#### Please Solve R. D. Sharma class 12 Chapter 5 determinants Exercise  5.4 Question 4 Maths textbook Solution.

Answer: $x=7 \text { and } y=-2$

Hint: Use Cramer’s rule to solve a system of two equations in two variables.

Given: \begin{aligned} &3 \mathrm{x}+\mathrm{y}=19 \\ &3 \mathrm{x}-\mathrm{y}=23 \end{aligned}

Solution:

First D: determinant of the coefficient matrix

$\mathrm{D}=\left|\begin{array}{cc} 3 & 1 \\ 3 & -1 \end{array}\right| \quad \because\left|\begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array}\right|=\left(a_{1} b_{2}-a_{2} b_{1}\right)$

\begin{aligned} &=(-1)(3)-(3)(1) \\ &=-3-3 \\ &=-6 \end{aligned}

Now, $D\neq 0$ . If we are solving for x, the x column is replaced with constant column i.e.

$\mathrm{D}_{1}=\left|\begin{array}{cc} 19 & 1 \\ 23 & -1 \end{array}\right|$

\begin{aligned} &=(19)(-1)-(23)(1) \\ &=-19-23 \\ &=-42 \end{aligned}

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

\begin{aligned} \mathrm{D}_{2} &=\left|\begin{array}{ll} 3 & 19 \\ 3 & 23 \end{array}\right| \\ &=(3)(23)-(19)(3) \\ &=69-57 \\ &=12 \end{aligned}

\begin{aligned} &\text { Now, } \mathrm{x}=\frac{D_{1}}{D}=\frac{-42}{-6}=7 \\ &\mathrm{y}=\frac{D_{2}}{D}=\frac{12}{-6}=-2 \end{aligned}

Hence $x=7 \text { and } y=-2$

Concept: Cramer’s rule for system of two equations.

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.