#### Need Solution for R.D.Sharma Maths Class 12 Chapter 5 determinants  Exercise 5.4 Question 10 Maths Textbook Solution.

Answer:$x=\frac{7}{5} \text { and } y=\frac{-1}{5}$

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

Given:

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

Solution:

First D: determinant of the coefficient matrix

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

\begin{aligned} &=(1)(1)-(3)(2) \\ &=1-6 \\ &=-5 \end{aligned}

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

\begin{aligned} \mathrm{D}_{1} &=\left|\begin{array}{ll} 1 & 2 \\ 4 & 1 \end{array}\right| \\ &=1-8 \\ &=-7 \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} 1 & 1 \\ 3 & 4 \end{array}\right| \\ &=4-3 \\ &=1 \end{aligned}

Now,\begin{aligned} &\mathrm{x}=\frac{D_{1}}{D}=\frac{-7}{-5}=\frac{7}{5} \\ &\mathrm{y}=\frac{D_{2}}{D}=\frac{1}{-5}=\frac{-1}{5} \end{aligned}

Hence,$x=\frac{7}{5} \text { and } y=\frac{-1}{5}$

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.