#### Please Solve R. D. Sharma class 12 Chapter 5 determinants Exercise  5.4 Question 18  Maths textbook Solution.

Answer: $x=-2, y=3 \text { and } z=-4$

Hint: Use Cramer’s rule to solve a system of linear equations

Given:

\begin{aligned} &x+y=1 \\ &x+z=-6 \\ &x-y-2 z=3 \end{aligned}

Solution:

First take coefficient of variables x, y and z.

$|\mathbf{A}|=\left|\begin{array}{ccc} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & -1 & -2 \end{array}\right|$                                                                $\because$(Taking first row for solving determinant)

\begin{aligned} &=1(0+1)-1(-2-1)+0(-1-0) \\ &=1+3 \\ &=4 \end{aligned}

Now for x, the x column is replaced with constant column i.e.

\begin{aligned} \mathrm{D}_{\mathrm{x}} &=\left|\begin{array}{ccc} 1 & 1 & 0 \\ -6 & 0 & 1 \\ 3 & -1 & -2 \end{array}\right| \\ &=1(0+1)-1(12-3)+0(6-0) \\ &=1-9 \\ &=-8 \end{aligned}

If we are solving for y, the y column is replaced with constant column i.e.

\begin{aligned} \mathrm{D}_{\mathrm{y}} &=\left|\begin{array}{ccc} 1 & 1 & 0 \\ 1 & -6 & 1 \\ 1 & 3 & -2 \end{array}\right| \\ &=1(12-3)-1(-2-1)+0(3+6) \\ &=9+3 \\ &=12 \end{aligned}

If we are solving for z, the z column is replaced with constant column i.e.

\begin{aligned} \mathrm{D}_{z} &=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 0 & -6 \\ 1 & -1 & 3 \end{array}\right| \\ &=1(0-6)-1(3+6)+1(-1+0) \\ &=-6-9-1 \\ &=-16 \end{aligned}

By Cramer’s rule,

\begin{aligned} &\Rightarrow x=\frac{D_{x}}{D}=\frac{-8}{4}=-2 \\ &\Rightarrow y=\frac{D_{y}}{D}=\frac{12}{4}=3 \\ &\Rightarrow z=\frac{D_{z}}{D}=\frac{-16}{4}=-4 \end{aligned}

Concept: Determinant solving of 3 x 3 matrix (Cramer’s rule)

Note: Cramer’s rule will give us unique solution to a system of equations, if it exists. However, if the system has no solution or an infinitive number of solutions that is determinant is zero.