#### Explain Solution R.D. Sharma Class 12 Chapter 5 deteminants Exercise 5.4 Question 23 maths Textbook Solution.

Answer:$x=\infty \text { and } y=\infty$

Hint: Solving determinant gives zero.

Given:

\begin{aligned} &3 x+y=5 \\ &-6 x-2 y=9 \end{aligned}

Solution:

\begin{aligned} &(3 x+y=5) \times 2\\ &6 x-2 y=10 \end{aligned}                                                                                                                            ....(1)

$\begin{gathered} 6 x+2 y=10 \\ -6 x-2 y=9 \\ \hline 0 \end{gathered}$

Hence, linear equations are inconsistent.

By Cramer’s rule:

Solving determinant,

\begin{aligned} &|\mathrm{A}|=\left|\begin{array}{cc} 3 & 1 \\ -6 & -2 \end{array}\right|=-6+6=0 \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) \\ &\mathrm{D}=0 \end{aligned}

\begin{aligned} &\mathrm{D}_{\mathrm{x}}=\left|\begin{array}{cc} 5 & 1 \\ 9 & -2 \end{array}\right|=-10-9=-19 \neq 0\\ &\mathrm{D}_{\mathrm{y}}=\left|\begin{array}{cc} 3 & 5 \\ -6 & 9 \end{array}\right|=27+30=57 \neq 0\\ &\text { By Cramer's rule, }\\ &\Rightarrow x=\frac{D_{x}}{D}=\frac{-19}{0}=\infty\\ &\Rightarrow y=\frac{D_{y}}{D}=\frac{57}{0}=\infty \end{aligned}

Since, $D=0$ and $D_{x}$ and $D_{y}\neq 0$

$\therefore$ Linear equations are inconsistent.

Concept: Solving matrix of order 2x2 by solving linear equations

Note: When D = 0, there is either no solution or infinite solutions.