#### Need Solution for R.D.Sharma Maths Class 12 Chapter 5 determinants  Exercise 5.4 Question 11 Maths Textbook Solution.

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

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

Given:

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

$4 x+y-3 z=-11$

Solution:

First take coefficient of variables x, y and z.

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

\begin{aligned} &=3(12-3)-1(-6-12)+1(2+16) \\ &=3(9)-1(-18)+(1)(18) \\ &=27+18+18 \\ &=63 \end{aligned}

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

$\mathrm{D}_{\mathbf{x}}=\left|\begin{array}{ccc} 2 & 1 & 1 \\ -1 & -4 & 3 \\ -11 & 1 & -3 \end{array}\right|$

\begin{aligned} &=2(12-3)-1(3+33)+1(-1-44) \\ &=2(9)-1(36)+1(-45) \\ &=18-36-45 \\ &=-63 \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} 3 & 2 & 1 \\ 2 & -1 & 3 \\ 4 & -11 & -3 \end{array}\right| \\ &=3(3+33)-2(-6-12)+1(-22+4) \\ &=3(36)-2(-18)+1(-18) \\ &=108+36-18 \\ &=126 \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} 3 & 1 & 2 \\ 2 & -4 & -1 \\ 4 & 1 & -11 \end{array}\right| \\ &=3(44+1)-1(-22+4)+2(2+16) \\ &=3(45)-1(-18)+2(18) \\ &=135+18+36 \\ &=189 \end{aligned}

By Cramer’s rule,

\begin{aligned} &\Rightarrow x=\frac{D_{x}}{D}=\frac{-63}{63}=-1 \\ &\Rightarrow y=\frac{D_{y}}{D}=\frac{126}{63}=2 \\ &\Rightarrow z=\frac{D_{z}}{D}=\frac{189}{63}=3 \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.