Question

If $\begin{bmatrix}3&-1\\ 0&6\end{bmatrix}\begin{bmatrix}3x\\ 1\end{bmatrix}+\begin{bmatrix}-2x\\ 3\end{bmatrix}=\begin{bmatrix}8\\ 9\end{bmatrix}$ the value of $x$ is

• A. $7$
• B. $-\frac{2}{9}$
• C. $-\frac{3}{8}$
• D. None of the options

Answer 1

Answer: (D)

Given, $\begin{bmatrix}3&-1\\ 0&6\end{bmatrix}\begin{bmatrix}3x\\ 1\end{bmatrix} +\begin{bmatrix}-2x\\ 3\end{bmatrix}= \begin{bmatrix}8\\ 9\end{bmatrix}$
$LH S =\begin{bmatrix}3 & -1 \\ 0 & 6\end{bmatrix}\begin{bmatrix}3 x \\ 1\end{bmatrix}+\begin{bmatrix}-2 x \\ 3\end{bmatrix}$
$=\begin{bmatrix}9 x+(-1) \\ 0+6\end{bmatrix}+\begin{bmatrix}-2 x \\ 3\end{bmatrix}$
$=\begin{bmatrix}9 x-1 \\ 6\end{bmatrix}+\begin{bmatrix}-2 x \\ 3\end{bmatrix} \\ =\begin{bmatrix}9 x-1-2 x \\ 6+3\end{bmatrix}=\begin{bmatrix}7 x-1 \\ 9\end{bmatrix}$
Now, $\begin{bmatrix}7 x-1 \\ 9\end{bmatrix}=\begin{bmatrix}8 \\ 9\end{bmatrix}$
$\therefore 7 \times-1=8$
$\Rightarrow 7 x=9$
$\Rightarrow x=\frac{9}{7}$