Question

If $A=\left[\begin{array}{cc}3& 2\\ 4& 5\end{array}\right]$ and $AC=\left[\begin{array}{cc}19& 24\\ 37& 46\end{array}\right]$ then $C=$

• A. $\left[\begin{array}{cc}3& 4\\ 5& 2\end{array}\right]$
• B. $\left[\begin{array}{cc}3& 4\\ 5& 3\end{array}\right]$
• C. $\left[\begin{array}{cc}3& 4\\ 5& 6\end{array}\right]$
• D. $\left[\begin{array}{cc}3& 4\\ 5& 5\end{array}\right]$

Answer 1

Answer: (C)

We have, $A=\left(\begin{array}{cc}3& 2\\ 4& 5\end{array}\right)AC=\left(\begin{array}{cc}19& 24\\ 37& 46\end{array}\right)$
Since , $|A|\ne 0\therefore A^{-1}$ exists
Multiplying by $A^{-1}$ on both sides of $AC$, we get
$A^{-1}AC=A^{-1}\left(\begin{array}{cc}19& 24\\ 37& 46\end{array}\right)$
$C=\frac{adjA}{|A|}\left(\begin{array}{cc}19& 24\\ 37& 46\end{array}\right)\left[\because A^{-1}=\frac{adjA}{|A|}\right]$
$\Rightarrow C=\frac{1}{7}\left(\begin{array}{cc}5& -2\\ -4& 3\end{array}\right)\left(\begin{array}{cc}19& 24\\ 37& 46\end{array}\right)$
$\Rightarrow C=\frac{1}{7}\left(\begin{array}{cc}95-74& 120-92\\ -76+111& -96+138\end{array}\right)$
$\Rightarrow C=\frac{1}{7}\left(\begin{array}{cc}21& 28\\ 35& 42\end{array}\right)\Rightarrow C=\left(\begin{array}{cc}3& 4\\ 5& 6\end{array}\right)$