#### Explain solution RD Sharma class 12 chapter 7 Solution of Simultaneous Linear Equation Exercise Fill in the blank Question 4 Maths Textbook Solution.

Answer$\rightarrow \lambda=-\frac{5}{3}$

Given $\rightarrow$ Given that the system of equations $2 x-y-z=12, x-2 y+z=-4, x+y+\lambda z=4$ has no solution.

To find $\rightarrow$ We have to find out the value of$\lambda$

Hint $\rightarrow$ If system has no solution then $(\operatorname{adj} A) B \neq 0 \text { and }|A|=0$

Solution $\rightarrow$  We have

\begin{aligned} &2 x-y-z=12 \\ &x-2 y+z=-4 \\ &x+y+\lambda z=4 \end{aligned}

$A=\left|\begin{array}{ccc} 2 & -1 & -1 \\ 1 & -2 & 1 \\ 1 & 1 & \lambda \end{array}\right|$

We know that the system of given equation has no solution when $\left | A \right |=0$

$\Rightarrow\left|\begin{array}{ccc} 2 & -1 & -1 \\ 1 & -2 & 1 \\ 1 & 1 & \lambda \end{array}\right|=0$

\begin{aligned} &\Rightarrow 2(-2 \lambda-1)+1(\lambda-1)-1(1+2)=0 \\\\ &\Rightarrow-4 \lambda-2+\lambda-1-2=0 \\ &\Rightarrow-3 \lambda=5 \\ &\Rightarrow \lambda=-\frac{5}{3} \end{aligned}