#### Explain Solution R.D. Sharma Class 12 Chapter 5 deteminants Exercise 5.4 Question 17 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} &2 x-3 y-4 z=29 \\ &-2 x+5 y-z=-15 \\ &3 x-y+5 z=-11 \end{aligned}

Solution:

First take coefficient of variables x, y and z.

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

\begin{aligned} &=2(25-1)+3(-10+3)-4(2-15) \\ &=2(24)+3(-7)-4(-13) \\ &=48-21+52 \\ &=79 \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} 29 & -3 & -4 \\ -15 & 5 & -1 \\ -11 & -1 & 5 \end{array}\right| \\ &=29(25-1)+3(-75-11)-4(15+55) \\ &=29(24)+3(-86)-4(70) \\ &=696-258-280 \\ &=158 \end{aligned}

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

\begin{aligned} \mathrm{D}_{\mathrm{x}} &=\left|\begin{array}{ccc} 2 & 29 & -4 \\ -2 & -15 & -1 \\ 3 & -11 & 5 \end{array}\right| \\ &=2(-75-11)-29(-10+3)-4(22+45) \\ &=2(-86)-29(-7)-4(67) \\ &=-172+203-268 \\ &=-237 \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} 2 & -3 & 29 \\ -2 & 5 & -15 \\ 3 & -1 & -11 \end{array}\right| \\ &=2(-55-15)+3(22+45)+29(2-15) \\ &=2(-70)+3(67)+29(-13) \\ &=-140+201-377 \\ &=-316 \end{aligned}

By Cramer’s rule,

\begin{aligned} &\Rightarrow x=\frac{D_{x}}{D}=\frac{158}{79}=2 \\ &\Rightarrow y=\frac{D_{y}}{D}=\frac{-237}{79}=-3 \\ &\Rightarrow z=\frac{D_{z}}{D}=\frac{-316}{79}=-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.