5 \\ (-2) \times 5 + 2 \times 5 & (-2) \times 5 + 2 \times 5 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 10 - 10 & 10 - 10 \\ -10 + 10 & -10 + 10 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*}$

So, AB = 0 and also from above example it is proved that AB can be equal to zero even if A ≠ 0 and B ≠ 0.

∴ Both the statements are true.

Hence, option 1 is the correct option.

Find x and y, if :

$\begin{bmatrix*}[r] 3 & -2 \\ -1 & 4 \end{bmatrix*}\begin{bmatrix*}[r] 2x \\ 1 \end{bmatrix*} + 2\begin{bmatrix*}[r] -4 \\ 5 \end{bmatrix*} = 4\begin{bmatrix*}[r] 2 \\ y \end{bmatrix*}$

Answer

Given,

$\Rightarrow \begin{bmatrix*}[r] 3 & -2 \\ -1 & 4 \end{bmatrix*}\begin{bmatrix*}[r] 2x \\ 1 \end{bmatrix*} + 2\begin{bmatrix*}[r] -4 \\ 5 \end{bmatrix*} = 4\begin{bmatrix*}[r] 2 \\ y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3 \times 2x + (-2) \times 1 \\ -1 \times 2x + 4 \times 1 \end{bmatrix*} + \begin{bmatrix*}[r] -8 \\ 10 \end{bmatrix*} = \begin{bmatrix*}[r] 8 \\ 4y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6x - 2 \\ -2x + 4 \end{bmatrix*} + \begin{bmatrix*}[r] -8 \\ 10 \end{bmatrix*} = \begin{bmatrix*}[r] 8 \\ 4y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6x - 2 + (-8) \\ -2x + 4 + 10 \end{bmatrix*} = \begin{bmatrix*}[r] 8 \\ 4y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6x - 10 \\ -2x + 14 \end{bmatrix*} = \begin{bmatrix*}[r] 8 \\ 4y \end{bmatrix*}$

By definition of equality of matrices we get,

6x - 10 = 8
⇒ 6x = 18
⇒ x = 3

-2x + 14 = 4y
⇒ -2(3) + 14 = 4y
⇒ -6 + 14 = 4y
⇒ 8 = 4y
⇒ y = 2.

Hence, x = 3 and y = 2.

Find x and y, if :

$\begin{bmatrix*}[r] 3x & 8 \end{bmatrix*}\begin{bmatrix*}[r] 1 & 4 \\ 3 & 7 \end{bmatrix*} - 3\begin{bmatrix*}[r] 2 & -7 \end{bmatrix*} = 5\begin{bmatrix*}[r] 3 & 2y \end{bmatrix*}$

Answer

Given,

$\Rightarrow \begin{bmatrix*}[r] 3x \times 1 + 8 \times 3 & 3x \times 4 + 8 \times 7 \end{bmatrix*} - \begin{bmatrix*}[r] 6 & -21 \end{bmatrix*} = \begin{bmatrix*}[r] 15 & 10y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3x + 24 & 12x + 56 \end{bmatrix*} - \begin{bmatrix*}[r] 6 & -21 \end{bmatrix*} = \begin{bmatrix*}[r] 15 & 10y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3x + 24 - 6 & 12x + 56 - (-21) \end{bmatrix*} = \begin{bmatrix*}[r] 15 & 10y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3x + 18 & 12x + 77 \end{bmatrix*} = \begin{bmatrix*}[r] 15 & 10y \end{bmatrix*}$

By definition of equality of matrices we get,

3x + 18 = 15
⇒ 3x = -3
⇒ x = -1.

