$inverse of matrices$

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$inverse of matrices$

can anyone help solve this problem with inverse of matrices?

Precalculus Matrices

Asialynn

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Since $AX+B=C$, we have

\[AX=C-B\]
\[ \Rightarrow X= A^{-1}(C-B). (1)\]
Next we compute $C-B$ and $A^{-1}.$

$C-B$ = $\begin{bmatrix} 3 & 0 \\ -6 & 3\end{bmatrix} $ -$\begin{bmatrix} -1 & -3 \\ -2 & 1 \end{bmatrix} $ =$\begin{bmatrix} 4 & 3 \\ -4 & 2 \end{bmatrix} $

Also

A^{-1}= $\frac{1}{-4 \times 3 - (1\times -2)}$ $\begin{bmatrix} 3 & 2 \\ -1& -4\end{bmatrix} =\frac{1}{-10}\begin{bmatrix} 3 & 2 \\ -1& -4\end{bmatrix}. $

Substititing in (1) we get
\[X=A^{-1}(C-B)=\frac{1}{-10}\begin{bmatrix} 3 & 2 \\ -1& -4\end{bmatrix}\begin{bmatrix} 4 & 3 \\ -4 & 2 \end{bmatrix}\]
\[=-\frac{1}{10}\begin{bmatrix} 4 & 13 \\ 12 & -11 \end{bmatrix}=\begin{bmatrix} -2/5 & -13/10 \\ -6/5 & 11/10 \end{bmatrix}.\]

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$inverse of matrices$

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