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

Answer $\rightarrow$ $k \neq 0, k=\pm 1, \pm 2, \pm 3, \pm 4, \ldots \ldots . \pm n$

Given $\rightarrow$ Given that the system of equations $x+y+z=2,2 x+y-z=3 \text { and } 3 x+2 y+k z=4$ has a unique solution.

To find $\rightarrow$ We have to find the real value of $k$

Hint $\rightarrow$ If the system of equations has a unique solution $\Rightarrow|A| \neq 0$

Solution $\rightarrow$ We have system of equation,

\begin{aligned} &x+y+z=2 \\ &2 x+y-z=3 \\ &3 x+2 y+k z=4 \end{aligned}

Then,

$\Rightarrow A=\left[\begin{array}{ccc} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 3 & 2 & k \end{array}\right]$

We know that the system of equations has a unique solution.

$\Rightarrow \left | A \right |$ should not be zero

$\text { i.e. }|A| \neq 0$

$\Rightarrow\left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 3 & 2 & k \end{array}\right| \neq 0$

\begin{aligned} &\Rightarrow 1(k+2)-1(2 k+3)+1(4-3) \neq 0 \\ &\Rightarrow k+2-2 k-3+1 \neq 0 \\ &\Rightarrow-k \neq 0 \\ &\Rightarrow k=\pm 1, \pm 2, \ldots \ldots \pm n \end{aligned}