12x + 77 = 10y
⇒ 12(-1) + 77 = 10y
⇒ 65 = 10y
⇒ y = 6.5

Hence, x = -1 and y = 6.5

If $\begin{bmatrix*}[r] x & y \end{bmatrix*}\begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 25 \end{bmatrix*} \text{ and } \begin{bmatrix*}[r] -x & y \end{bmatrix*}\begin{bmatrix*}[r] 2x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] -2 \end{bmatrix*}$;

find x and y, if :

(i) x, y ∈ W (whole numbers)

(ii) x, y ∈ Z (integers)

Answer

Given,

$\Rightarrow \begin{bmatrix*}[r] x & y \end{bmatrix*}\begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 25 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x \times x + y \times y \end{bmatrix*} = \begin{bmatrix*}[r] 25 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x^2 + y^2 \end{bmatrix*} = \begin{bmatrix*}[r] 25 \end{bmatrix*}$

By definition of equality of matrices we get,

x² + y² = 25
⇒ x² = 25 - y² ......(i)

Given,

$\Rightarrow \begin{bmatrix*}[r] -x & y \end{bmatrix*}\begin{bmatrix*}[r] 2x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] -2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -x \times 2x + y \times y \end{bmatrix*} = \begin{bmatrix*}[r] -2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -2x^2 + y^2 \end{bmatrix*} = \begin{bmatrix*}[r] -2 \end{bmatrix*}$

By definition of equality of matrices we get,

-2x² + y² = -2 ......(ii)

Substituting value of x² from (i) in (ii) we get,

⇒ -2(25 - y²) + y² = -2
⇒ -50 + 2y² + y² = -2
⇒ 3y² = -2 + 50
⇒ 3y² = 48
⇒ y² = 16
⇒ y = ± 4.

⇒ x² = 25 - y²
⇒ x² = 25 - 16
⇒ x² = 9
⇒ x = ± 3.

(i) Since, x, y ∈ W

∴ x = 3, y = 4.

Hence, x = 3 and y = 4.

(ii) Since, x, y ∈ Z

∴ x = ±3, y = ±4.

Hence, x = ±3 and y = ±4.

Evaluate :

$\begin{bmatrix*}[r] cos 45° & sin 30° \\ \sqrt{2}cos 0° & sin 0° \end{bmatrix*}\begin{bmatrix*}[r] sin 45° & cos 90° \\ sin 90° & cot 45° \end{bmatrix*}$

Answer

Given,

$\Rightarrow \begin{bmatrix*}[r] cos 45° & sin 30° \\ \sqrt{2}cos 0° & sin 0° \end{bmatrix*}\begin{bmatrix*}[r] sin 45° & cos 90° \\ sin 90° & cot 45° \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] \dfrac{1}{\sqrt{2}} & \dfrac{1}{2} \\ \sqrt{2}(1) & 0 \end{bmatrix*}\begin{bmatrix*}[r] \dfrac{1}{\sqrt{2}} & 0 \\ 1 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] \dfrac{1}{\sqrt{2}} & \dfrac{1}{2} \\ \sqrt{2} & 0 \end{bmatrix*}\begin{bmatrix*}[r] \dfrac{1}{\sqrt{2}} & 0 \\ 1 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] \dfrac{1}{\sqrt{2}} \times \dfrac{1}{\sqrt{2}} + \dfrac{1}{2} \times 1 & \dfrac{1}{\sqrt{2}} \times 0 + \dfrac{1}{2} \times 1 \\ \sqrt{2} \times \dfrac{1}{\sqrt{2}} + 0 \times 1 & \sqrt{2} \times 0 + 0 \times 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] \dfrac{1}{2} + \dfrac{1}{2} & 0 + \dfrac{1}{2} \\ 1 + 0 & 0 + 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 1 & \dfrac{1}{2} \\ 1 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 1 & 0.5 \\ 1 & 0 \end{bmatrix*}$

Hence, $\begin{bmatrix*}[r] cos 45° & sin 30° \\ \sqrt{2}cos 0° & sin 0° \end{bmatrix*}\begin{bmatrix*}[r] sin 45° & cos 90° \\ sin 90° & cot 45° \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0.5 \\ 1 & 0 \end{bmatrix*}.$

If A = $\begin{bmatrix*}[r] 0 & -1 \\ 4 & -3 \end{bmatrix*}, B = \begin{bmatrix*}[r] -5 \\ 6 \end{bmatrix*}$ and 3A × M = 2B; find matrix M.

