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### Answers (1)

Answer

$\rightarrow x=2, y=3, z=-1$

Given
$\rightarrow\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 2 \\ -1 \\ 3 \end{array}\right]$

Explanation

$\rightarrow\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 2 \\ -1 \\ 3 \end{array}\right]$
\begin{aligned} &{\left[\begin{array}{l} 1 \times x+0 \times y+0 \times z \\ 0 \times x+0 \times y+1 \times z \\ 0 \times x+1 \times y+0 \times z \end{array}\right]=\left[\begin{array}{c} 2 \\ -1 \\ 3 \end{array}\right]} \\\\ &{\left[\begin{array}{l} x \\ z \\ y \end{array}\right]=\left[\begin{array}{c} 2 \\ -1 \\ 3 \end{array}\right]} \end{aligned}

Comparing both sides we get $x=2, y=3, z=-1$

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