#### Explain Solution R.D.Sharma Class 12 Chapter 5 deteminants Exercise 5.4 Question 15 maths Textbook Solution.

Answer:$\mathrm{x}=5, \mathrm{y}=-3 \text { and } \mathrm{z}=-2$

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

Given:

\begin{aligned} &2 y-3 z=0 \\ &x+3 y=-4 \\ &3 x+4 y=3 \end{aligned}

Solution:

First take coefficient of variables x, y and z.

\begin{aligned} |\mathrm{A}| &=\left|\begin{array}{ccc} 0 & 2 & -3 \\ 1 & 3 & 0 \\ 3 & 4 & 0 \end{array}\right| \\ &=0(0)-2(0-0)-3(4-9) \\ &=-3(-5) \\ &=15 \end{aligned}                                                                                $\because$(Taking first row for solving determinant)

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

\begin{aligned} \mathrm{D}_{\mathrm{x}} &=\left|\begin{array}{ccc} 0 & 2 & -3 \\ -4 & 3 & 0 \\ 3 & 4 & 0 \end{array}\right| \\ &=0(0)-2(0)-3(-16-9) \\ &=-3(-25) \\ &=75 \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} 0 & 0 & -3 \\ 1 & -4 & 0 \\ 3 & 3 & 0 \end{array}\right| \\ &=0(0)-0(0)-3(3+12) \\ &=-3(15) \\ &=-45 \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} 0 & 2 & 0 \\ 1 & 3 & -4 \\ 3 & 4 & 3 \end{array}\right| \\ &=0(9+16)-2(3+12)+0(4-9) \\ &=0-2(15)+0 \\ &=-30 \end{aligned}

By Cramer’s rule,

\begin{aligned} &\Rightarrow x=\frac{D_{x}}{D}=\frac{75}{15}=5 \\ &\Rightarrow y=\frac{D_{y}}{D}=\frac{-45}{15}=-3 \\ &\Rightarrow z=\frac{D_{z}}{D}=\frac{-30}{15}=-2 \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.