#### Explain solution RD Sharma class 12 chapter 7 Solution of Simultaneous Linear Equation Exercise Very short answer Question 3 Maths Textbook Solution.

Answer    $\rightarrow x=1,y=0,z=1$

Given

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

Explanation

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

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

Comparing both sides we get  $x=1,y=0,z=1$