Short Answer Questions 2 — Question 9

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.

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

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.