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#### Need Solution for R.D.Sharma Maths Class 12 Chapter 5 determinants  Exercise 5.4 Question 12 Maths Textbook Solution.

Answer:$x=-1, y=-5 \text { and } z=8$

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

Given:

\begin{aligned} &x-4 y-z=11 \\ &2 x-5 y+2 z=39 \\ &-3 x+2 y+z=1 \end{aligned}

Solution:

First take coefficient of variables x, y and z.

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

\begin{aligned} &=1(-5-4)+4(2+6)-1(4-15) \\ &=1(-9)+4(8)-1(-11) \\ &=-9+32+11 \\ &=34 \end{aligned}

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

\begin{aligned} D_{x} &=\left|\begin{array}{ccc} 11 & -4 & -1 \\ 39 & -5 & 2 \\ 1 & 2 & 1 \end{array}\right| \\ &=11(-5-4)+4(39-2)-1(78+5) \\ &=11(-9)+4(37)-1(83) \\ &=-99+148-83 \\ &=34 \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} 1 & 11 & -1 \\ 2 & 39 & 2 \\ -3 & 1 & 1 \end{array}\right| \\ &=1(39-2)-11(2+6)-1(2+117) \\ &=1(37)-11(8)-1(119) \\ &=37-88-119 \\ &=-170 \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 & -4 & 11 \\ 2 & -5 & 39 \\ -3 & 2 & 1 \end{array}\right| \\ &=1(-5-78)+4(2+117)+11(4-15) \end{aligned}

\begin{aligned} &=-83+4(119)+11(-11) \\ &=272 \end{aligned}

By Cramer’s rule,

\begin{aligned} &\Rightarrow x=\frac{D_{x}}{D}=\frac{-34}{34}=-1 \\ &\Rightarrow y=\frac{D_{y}}{D}=\frac{-170}{34}=-5 \\ &\Rightarrow z=\frac{D_{z}}{D}=\frac{272}{34}=8 \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.