Question

If A = $\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$, B = $\left[\begin{array}{cc}2& 3\\ 4& 5\end{array}\right]$ and $4A-3B+C=0,$ then $C=$

• A. $\left[\begin{array}{cc}2& -1\\ 0& 1\end{array}\right]$
• B. $\left[\begin{array}{cc}2& 1\\ 0& -1\end{array}\right]$
• C. $\left[\begin{array}{cc}-2& 1\\ 0& -1\end{array}\right]$
• D. None

Answer 1

Answer: (D)

None

Since, $A=\left[\begin{array}{cc}1& 2\\ 2& 4\end{array}\right]$

$B=\left[\begin{array}{cc}2& 3\\ 4& 5\end{array}\right]$

Also, $4A-3B+C=0$

$\Rightarrow 4\left[\begin{array}{cc}1& 2\\ 2& 4\end{array}\right]-3\left[\begin{array}{cc}2& 3\\ 4& 5\end{array}\right]+C=0$

$\Rightarrow \left[\begin{array}{cc}4& 8\\ 8& 16\end{array}\right]-\left[\begin{array}{cc}6& 9\\ 12& 15\end{array}\right]+C=0$

$\Rightarrow \left[\begin{array}{cc}-2& -1\\ -4& 1\end{array}\right]+C=0$

$\Rightarrow C=\left[\begin{array}{cc}2& 1\\ 4& -1\end{array}\right]$