Question

For how many values of $x$ in the closed interval [-4, -4] the matrix $\left[ {\begin{array}{*{20}{c}} 3&{ - x - 1}&2 \\ 3&{ - 1}&{x + 2} \\ {x + 3}&{ - 1}&2 \end{array}} \right]$ is singular.A) 1B) 2C) 4D) 3

Hint: Convert the matrix into determinant. Apply the condition of singularity. Solve the determinant, make the algebraic equation and then proceed.

∴ $\left| {\begin{array}{*{20}{c}} 3&{ - x - 1}&2 \\ 3&{ - 1}&{x + 2} \\ {x + 3}&{ - 1}&2 \end{array}} \right| = 0$
$\Rightarrow 3( - 2 + x + 2) + (x + 1)(6 - (x + 2)(x + 3)) + 2( - 3 + x + 3) = 0$
$\Rightarrow 3x + (x + 1)( - {x^2} - 5x) + 2x = 0$
$\Rightarrow 5x - {x^3} -5{x^2} - {x^2} - 5x = 0$
$\Rightarrow {x^3} +6{x^2} =0$
$\Rightarrow x = 0, 0, - 6$
$0 \in \left[ { - 4, - 1} \right]$