Question

If $A=\left[\begin{array}{c}410\\ 1-22\end{array}\right],$ $B=\left[\begin{array}{c}20-1\\ 314\end{array}\right],$ $C=\left[\begin{array}{c}1\\ 2\\ -1\end{array}\right]$ and $\left(3B-2A\right)C+2X=0,$ then $X$=

• 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

Answer: (B)

$\frac{1}{2}\left[\begin{array}{c}3\\ -13\end{array}\right]$

Given, $A=\left[\begin{array}{ccc}4& 1& 0\\ 1& -2& 2\end{array}\right]B=\left[\begin{array}{ccc}2& 0& -1\\ 3& 1& 4\end{array}\right]C=\left[\begin{array}{c}1\\ 2\\ -1\end{array}\right]X=\left[\begin{array}{c}x\\ y\end{array}\right]$

Consider, $3B-2A=\left[\begin{array}{ccc}6& 0& -3\\ 9& 3& 12\end{array}\right]-\left[\begin{array}{ccc}8& 2& 0\\ 2& -4& 4\end{array}\right]=\left[\begin{array}{ccc}-2& -2& -3\\ 7& 7& 8\end{array}\right]$

Now, $(3B-2A)C=\left[\begin{array}{ccc}-2& -2& -3\\ 7& 7& 8\end{array}\right]\left[\begin{array}{c}1\\ 2\\ -1\end{array}\right]=\left[\begin{array}{c}-3\\ 13\end{array}\right]$

Also, $-2X=\left[\begin{array}{c}-3\\ 13\end{array}\right]X=\frac{1}{2}\left[\begin{array}{c}3\\ -13\end{array}\right]$

$\therefore B$ is correct