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

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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}

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