#### Explain Solution R.D. Sharma Class 12 Chapter 5 deteminants Exercise 5.4 Question 8 maths Textbook Solution.

Answer: $x=\frac{9}{2} \text { and } y=\frac{-7}{2}$

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

Given:

\begin{aligned} &5 x+7 y=-2 \\ &4 x+6 y=-3 \end{aligned}

Solution:

First D: determinant of the coefficient matrix

$\mathrm{D}=\left|\begin{array}{ll} 5 & 7 \\ 4 & 6 \end{array}\right| \quad \because\left|\begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array}\right|=\left(a_{1} b_{2}-a_{2} b_{1}\right)$

\begin{aligned} &=(5)(6)-(7)(4) \\ &=30-28 \\ &=2 \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}{rr} -2 & 7 \\ -3 & 6 \end{array}\right| \\ &=-12+21 \\ &=9 \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} 5 & -2 \\ 4 & -3 \end{array}\right| \\ &\quad=-15+8 \\ &=-7 \\ &\text { Now, } \mathrm{x}=\frac{D_{1}}{D}=\frac{9}{2} \\ &\mathrm{y}=\frac{D_{2}}{D}=\frac{-7}{2} \end{aligned}

Hence,$x=\frac{9}{2} \text { and } y=\frac{-7}{2}$

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.