Question

$1fA=\left[\begin{array}{ccc}4& 1& 0\\ 1& -2& 2\end{array}\right],B=\left[\begin{array}{ccc}2& 0& -1\\ 3& 1& 4\end{array}\right]$, $C=\left[\begin{array}{c}1\\ 2\\ -1\end{array}\right]$ and $(3B-2A)C+2X=0$ then $X$ is equal to

• A. $\frac{1}{2}\left[\begin{array}{c}3\\ 13\end{array}\right]$
• B. $\frac{1}{2}\left[\begin{array}{c}3\\ -13\end{array}\right]$
• C. $\frac{1}{2}\left[\begin{array}{c}-3\\ 13\end{array}\right]$
• D. $\left[\begin{array}{c}3\\ -13\end{array}\right]$

Answer 1

The correct option is B $\frac{1}{2}\left[\begin{array}{c}3\\ -13\end{array}\right]$

Given, $A=\left[\begin{array}{ccc}4& 1& 0\\ 1& -2& 2\end{array}\right],B=\left[\begin{array}{ccc}2& 0& -1\\ 3& 1& 4\end{array}\right],C=\left[\begin{array}{c}1\\ 2\\ -1\end{array}\right]$
Now, $(3B-2A)C+2X=0$
$\Rightarrow (\left[\begin{array}{ccc}6& 0& -3\\ 9& 3& 12\end{array}\right]-\left[\begin{array}{ccc}8& 2& 0\\ 2& -4& 4\end{array}\right])\left[\begin{array}{c}1\\ 2\\ -1\end{array}\right]+2X=0$
$\left[\begin{array}{ccc}-2& -2& -3\\ 7& 7& 8\end{array}\right]\left[\begin{array}{c}1\\ 2\\ -1\end{array}\right]+2X=0$
$\left[\begin{array}{c}-3\\ 13\end{array}\right]+2X=0$
$\Rightarrow 2X=\left[\begin{array}{c}3\\ -13\end{array}\right]$
$\Rightarrow X=\frac{1}{2}\left[\begin{array}{c}3\\ -13\end{array}\right]$