Question

The number of values of $x$ in $\left[- 4 , - 1\right]$ , for which the matrix $\begin{bmatrix} 3 & -1+x & 2 \\ 3 & -1 & x+2 \\ x+3 & -1 & 2 \end{bmatrix}$ is singular, is

• A. $1$
• B. $3$
• C. $4$
• D. $5$

Answer 1

Answer: (A)

For the matrix to be singular, its determinant value should be zero
$\begin{vmatrix} 3 & -1+x & 2 \\ 3 & -1 & x+2 \\ x+3 & -1 & 2 \end{vmatrix}=0$
$\Rightarrow 3\left(- 2 + x + 2\right)-\left(x - 1\right)\left[6 - \left(x + 2\right) \left(x + 3\right)\right]+2\left(- 3 + x + 3\right)=0$
$\Rightarrow 3x+\left(x - 1\right)x\left(x + 5\right)+2x=0$
$\Rightarrow x\left[5 + x^{2} + 4 x - 5\right]=0\Rightarrow x=0,-4$
But, $x=0$ is rejected
Hence, only one value in $\left[- 4 , - 1\right]$