If $A = ⎡ ⎢ ⎣ \begin{matrix} 1 & - 2 & 1 0 & - 1 & 1 2 & 0 & - 3 \end{matrix} ⎤ ⎥ ⎦$ then find $A^{- 1}$ and hence, solve the system equations $x - 2 y + z = 0, - y + z = - 2$ and $2 x - 3 z = 10$ .

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Solution

$A = ⎡ ⎢ ⎣ \begin{matrix} 1 & - 2 & 1 0 & - 1 & 1 2 & 0 & - 3 \end{matrix} ⎤ ⎥ ⎦$
$A^{- 1} = \frac{1}{| A |} . a d j A$
$| A | = 1 ∣ ∣ ∣ \begin{matrix} - 1 & 1 0 & - 3 \end{matrix} ∣ ∣ ∣ + 2 ∣ ∣ ∣ \begin{matrix} 0 & 1 2 & - 3 \end{matrix} ∣ ∣ ∣ + 1 ∣ ∣ ∣ \begin{matrix} 0 & - 1 2 & 0 \end{matrix} ∣ ∣ ∣$
$= 1 [3 - 0] + 2 [0 - 2] + 1 [0 + 2]$
$= 3 - 4 + 2 = 1$
$a d j A = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} + ∣ ∣ ∣ \begin{matrix} - 1 & 1 0 & - 3 \end{matrix} ∣ ∣ ∣ & - ∣ ∣ ∣ \begin{matrix} 0 & 1 2 & - 3 \end{matrix} ∣ ∣ ∣ & + ∣ ∣ ∣ \begin{matrix} 0 & - 1 2 & 0 \end{matrix} ∣ ∣ ∣ - ∣ ∣ ∣ \begin{matrix} - 2 & 1 0 & - 3 \end{matrix} ∣ ∣ ∣ & + ∣ ∣ ∣ \begin{matrix} 1 & 1 2 & - 3 \end{matrix} ∣ ∣ ∣ & - ∣ ∣ ∣ \begin{matrix} 1 & 2 2 & 0 \end{matrix} ∣ ∣ ∣ + ∣ ∣ ∣ \begin{matrix} - 2 & 1 - 1 & 1 \end{matrix} ∣ ∣ ∣ & - ∣ ∣ ∣ \begin{matrix} 1 & 1 0 & 1 \end{matrix} ∣ ∣ ∣ & + ∣ ∣ ∣ \begin{matrix} 1 & 2 0 & - 1 \end{matrix} ∣ ∣ ∣ \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$
$= ⎡ ⎢ ⎣ \begin{matrix} 3 & 2 & 2 - 6 & - 5 & - 4 - 1 & - 1 & - 1 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 3 & - 6 & - 1 2 & - 5 & - 1 2 & - 4 & - 1 \end{matrix} ⎤ ⎥ ⎦$
$A^{- 1} = ⎡ ⎢ ⎣ \begin{matrix} 3 & - 6 & - 1 2 & - 5 & - 1 2 & - 4 & - 1 \end{matrix} ⎤ ⎥ ⎦$
$x - 2 y + z = 0, - y + z = - 2, 2 x - 3 z = 10$
$y = z + 2; 2 x = 10 + 3 z$
$x = \frac{10 + 3 z}{2}$
$x - 2 y + z = 0$
$\frac{10 + 3 z}{2} - 2 (z + 2) + z = 0$ $x = \frac{10 + 3 (- z)}{2}$
$10 + 3 z - 4 z - 8 + 22 = 0$ $x = \frac{4}{2}$
$z + 2 = 0$
$z = - 2$ $x = 2$
So :- $y = z + 2$
$y = - 2 + 2$
$y = 0$

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