Answer

Let order of matrix M be a × b.

Given,

$\Rightarrow 3A \times M = 2B .......(i) \\[1em] \Rightarrow 3\begin{bmatrix*}[r] 0 & -1 \\ 4 & -3 \end{bmatrix*}_{2 \times 2} \times M_{a \times b} = 2\begin{bmatrix*}[r] -5 \\ 6 \end{bmatrix*}_{2 \times 1}$

Since product of matrix is possible, only when the number of columns in the first matrix is equal to no. of rows in second.

∴ a = 2.

Also the no. of columns of product (resulting matrix) is equal to no. of columns of second matrix.

∴ b = 1.

Hence, order of matrix M = 2 × 1.

Let M = $\begin{bmatrix*}[r] a \\ b \end{bmatrix*}$

Substituting value of A, M and B in (i) we get,

$\Rightarrow 3\begin{bmatrix*}[r] 0 & -1 \\ 4 & -3 \end{bmatrix*} \times \begin{bmatrix*}[r] a \\ b \end{bmatrix*} = 2\begin{bmatrix*}[r] -5 \\ 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 & -3 \\ 12 & -9 \end{bmatrix*} \times \begin{bmatrix*}[r] a \\ b \end{bmatrix*} = \begin{bmatrix*}[r] -10 \\ 12 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 \times a + (-3) \times b \\ 12 \times a + (-9) \times b \end{bmatrix*} = \begin{bmatrix*}[r] -10 \\ 12 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -3b \\ 12a - 9b \end{bmatrix*} = \begin{bmatrix*}[r] -10 \\ 12 \end{bmatrix*}$

By definition of equality of matrices we get,

-3b = -10
⇒ b = $\dfrac{10}{3}$ ....(i)

12a - 9b = 12

Substituting value of b from (i) in above equation we get,

$\Rightarrow 12a - 9 \times \dfrac{10}{3} = 12 \\[1em] \Rightarrow 12a - 30 = 12 \\[1em] \Rightarrow 12a = 42 \\[1em] \Rightarrow a = \dfrac{7}{2}.$

$\therefore \begin{bmatrix*}[r] a \\ b \end{bmatrix*} = \begin{bmatrix*}[r] \dfrac{7}{2} \\ \dfrac{10}{3} \end{bmatrix*}.$

Hence, M = $\begin{bmatrix*}[r] \dfrac{7}{2} \\ \dfrac{10}{3} \end{bmatrix*}.$

Find x and y if : $\begin{bmatrix*}[r] x & 3x \\ y & 4y \end{bmatrix*}\begin{bmatrix*}[r] 2 \\ 1 \end{bmatrix*} = \begin{bmatrix*}[r] 5 \\ 12 \end{bmatrix*}$.

Answer

$\Rightarrow \begin{bmatrix*}[r] x & 3x \\ y & 4y \end{bmatrix*}\begin{bmatrix*}[r] 2 \\ 1 \end{bmatrix*} = \begin{bmatrix*}[r] 5 \\ 12 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x \times 2 + 3x \times 1 \\ y \times 2 + 4y \times 1 \end{bmatrix*}= \begin{bmatrix*}[r] 5 \\ 12 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 5x \\ 6y \end{bmatrix*}= \begin{bmatrix*}[r] 5 \\ 12 \end{bmatrix*} \\[1em]$

By definition of equality of matrices we get,

5x = 5
⇒ x = 1.

6y = 12
⇒ y = 2.

Hence, x = 1 and y = 2.

If matrix X = $\begin{bmatrix*}[r] -3 & 4 \\ 2 & -3 \end{bmatrix*}\begin{bmatrix*}[r] 2 \\ -2 \end{bmatrix*} \text{ and 2X - 3Y} = \begin{bmatrix*}[r] 10 \\ -8 \end{bmatrix*}$, find the matrix 'X' and matrix 'Y'.

