Question

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

• A. 2
• B. 0
• C. 3
• D. 1

Answer 1

Answer: (D)

1

Answer 2

$\left| \begin{array} { c c c } 3 & x - 1 & 2 \\ 3 & - 1 & x + 2 \\ x + 3 & - 1 & 2 \end{array} \right| = 0$ $\left| \begin{matrix} 0 & x & - x \\ 3 & - 1 & x + 2 \\ x + 3 & - 1 & 2 \end{matrix} \right| = 0$ $\lbrack R_{1} \rightarrow R_{1} - R_{2}\rbrack$, $\left| \begin{matrix} 0 & x & - x \

• x & 0 & x \ x + 3 & - 1 & 2 \end{matrix} \right| = 0$$\lbrack R_{2} \rightarrow R_{2} - R_{3}\rbrack$

$\left| \begin{matrix} 0 & x & - x \\ 3 & - 1 & x + 2 \\ x + 3 & - 1 & 2 \end{matrix} \right| = 0$ $\lbrack C_{2} \rightarrow C_{2} + C_{3}\rbrack$

$- x\lbrack( - x) - x(x + 3)\rbrack = 0$⇒$x(x^{2} + 4x) = 0$ ⇒ $x = 0, - 4$

Hence only one value of x in closed interval [–4,–1] i.e. $x = - 4$