If $A = ⎡ ⎢ ⎣ \begin{matrix} 1 & - 1 & 1 2 & 1 & 3 1 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦$ , Find $A^{-} 1$ . Using $A^{-} 1$ , solve the following system or liner equation.
$x + 2 y + z = 4; - z + y + z = 0; x - 3 y + z = 2.$

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Solution

$⎡ ⎢ ⎣ \begin{matrix} - 2 & 1 & 1 2 & 0 & - 2 - 4 & - 1 & 3 \end{matrix} ⎤ ⎥ ⎦$
$A^{- 1} = ⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 1 / 2 & - 1 / 2 - 1 & 0 & 1 2 & \frac{1}{2} & \frac{- 3}{2} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦$
$A^{- 1} = ⎡ ⎢ ⎣ \begin{matrix} 1 & - 1 & 2 - 1 / 2 & 0 & 1 / 2 - 1 / 2 & 1 & - 3 / 2 \end{matrix} ⎤ ⎥ ⎦$

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