Answer

Given,

$X = \begin{bmatrix*}[r] -3 & 4 \\ 2 & -3 \end{bmatrix*}\begin{bmatrix*}[r] 2 \\ -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -3 \times 2 + 4 \times (-2) \\ 2 \times 2 + (-3) \times (-2) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -6 + (-8) \\ 4 + 6 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -14 \\ 10 \end{bmatrix*}.$

Given,

$\Rightarrow 2X - 3Y = \begin{bmatrix*}[r] 10 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow 2\begin{bmatrix*}[r] -14 \\ 10 \end{bmatrix*} - 3Y = \begin{bmatrix*}[r] 10 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -28 \\ 20 \end{bmatrix*} - 3Y = \begin{bmatrix*}[r] 10 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow 3Y = \begin{bmatrix*}[r] -28 \\ 20 \end{bmatrix*} - \begin{bmatrix*}[r] 10 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow 3Y = \begin{bmatrix*}[r] -28 - 10 \\ 20 - (-8) \end{bmatrix*} \\[1em] \Rightarrow 3Y = \begin{bmatrix*}[r] -38 \\ 28 \end{bmatrix*} \\[1em] \Rightarrow Y = \dfrac{1}{3}\begin{bmatrix*}[r] -38 \\ 28 \end{bmatrix*}.$

Hence, X = $\begin{bmatrix*}[r] -14 \\ 10 \end{bmatrix*} \text{ and Y} = \dfrac{1}{3}\begin{bmatrix*}[r] -38 \\ 28 \end{bmatrix*}$.

If A = $\begin{bmatrix*}[r] 2 & 5 \\ 1 & 3 \end{bmatrix*}, B = \begin{bmatrix*}[r] 4 & -2 \\ -1 & 3 \end{bmatrix*}$ and I is the identity matric of same order and A^t is transpose of matrix A, find A^t.B + BI.

Answer

$A = \begin{bmatrix*}[r] 2 & 5 \\ 1 & 3 \end{bmatrix*}, A^t = \begin{bmatrix*}[r] 2 & 1 \\ 5 & 3 \end{bmatrix*}, I = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}$.

$\Rightarrow A^t.B + BI = \begin{bmatrix*}[r] 2 & 1 \\ 5 & 3 \end{bmatrix*}\begin{bmatrix*}[r] 4 & -2 \\ -1 & 3 \end{bmatrix*} + \begin{bmatrix*}[r] 4 & -2 \\ -1 & 3 \end{bmatrix*} \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 \times 4 + 1 \times (-1) & 2 \times (-2) + 1 \times 3 \\ 5 \times 4 + 3 \times (-1) & 5 \times (-2) + 3 \times 3 \end{bmatrix*} + \begin{bmatrix*}[r] 4 \times 1 + (-2) \times 0 & 4 \times 0 + (-2) \times 1 \\ -1 \times 1 + 3 \times 0 & -1 \times 0 + 3 \times 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 8 - 1 & -4 + 3 \\ 20 - 3 & -10 + 9 \end{bmatrix*} + \begin{bmatrix*}[r] 4 + 0 & 0 - 2 \\ -1 + 0 & 0 + 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 7 & -1 \\ 17 & -1 \end{bmatrix*} + \begin{bmatrix*}[r] 4 & -2 \\ -1 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 7 + 4 & -1 + (-2) \\ 17 + (-1) & -1 + 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 11 & -3 \\ 16 & 2 \end{bmatrix*}.$

Hence, $A^t.B + BI = \begin{bmatrix*}[r] 11 & -3 \\ 16 & 2 \end{bmatrix*}.$

Let A = $\begin{bmatrix*}[r] 1 & 0 \\ 2 & 1 \end{bmatrix*}, B = \begin{bmatrix*}[r] 2 & 3 \\ -1 & 0 \end{bmatrix*}.$ Find A² + AB + B².

