#### Please solve RD Sharma class 12 chapter 7 Solution of Simultaneous Linear Equation Exercise Fill in the blank Question 6 Maths Textbook Solution.

Answer $\rightarrow$

Given  $\rightarrow$ The given system of equations $\lambda x+y+z=0,-x+\lambda y+z=0 \text { and }-x-y+\lambda z=0$has a non-zero solution.

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

Hint $\rightarrow$ The system has a non-zero solution if $|A|=0$

Solution $\rightarrow$  Here system of equations are

\begin{aligned} &\lambda x+y+z=0 \\ &-x+\lambda y+z=0 \\ &-x-y+\lambda z=0 \end{aligned}

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

We know, if system of equation has a non-zero solution then $\left | A \right |=0$

$\text { Now, }|A|=\left|\begin{array}{ccc} \lambda & 1 & 1 \\ -1 & \lambda & 1 \\ -1 & -1 & \lambda \end{array}\right|=0$

\begin{aligned} &\Rightarrow \lambda\left(\lambda^{2}+1\right)-1(-\lambda+1)+1(1+\lambda)=0 \\ &\Rightarrow \lambda^{3}+\lambda+\lambda-1+1+\lambda=0 \\ &\Rightarrow \lambda^{3}+3 \lambda=0 \\ &\Rightarrow \lambda\left(\lambda^{2}+3\right)=0 \\ &\Rightarrow \lambda^{2}+3=0 \text { or } \quad \lambda=0 \end{aligned}

In this case $\lambda$ is an imaginary number which is non-existent.

$\Rightarrow \lambda=0$

Hence, $\lambda=0$ is required answer.