Matrices — Analytical & Application Based Questions

$A = \begin{bmatrix} -6 & 0 \\ 4 & 2 \end{bmatrix} \text{ and } B = \begin{bmatrix} 1 & 0 \\ 1 & 3 \end{bmatrix}$.

Find matrix M, if M = $\dfrac{1}{2}A - 2B + 5l$, where l is the identity matrix.

Answer

Given,

$\Rightarrow M = \dfrac{1}{2}A - 2B + 5l \\[1em] = \dfrac{1}{2}\begin{bmatrix} -6 & 0 \\ 4 & 2 \end{bmatrix} - 2\begin{bmatrix} 1 & 0 \\ 1 & 3 \end{bmatrix} + 5\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix} -3 & 0 \\ 2 & 1 \end{bmatrix} - \begin{bmatrix} 2 & 0 \\ 2 & 6 \end{bmatrix} + \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} \\[1em] = \begin{bmatrix} -3 - 2 + 5 & 0 - 0 + 0 \\ 2 - 2 + 0 & 1 - 6 + 5 \end{bmatrix} \\[1em] = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}.$

Hence, M = $\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}.$

Given, matrix A = $\begin{bmatrix} x & 1 \\ y & 2 \end{bmatrix}$ and B = $\begin{bmatrix} x \\ x - 2 \end{bmatrix}$ such that AB is a null matrix. Find:

(a) order of the null matrix.

(b) possible values of x and y.

Answer

(a) Given,

AB is a null matrix.

Let AB = X, where X is a null matrix of order a × b.

⇒ AB = X

⇒ A_{2 × 2} × B_{2 × 1} = X_{a × b}

We know that,

The resultant matrix has no. of rows equal to the rows in the first matrix and no. of columns equal to the no. of columns in the second matrix.

∴ a = 2 and b = 1.

Hence, the order of null matrix is 2 × 1.

(b) Calculating,

$\therefore \begin{bmatrix} x & 1 \\ y & 2 \end{bmatrix}_{2 \times 2} \begin{bmatrix} x \\ x - 2 \end{bmatrix}_{2 \times 1} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} x^2 + 1(x - 2) \\ xy + 2(x - 2) \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} x^2 + x - 2 \\ xy + 2x - 4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \text{ ………….(1)}$

From equation (1) :

⇒ x² + x - 2 = 0

⇒ x² + 2x - x - 2 = 0

⇒ x(x + 2) - 1(x + 2) = 0

⇒ (x - 1)(x + 2) = 0

⇒ x - 1 = 0 or x + 2 = 0

⇒ x = 1 or x = -2.

From equation (1) :

⇒ xy + 2x - 4 = 0

Substituting x = 1, we get :

⇒ 1.y + 2.1 - 4 = 0

⇒ y + 2 - 4 = 0

⇒ y - 2 = 0

⇒ y = 2.

Substituting x = -2, we get :

⇒ (-2).y + 2.(-2) - 4 = 0

⇒ -2y - 4 - 4 = 0

⇒ -2y - 8 = 0

⇒ -2y = 8

⇒ y = $\dfrac{8}{-2}$ = -4.

Hence, x = 1, y = 2 or x = -2, y = -4.

If A = $\begin{bmatrix} -1 & 3 \\ 2 & 0 \end{bmatrix}$, B = $\begin{bmatrix} 1 & -2 \\ 0 & 3 \end{bmatrix}$, C = $\begin{bmatrix} 1 & -4 \end{bmatrix}$ and D = $\begin{bmatrix} 4 \\ 1 \end{bmatrix}$.

(a) Is the product AC possible? Justify your answer.

(b) Find the matrix X, such that X = AB + B² - DC.

Answer

(a) Order of matrix A = 2 × 2, Order of matrix C = 1 × 2

The product AC is not possible as the no. of columns in A is not equal to the no. of rows in C.

Hence, product AC is not possible.

(b) Given,

$\Rightarrow X = AB + B^2 - DC \\[1em] \Rightarrow X = \begin{bmatrix} -1 & 3 \\ 2 & 0 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ 0 & 3 \end{bmatrix} + \begin{bmatrix} 1 & -2 \\ 0 & 3 \end{bmatrix}\begin{bmatrix} 1 & -2 \\ 0 & 3 \end{bmatrix} - \begin{bmatrix} 4 \\ 1 \end{bmatrix}\begin{bmatrix} 1 & -4 \end{bmatrix} \\[1em] \Rightarrow X = \begin{bmatrix} -1 \times 1 + 3 \times 0 & -1 \times -2 + 3 \times 3 \\ 2 \times 1 + 0 \times 0 & 2 \times -2 + 0 \times 3 \end{bmatrix} + \begin{bmatrix} 1 \times 1 + -2 \times 0 & 1 \times -2 + -2 \times 3 \\ 0 \times 1 + 3 \times 0 & 0 \times -2 + 3 \times 3 \end{bmatrix} - \begin{bmatrix} 4 \times 1 & 4 \times -4 \\ 1 \times 1 & 1 \times -4 \end{bmatrix} \\[1em] \Rightarrow X = \begin{bmatrix} -1 + 0 & 2 + 9 \\ 2 + 0 & -4 + 0 \end{bmatrix} + \begin{bmatrix} 1 + 0 & -2 - 6 \\ 0 + 0 & 0 + 9 \end{bmatrix} - \begin{bmatrix} 4 & -16 \\ 1 & -4 \end{bmatrix} \\[1em] \Rightarrow X = \begin{bmatrix} -1 & 11 \\ 2 & -4 \end{bmatrix} + \begin{bmatrix} 1 & -8 \\ 0 & 9 \end{bmatrix} - \begin{bmatrix} 4 & -16 \\ 1 & -4 \end{bmatrix} \\[1em] \Rightarrow X = \begin{bmatrix} -1 + 1 - 4 & 11 + (-8) - (-16) \\ 2 + 0 - 1 & (-4) + 9 - (-4) \end{bmatrix} \\[1em] \Rightarrow X = \begin{bmatrix} -4 & 11 - 8 + 16 \\ 1 & -4 + 9 + 4 \end{bmatrix} \\[1em] \Rightarrow X = \begin{bmatrix} -4 & 19 \\ 1 & 9 \end{bmatrix}$

Hence, X = $\begin{bmatrix} -4 & 19 \\ 1 & 9 \end{bmatrix}$.