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

Suppose that given matrix is A.,

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

For singularity,

$$\begin{gathered} \Rightarrow \left|A\right|=0 \\ \Rightarrow \left|\begin{array}{ccc}3& -1+x& 2\\ 3& -1& x+2\\ x+3& -1& 2\end{array}\right|=0 \\ \Rightarrow 3\left\{-\cancel{2}+\left(x+\cancel{2}\right)\right\}-3\left\{-\cancel{2}+2X+\cancel{2}\right\}+\left(X+3\right)\left\{\left(1+X\right)\left(X+2\right)+2\right\}=0 \\ \Rightarrow -3\cancel{X}+X^3+X^2+3X^2+3\cancel{X}=0 \\ \Rightarrow x^3+x^2+3x^2=0 \\ roots are x^2\left(x+4\right)=0 \\ \Rightarrow x=0,-4 \end{gathered}$$

Since $0\in [-4,-1]$

$-4\in [-4,-1]$

Here, x=-4 is only vailed value.

Hence, the number of values of x in the given closed interval is only 1.