# Use the given of the coefficient matrix to solve the

Use the given of the coefficient matrix to solve the following system.
$$\displaystyle{7}{x}_{{1}}+{3}{x}_{{2}}={6}$$
$$\displaystyle-{6}{x}_{{1}}-{3}{x}_{{2}}={4}$$
$A^{-1}=\begin{bmatrix}1&1\\-2&-\frac{7}{3}\end{bmatrix}$

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Michele Grimsley

$$\displaystyle{7}{x}_{{1}}+{3}{x}_{{2}}={6}$$
$$\displaystyle-{6}{x}_{{1}}-{3}{x}_{{2}}={4}$$
Ax=B
$$\displaystyle\Rightarrow{x}={A}^{{-{1}}}{B}$$
Here $A=\begin{bmatrix}7&3\\-6&-3\end{bmatrix}B=\begin{bmatrix}6\\4\end{bmatrix}$
$x=A^{-1}B=\begin{bmatrix}1&1\\-2&-\frac{7}{3}\end{bmatrix}\begin{bmatrix}6\\4\end{bmatrix}$
$$\displaystyle{7}{x}_{{1}}+{3}{x}_{{2}}={6}$$
$$\displaystyle-{6}{x}_{{1}}-{3}{x}_{{2}}={4}$$
$$\displaystyle{A}{x}={B}$$
$$\displaystyle\Rightarrow{x}={A}^{{-{1}}}{B}$$
Here $A=\begin{bmatrix}7&3\\-6&-3\end{bmatrix}B=\begin{bmatrix}6\\4\end{bmatrix}$
$=\begin{bmatrix}1&1\\-2&-\frac{7}{3}\end{bmatrix}\begin{bmatrix}6\\4\end{bmatrix}$
$=\begin{bmatrix}6+4\\-12-\frac{28}{3}\end{bmatrix}$
$=\begin{bmatrix}10\\-64/3\end{bmatrix}$