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#### karush

##### Well-known member

- Jan 31, 2012

- 2,886

$A^{-1}=\begin{bmatrix}1&3\\2&5 \end{bmatrix}$

Solve the matrix equation $AX=B$ to find $x$ and $y$ where

$X=\begin{bmatrix}x\\y \end{bmatrix}\& \quad B=\begin{bmatrix}1\\3 \end{bmatrix}$

ok well first find A

$A=\begin{bmatrix}1&3\\2&5 \end{bmatrix}^{-1}

=\left[ \begin{array}{rr|rr}1&3&1&0 \\ 2&5&0&1\end{array}\right]

=\left[ \begin{array}{rr|rr}1&0&-5&3 \\ 0&1&2&-1\end{array}\right]

=\left[ \begin{array}{rr} -5 & 3 \\ 2 & -1 \end{array} \right]$

then we have

$\left[ \begin{array}{rr} -5 & 3 \\ 2 & -1 \end{array} \right]

\begin{bmatrix}x\\y \end{bmatrix}

=\begin{bmatrix}1\\3 \end{bmatrix}$

ok just seeing If I am going the right direction on this .. if so the rest would be a simultaneous equation