Question

If $A = \begin{bmatrix}3&2\\\ 4&5\end{bmatrix}$ and $AC = \begin{bmatrix}19&24\\\ 37&46\end{bmatrix}$ then $C=$

• A. $\begin{bmatrix}3&4\\\ 5&2\end{bmatrix}$
• B. $\begin{bmatrix}3&4\\\ 5&3\end{bmatrix}$
• C. $\begin{bmatrix}3&4\\\ 5&6\end{bmatrix}$
• D. $\begin{bmatrix}3&4\\\ 5&5\end{bmatrix}$

Answer 1

Answer: (C)

$\begin{bmatrix}3&4\\\ 5&6\end{bmatrix}$

Answer 2

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