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  4. If A = [3&2 4&5] and AC = [19&24 37&46] then C=

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

COMEDKCOMEDK 2011Determinants

Solution:

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}$
$\Rightarrow C = \frac{1}{7} \begin{pmatrix}95-74&120-92\\ -76+111&-96+138\end{pmatrix}$
$\Rightarrow C = \frac{1}{7} \begin{pmatrix}21&28\\ 35&42\end{pmatrix} \Rightarrow C = \begin{pmatrix}3&4\\ 5&6\end{pmatrix}$