For how many value(s) of x in the closed inte | vedclass

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(D) For the matrix to be singular,its determinant must be zero: $\begin{vmatrix} 3 & -1+x & 2 \ 3 & -1 & x+2 \ x+3 & -1 & 2 \end{vmatrix} = 0$. Appl

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(D) For the matrix to be singular,its determinant must be zero: $\begin{vmatrix} 3 & -1+x & 2 \ 3 & -1 & x+2 \ x+3 & -1 & 2 \end{vmatrix} = 0$. Applying row operations $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_1$: $\begin{vmatrix} 3 & -1+x & 2 \ 0 & -x & x \ x & -x & 0 \end{vmatrix} = 0$. Applying column operation $C_1 	o C_1 + C_2 + C_3$: $\begin{vmatrix} x+4 & -1+x & 2 \ 0 & -x & x \ 0 & -x & 0 \end{vmatrix} = 0$. Expanding along the first column: $(x+4) \cdot [(-x)(0) - (x)(-x)] = 0$ $(x+4)(x^2) = 0$. This gives $x = -4$ or $x = 0$. We are given the interval $[-4, -1]$. Since $-4 \in [-4, -1]$ and $0 
otin [-4, -1]$,there is only $1$ value of $x$ that satisfies the condition.

For how many value$(s)$ of $x$ in the closed interval $[-4, -1]$ is the matrix $\begin{bmatrix} 3 & -1+x & 2 \\ 3 & -1 & x+2 \\ x+3 & -1 & 2 \end{bmatrix}$ singular?

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