Given, matrix $$A = \begin{bmatrix} x & 1 \\ 2 & 2 \end{bmatrix}$$ and $$B = \begin{bmatrix} x \\ x - 2 \end{bmatrix}$$, such that AB is a null matrix. Find the :
(a) order of the null matrix.
(b) value of $x$, given that $x \gt 0$.
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(a) Given,
Matrix A is of order 2 × 2 and matrix B is of order 2 × 1.
We know that,
If matrix A is of order $m \times n$ and matrix B is of order $n \times p$, then the order of the product AB is $m \times p$.
∴ Order of AB = 2 × 1
Since, AB is a null matrix, the order of the null matrix is the same as the order of AB.
Hence, the order of the null matrix is 2 × 1.
(b) Calculating AB,
$$\Rightarrow AB = \begin{bmatrix*}[r] x & 1 \\ 2 & 2 \end{bmatrix*}\begin{bmatrix*}[r] x \\ x - 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] x \times x + 1 \times (x - 2) \\ 2 \times x + 2 \times (x - 2) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] x^2 + x - 2 \\ 2x + 2x - 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] x^2 + x - 2 \\ 4x - 4 \end{bmatrix*}.$$
Since, AB is a null matrix,
$$\Rightarrow \begin{bmatrix*}[r] x^2 + x - 2 \\ 4x - 4 \end{bmatrix*} = \begin{bmatrix*}[r] 0 \\ 0 \end{bmatrix*}$$
Comparing the corresponding elements, we get :
$$\begin{aligned}\Rightarrow 4x - 4 = 0 \\[1em] \Rightarrow 4x = 4 \\[1em] \Rightarrow x = 1\end{aligned}$$
Also,
$$\begin{aligned}\Rightarrow x^2 + x - 2 = 0 \\[1em] \Rightarrow x^2 + 2x - x - 2 = 0 \\[1em] \Rightarrow x(x + 2) - 1(x + 2) = 0 \\[1em] \Rightarrow (x + 2)(x - 1) = 0 \\[1em] \Rightarrow (x + 2) = 0 \text{ or } (x - 1) = 0 \qquad [\text{Using zero product rule}] \\[1em] \Rightarrow x = -2 \text{ or } x = 1\end{aligned}$$
The value of $x$ which satisfies both the equations and is greater than 0 is 1.
Hence, $\bm{x}$ = 1.