Answer

$A^2 = \begin{bmatrix*}[r] 1 & 0 \\ 2 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 1 & 0 \\ 2 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 1 + 0 \times 2 & 1 \times 0 + 0 \times 1 \\ 2 \times 1 + 1 \times 2 & 2 \times 0 + 1 \times 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 + 0 & 0 + 0 \\ 2 + 2 & 0 + 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 0 \\ 4 & 1 \end{bmatrix*} \\[1em] B^2 = \begin{bmatrix*}[r] 2 & 3 \\ -1 & 0 \end{bmatrix*}\begin{bmatrix*}[r] 2 & 3 \\ -1 & 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 \times 2 + 3 \times (-1) & 2 \times 3 + 3 \times 0 \\ -1 \times 2 + 0 \times (-1) & -1 \times 3 + 0 \times 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 - 3 & 6 + 0 \\ -2 + 0 & -3 + 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 6 \\ -2 & -3 \end{bmatrix*}.$

Substituting value of A² and B² in A² + AB + B² we get,

$A^2 + AB + B^2 = \begin{bmatrix*}[r] 1 & 0 \\ 4 & 1 \end{bmatrix*} + \begin{bmatrix*}[r] 1 & 0 \\ 2 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 2 & 3 \\ -1 & 0 \end{bmatrix*} + \begin{bmatrix*}[r] 1 & 6 \\ -2 & -3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 0 \\ 4 & 1 \end{bmatrix*} + \begin{bmatrix*}[r] 1 \times 2 + 0 \times (-1) & 1 \times 3 + 0 \times 0 \\ 2 \times 2 + 1 \times (-1) & 2 \times 3 + 1 \times 0 \end{bmatrix*} + \begin{bmatrix*}[r] 1 & 6 \\ -2 & -3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 0 \\ 4 & 1 \end{bmatrix*} + \begin{bmatrix*}[r] 2 + 0 & 3 + 0 \\ 4 - 1 & 6 + 0 \end{bmatrix*} + \begin{bmatrix*}[r] 1 & 6 \\ -2 & -3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 0 \\ 4 & 1 \end{bmatrix*} + \begin{bmatrix*}[r] 2 & 3 \\ 3 & 6 \end{bmatrix*} + \begin{bmatrix*}[r] 1 & 6 \\ -2 & -3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 + 2 + 1 & 0 + 3 + 6 \\ 4 + 3 + (-2) & 1 + 6 + (-3) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & 9 \\ 5 & 4 \end{bmatrix*}.$

Hence, A² + AB + B² = $\begin{bmatrix*}[r] 4 & 9 \\ 5 & 4 \end{bmatrix*}.$

If A = $\begin{bmatrix*}[r] 3 & a \\ -4 & 8 \end{bmatrix*}, B = \begin{bmatrix*}[r] c & 4 \\ -3 & 0 \end{bmatrix*}, C = \begin{bmatrix*}[r] -1 & 4 \\ 3 & b \end{bmatrix*}$ and 3A - 2C = 6B, find the values of a, b and c.

Answer

Given,

$\Rightarrow 3A - 2C = 6B \\[1em] \Rightarrow 3\begin{bmatrix*}[r] 3 & a \\ -4 & 8 \end{bmatrix*} - 2\begin{bmatrix*}[r] -1 & 4 \\ 3 & b \end{bmatrix*} = 6\begin{bmatrix*}[r] c & 4 \\ -3 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 9 & 3a \\ -12 & 24 \end{bmatrix*} - \begin{bmatrix*}[r] -2 & 8 \\ 6 & 2b \end{bmatrix*} = \begin{bmatrix*}[r] 6c & 24 \\ -18 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 9 - (-2) & 3a - 8 \\ -12 - 6 & 24 - 2b \end{bmatrix*} = \begin{bmatrix*}[r] 6c & 24 \\ -18 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 11 & 3a - 8 \\ -18 & 24 - 2b \end{bmatrix*} = \begin{bmatrix*}[r] 6c & 24 \\ -18 & 0 \end{bmatrix*}$

