#### Provide Solution For  R.D. Sharma Maths Class 12 Chapter 5 determinants Exercise 5.4 Question 2 Maths Textbook Solution.

Answer:$x=-3 \text { and } y=-7$

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

Given:\begin{aligned} &2 x-y=1 \\ &7 x-2 y=-7 \end{aligned}

Solution: First D: determinant of the coefficient matrix

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

\begin{aligned} &=(2)(-2)-(-1)(7) \\ &=-4+7 \\ &=-3 \end{aligned}

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

$\mathrm{D}_{1}=\left|\begin{array}{cc} 1 & -1 \\ -7 & -2 \end{array}\right|$

\begin{aligned} &=(1)(-2)-(-7)(-1) \\ &=-2-7 \\ &=-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}{cc} 2 & 1 \\ 7 & -7 \end{array}\right| \\ &=(2)(-7)-(7)(1) \\ &=-14-7 \\ &=-21 \end{aligned}

Now, $\mathrm{x}=\frac{D_{1}}{D}=\frac{-9}{3}=-3$

$\mathrm{y}=\frac{D_{2}}{D}=\frac{-21}{3}=-7$

$Hence, x = -3 and y= -7.$

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.