Solve the following system of linear equations using matrix method: $3 x + y + z = 1, 2 x + 2 z = 0, 5 x + 5 y + 2 z = 2$ .

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Solution

The system of linear equations are
$3 x + y + z = 1$
$2 x + 2 z = 0$
$5 x + 5 y + 2 x = 2$
This corresponds to $⎡ ⎢ ⎣ \begin{matrix} 3 & 1 & 1 2 & 0 & 2 5 & 5 & 2 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 102 \end{matrix} ⎤ ⎥ ⎦$
The augmented matrix is $⎡ ⎢ ⎣ \begin{matrix} 3 & 1 & 1 & 1 2 & 0 & 2 & 0 5 & 5 & 2 & 2 \end{matrix} ⎤ ⎥ ⎦$
Applying elementary row operations,
$⎡ ⎢ ⎣ \begin{matrix} 3 & 1 & 1 & 1 2 & 0 & 2 & 0 5 & 5 & 2 & 2 \end{matrix} ⎤ ⎥ ⎦ R_{1} \to R_{1} - R_{2} \to ⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & - 1 & 1 2 & 0 & 2 & 0 5 & 5 & 2 & 2 \end{matrix} ⎤ ⎥ ⎦$
$↓ R_{2} \to R_{2} - 2 R_{1}$
$R_{3} \to R_{3} - 5 R_{1}$
$⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & - 1 & 1 0 & 1 & - 2 & 1 0 & 0 & 7 & - 3 \end{matrix} ⎤ ⎥ ⎦ R_{2} \to \frac{R_{2}}{- 2} \leftarrow ⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 & - 1 0 & - 2 & 4 & - 2 0 & 0 & 7 & - 3 \end{matrix} ⎤ ⎥ ⎦$
$↓ R_{1} \to R_{1} - R_{2}$
$⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 1 & 0 0 & 1 & - 2 & 1 0 & 0 & 7 & - 3 \end{matrix} ⎤ ⎥ ⎦ R_{3} \to \frac{R_{3}}{7} \to ⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 0 & 1 & 0 0 & 1 & - 2 & 1 0 & 0 & 1 & \frac{- 3}{7} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦$
$↓ R_{2} \to R_{2} + 2 R_{3}$
$R_{1} \to R_{1} - R_{3}$
$⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 & \frac{3}{7} 0 & 1 & 0 & \frac{1}{7} 0 & 0 & 1 & \frac{- 3}{7} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$
So, $x = \frac{3}{7}, y = \frac{1}{7}, z = \frac{- 3}{7}$

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