By definition of equality of matrices we get,

6c = 11
⇒ c = $\dfrac{11}{6} = 1\dfrac{5}{6}.$

3a - 8 = 24
⇒ 3a = 32
⇒ a = $\dfrac{32}{3} = 10\dfrac{2}{3}$

24 - 2b = 0
⇒ 2b = 24
⇒ b = 12.

Hence, a = $10\dfrac{2}{3}, b = 12, c = 1\dfrac{5}{6}.$

Given A = $\begin{bmatrix*}[r] p & 0 \\ 0 & 2 \end{bmatrix*}, B = \begin{bmatrix*}[r] 0 & -q \\ 1 & 0 \end{bmatrix*}, C = \begin{bmatrix*}[r] 2 & -2 \\ 2 & 2 \end{bmatrix*}$ and BA = C². Find the values of p and q.

Answer

Given,

$\Rightarrow BA = C^2 \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 & -q \\ 1 & 0 \end{bmatrix*}\begin{bmatrix*}[r] p & 0 \\ 0 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] 2 & -2 \\ 2 & 2 \end{bmatrix*}\begin{bmatrix*}[r] 2 & -2 \\ 2 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 \times p + (-q) \times 0 & 0 \times 0 + (-q) \times 2 \\ 1 \times p + 0 \times 0 & 1 \times 0 + 0 \times 2 \end{bmatrix*} = \begin{bmatrix*}[r] 2 \times 2 + (-2) \times 2 & 2 \times (-2) + (-2) \times 2 \\ 2 \times 2 + 2 \times 2 & 2 \times (-2) + 2 \times 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 & -2q \\ p & 0 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & -8 \\ 8 & 0 \end{bmatrix*}$

By definition of equality of matrices we get,

-2q = -8
⇒ q = 4.

p = 8.

Hence, p = 8 and q = 4.

Evaluate : $\begin{bmatrix*}[r] 4 sin 30° & 2cos 60° \\ sin 90° & 2 cos 0° \end{bmatrix*}\begin{bmatrix*}[r] 4 & 5 \\ 5 & 4 \end{bmatrix*}$.

Answer

$\Rightarrow \begin{bmatrix*}[r] 4 sin 30° & 2cos 60° \\ sin 90° & 2 cos 0° \end{bmatrix*}\begin{bmatrix*}[r] 4 & 5 \\ 5 & 4 \end{bmatrix*} = \begin{bmatrix*}[r] 4 \times \dfrac{1}{2} & 2 \times \dfrac{1}{2} \\ 1 & 2 \times 1 \end{bmatrix*}\begin{bmatrix*}[r] 4 & 5 \\ 5 & 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 & 1 \\ 1 & 2 \end{bmatrix*}\begin{bmatrix*}[r] 4 & 5 \\ 5 & 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 \times 4 + 1 \times 5 & 2 \times 5 + 1 \times 4 \\ 1 \times 4 + 2 \times 5 & 1 \times 5 + 2 \times 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 8 + 5 & 10 + 4 \\ 4 + 10 & 5 + 8 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 13 & 14 \\ 14 & 13 \end{bmatrix*}.$

Hence, $\begin{bmatrix*}[r] 4 sin 30° & 2cos 60° \\ sin 90° & 2 cos 0° \end{bmatrix*}\begin{bmatrix*}[r] 4 & 5 \\ 5 & 4 \end{bmatrix*} = \begin{bmatrix*}[r] 13 & 14 \\ 14 & 13 \end{bmatrix*}.$

Given A = $\begin{bmatrix*}[r] 2 & 0 \\ -1 & 7 \end{bmatrix*} \text{ and } I = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}$ and A² = 9A + mI. Find m.

