#### Explain Solution R.D.Sharma Class 12 Chapter 5 deteminants Exercise 5.4 Question 13 maths Textbook Solution.

Answer: $\mathrm{x}=1, \mathrm{y}=2 \text { and } \mathrm{z}=1$

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

Given:

\begin{aligned} &6 x+y-3 z=5 \\ &x+3 y-2 z=5 \\ &2 x+y+4 z=8 \end{aligned}

Solution:

First take coefficient of variables x, y and z.

$|\mathrm{A}|=\left|\begin{array}{ccc} 6 & 1 & -3 \\ 1 & 3 & -2 \\ 2 & 1 & 4 \end{array}\right|$                                                            $\because$(Taking first row for solving determinant)

\begin{aligned} &=6(12+2)-1(4+4)-3(1-6) \\ &=6(14)-1(8)-3(-5) \\ &=84-8+15 \\ &=91 \end{aligned}

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

$\mathrm{D}_{\bar{x}}=\left|\begin{array}{ccc} 5 & 1 & -3 \\ 5 & 3 & -2 \\ 8 & 1 & 4 \end{array}\right|$

\begin{aligned} &=5(12+2)-1(20+16)-3(5-24) \\ &=5(14)-1(36)-3(-19) \\ &=70-36+57 \\ &=91 \end{aligned}

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

$\mathrm{D}_{\mathrm{x}}=\left|\begin{array}{ccc} 6 & 5 & -3 \\ 1 & 5 & -2 \\ 2 & 8 & 4 \end{array}\right|$

\begin{aligned} &=6(20+16)-5(4+4)-3(8+10) \\ &=6(36)+3(-2)-5(8) \\ &=216-6-40 \\ &=182 \end{aligned}

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

\begin{aligned} &D_{z}=\left|\begin{array}{lll} 6 & 1 & 5 \\ 1 & 3 & 5 \\ 2 & 1 & 8 \end{array}\right| \\ &=6(24-5)-1(8-10)+5(1-6) \\ &=6(19)-1(-2)+5(-5) \\ &=114+2-25 \\ &=91 \end{aligned}

By Cramer’s rule,

\begin{aligned} &\Rightarrow x=\frac{D_{x}}{D}=\frac{91}{91}=1 \\ &\Rightarrow y=\frac{D_{y}}{D}=\frac{182}{91}=2 \\ &\Rightarrow z=\frac{D_{z}}{D}=\frac{91}{91}=1 \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.