Solve by matrix method: $(x + 3 y + z = 5), (2 x - y + 3 z = 4), (x - 2 y + 5 z = 4)$ .

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Solution

$A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 3 & 1 2 & - 1 & 3 1 & - 2 & 5 \end{matrix} ⎤ ⎥ ⎦$ , $X = ⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦$ and $B = ⎡ ⎢ ⎣ \begin{matrix} 544 \end{matrix} ⎤ ⎥ ⎦$

$A X = B$ $\Rightarrow$ $X = A^{- 1} B$

But we know $A^{- 1} = \frac{a d j (A)}{| A |}$ .

So, $X = \frac{(a d j A) B}{| A |}$ .

$a d j (A) = ⎡ ⎢ ⎣ \begin{matrix} 1 & - 17 & 10 - 7 & 4 & - 1 - 3 & 5 & - 7 \end{matrix} ⎤ ⎥ ⎦$ and $| A | = - 23$

$\Rightarrow X = \frac{- 1}{23} ⎡ ⎢ ⎣ \begin{matrix} 1 & - 17 & 10 - 7 & 4 & - 1 - 3 & 5 & - 7 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 544 \end{matrix} ⎤ ⎥ ⎦ = \frac{- 1}{23} ⎡ ⎢ ⎣ \begin{matrix} - 23 - 23 - 23 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 111 \end{matrix} ⎤ ⎥ ⎦$

$∴$ $x = y = z = 1$

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