Explain Solution R.D.Sharma Class 12 Chapter 5 deteminants Exercise 5.4 Question 14 maths Textbook Solution.

Answer:$\mathrm{x}=3, \mathrm{y}=2 \text { and } \mathrm{z}=1$

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

Given:

\begin{aligned} &x+y=5 \\ &y+z=3 \\ &x+z=4 \end{aligned}

Solution:

First take coefficient of variables x, y and z.

\begin{aligned} |\mathrm{A}| &=\left|\begin{array}{lll} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{array}\right| \\ &=1(1)-1(-1)+0(-1) \\ &=1+1 \end{aligned}                                                                                $\because$(Taking first row for solving determinant)

$= 2$

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

\begin{aligned} &\mathrm{D}_{\overline{\mathrm{y}}}=\left|\begin{array}{lll} 5 & 1 & 0 \\ 3 & 1 & 1 \\ 4 & 0 & 1 \end{array}\right| \\ &=5(1)-1(3-4)+0(0-4) \\ &=5+1 \\ &=6 \end{aligned}

\begin{aligned} \mathrm{D}_{\mathrm{y}} &=\left|\begin{array}{lll} 1 & 5 & 0 \\ 0 & 3 & 1 \\ 1 & 4 & 1 \end{array}\right| \\ &=1(3-4)-5(0-1)+0(0-3) \\ &=-1+5 \\ &=4 \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}{lll} 1 & 1 & 5 \\ 0 & 1 & 3 \\ 1 & 0 & 4 \end{array}\right| \\ &=1(4-0)-1(0-3)+5(0-1) \\ &=4+3-5 \\ &=2 \end{aligned}

By Cramer’s rule,

\begin{aligned} &\Rightarrow x=\frac{D_{x}}{D}=\frac{6}{2}=3 \\ &\Rightarrow y=\frac{D_{y}}{D}=\frac{4}{2}=2 \\ &\Rightarrow z=\frac{D_{z}}{D}=\frac{2}{2}=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.