Solve by matrix method $2 x + y + 2 z = 5, x - y - z = 0, x + 2 y + 3 z = 5$ .

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Solution

$⎡ ⎢ ⎣ \begin{matrix} 2 & 1 & 2 1 & - 1 & - 1 1 & 2 & 3 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 505 \end{matrix} ⎤ ⎥ ⎦$
$⎡ ⎢ ⎣ \begin{matrix} 2 & 1 & 2 1 & - 1 & - 1 1 & 2 & 3 \end{matrix} \begin{matrix} : : : \end{matrix} \begin{matrix} 505 \end{matrix} ⎤ ⎥ ⎦$

$R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1}$
$= ⎡ ⎢ ⎣ \begin{matrix} 1 & 1 / 2 & 1 0 & - 3 / 2 & - 2 0 & 3 / 2 & 2 \end{matrix} \begin{matrix} : : : \end{matrix} \begin{matrix} 5 / 2 - 5 / 2 5 / 2 \end{matrix} ⎤ ⎥ ⎦$

$R_{1} \to R_{1} + \frac{1}{3} R_{2}$
$= ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 1 / 3 0 & - 3 / 2 & - 2 0 & 3 / 2 & 2 \end{matrix} \begin{matrix} : : : \end{matrix} \begin{matrix} 5 / 3 - 5 / 2 5 / 2 \end{matrix} ⎤ ⎥ ⎦$

$R_{2} \to \frac{- 2}{3} R_{2}$
$= ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 1 / 3 0 & 1 & 4 / 3 0 & 3 / 2 & - 2 \end{matrix} \begin{matrix} : : : \end{matrix} \begin{matrix} 5 / 3 5 / 3 5 / 2 \end{matrix} ⎤ ⎥ ⎦$

$R_{3} \to R_{3} - \frac{3}{2} R_{2}$
$= ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 1 / 3 0 & 1 & 4 / 3 0 & 0 & - 2 \end{matrix} \begin{matrix} : : : \end{matrix} \begin{matrix} 5 / 3 5 / 3 0 \end{matrix} ⎤ ⎥ ⎦$
$z = 0$ $y = \frac{5}{3}$ $x = \frac{5}{3}$
Hence, solved.

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