Question

If $\left[\begin{array}{cc}3& -1\\ 0& 6\end{array}\right]\left[\begin{array}{c}3x\\ 1\end{array}\right]+\left[\begin{array}{c}-2x\\ 3\end{array}\right]=\left[\begin{array}{c}8\\ 9\end{array}\right]$, then the value of $x$ is

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

Answer 1

Answer: (D)

None of these

Given, $\left[\begin{array}{cc}3& -1\\ 0& 6\end{array}\right]\left[\begin{array}{c}3x\\ 1\end{array}\right]+\left[\begin{array}{c}-2x\\ 3\end{array}\right]=\left[\begin{array}{c}8\\ 9\end{array}\right]$
LHS $=\left[\begin{array}{cc}3& -1\\ 0& 6\end{array}\right]\left[\begin{array}{c}3x\\ 1\end{array}\right]+\left[\begin{array}{c}-2x\\ 3\end{array}\right]$
$=\left[\begin{array}{c}9x+(-1)\\ 0+6\end{array}\right]+\left[\begin{array}{c}-2x\\ 3\end{array}\right]=\left[\begin{array}{c}9x-1\\ 6\end{array}\right]+\left[\begin{array}{c}-2x\\ 3\end{array}\right]$
$=\left[\begin{array}{c}9x-1-2x\\ 6+3\end{array}\right]=\left[\begin{array}{c}7x-1\\ 9\end{array}\right]$
Now, $\left[\begin{array}{c}7x-1\\ 9\end{array}\right]=\left[\begin{array}{c}8\\ 9\end{array}\right]$
$\therefore 7x-1=8$
$\Rightarrow 7x=9$
$\Rightarrow x=\frac{9}{7}$.