Answer

Given,

$\Rightarrow A^2 = 9A + mI \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 & 0 \\ -1 & 7 \end{bmatrix*}\begin{bmatrix*}[r] 2 & 0 \\ -1 & 7 \end{bmatrix*} = 9\begin{bmatrix*}[r] 2 & 0 \\ -1 & 7 \end{bmatrix*} + m\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \times 2 + 0 \times (-1) & 2 \times 0 + 0 \times 7 \\ -1 \times 2 + 7 \times (-1) & -1 \times 0 + 7 \times 7 \end{bmatrix*} = \begin{bmatrix*}[r] 18 & 0 \\ -9 & 63 \end{bmatrix*} + \begin{bmatrix*}[r] m & 0 \\ 0 & m \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & 0 \\ -9 & 49 \end{bmatrix*} = \begin{bmatrix*}[r] 18 & 0 \\ -9 & 63 \end{bmatrix*} + \begin{bmatrix*}[r] m & 0 \\ 0 & m \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] m & 0 \\ 0 & m \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 0 \\ -9 & 49 \end{bmatrix*} - \begin{bmatrix*}[r] 18 & 0 \\ -9 & 63 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] m & 0 \\ 0 & m \end{bmatrix*} = \begin{bmatrix*}[r] 4 - 18 & 0 - 0 \\ -9 - (-9) & 49 - 63 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] m & 0 \\ 0 & m \end{bmatrix*} = \begin{bmatrix*}[r] -14 & 0 \\ 0 & -14 \end{bmatrix*}$

By definition of equality of matrices we get,

⇒ m = -14.

Hence, m = -14.

Given matrix A = $\begin{bmatrix*}[r] 4 sin 30° & cos 0° \\ cos 0° & 4 sin 30° \end{bmatrix*}\text{ and } B = \begin{bmatrix*}[r] 4 \\ 5 \end{bmatrix*}$. If AX = B,

(i) write the order of matrix X.

(ii) find the matrix 'X'.

Answer

(i) Let order of matrix X be a × b.

i.e. $\begin{bmatrix*}[r] 4 sin 30° & cos 0° \\ cos 0° & 4 sin 30° \end{bmatrix*}_{2 \times 2} \times X_{a \times b} = \begin{bmatrix*}[r] 4 \\ 5 \end{bmatrix*}_{2 \times 1}$

Since product of matrix is possible, only when the number of columns in the first matrix is equal to no. of rows in second.

∴ a = 2.

Also the no. of columns of product (resulting matrix) is equal to no. of columns of second matrix.

∴ b = 1.

Hence, order of matrix X = 2 × 1.

(ii) Let matrix X = $\begin{bmatrix*}[r] x \\ y \end{bmatrix*}$

Given,

$\Rightarrow AX = B \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 sin 30° & cos 0° \\ cos 0° & 4 sin 30° \end{bmatrix*}\begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 4 \\ 5 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 \times \dfrac{1}{2} & 1 \\ 1 & 4 \times \dfrac{1}{2} \end{bmatrix*}\begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 4 \\ 5 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 & 1 \\ 1 & 2 \end{bmatrix*}\begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 4 \\ 5 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x + y \\ x + 2y \end{bmatrix*} = \begin{bmatrix*}[r] 4 \\ 5 \end{bmatrix*}$

By definition of equality of matrices we get,

2x + y = 4
⇒ y = 4 - 2x .......(i)

x + 2y = 5

Substituting value of y from (i) in above equation we get,

⇒ x + 2(4 - 2x) = 5
⇒ x + 8 - 4x = 5
⇒ -3x = 5 - 8
⇒ -3x = -3
⇒ x = 1.

⇒ y = 4 - 2x = 4 - 2(1) = 2.

$\therefore X = \begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 1 \\ 2 \end{bmatrix*}$

Hence, X = $\begin{bmatrix*}[r] 1 \\ 2 \end{bmatrix*}.$