# 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) 1

B) 2

C) 4

D) 3

Last updated date: 13th Mar 2023

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Answer

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303.3k+ views

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

Complete step-by-step answer:

A matrix is singular only if its determinant is zero.

∴ \[\left| {\begin{array}{*{20}{c}}

3&{ - x - 1}&2 \\

3&{ - 1}&{x + 2} \\

{x + 3}&{ - 1}&2

\end{array}} \right| = 0\]

\[ \Rightarrow 3( - 2 + x + 2) + (x + 1)(6 - (x + 2)(x + 3)) + 2( - 3 + x + 3) = 0\]

\[ \Rightarrow 3x + (x + 1)( - {x^2} - 5x) + 2x = 0\]

\[ \Rightarrow 5x - {x^3} -5{x^2} - {x^2} - 5x = 0\]

\[ \Rightarrow {x^3} +6{x^2} =0\]

\[ \Rightarrow x = 0, 0, - 6\]

\[0 \in \left[ { - 4, - 1} \right]\]

∴ The correct option is ‘a’.

Note: A matrix is said to be singular if and only if the value of its determinant is zero. If the value of determinant is not zero, the matrix is said to be non singular. In the above question -6 is neglected because it does not belong to the range [-4, -1].

Complete step-by-step answer:

A matrix is singular only if its determinant is zero.

∴ \[\left| {\begin{array}{*{20}{c}}

3&{ - x - 1}&2 \\

3&{ - 1}&{x + 2} \\

{x + 3}&{ - 1}&2

\end{array}} \right| = 0\]

\[ \Rightarrow 3( - 2 + x + 2) + (x + 1)(6 - (x + 2)(x + 3)) + 2( - 3 + x + 3) = 0\]

\[ \Rightarrow 3x + (x + 1)( - {x^2} - 5x) + 2x = 0\]

\[ \Rightarrow 5x - {x^3} -5{x^2} - {x^2} - 5x = 0\]

\[ \Rightarrow {x^3} +6{x^2} =0\]

\[ \Rightarrow x = 0, 0, - 6\]

\[0 \in \left[ { - 4, - 1} \right]\]

∴ The correct option is ‘a’.

Note: A matrix is said to be singular if and only if the value of its determinant is zero. If the value of determinant is not zero, the matrix is said to be non singular. In the above question -6 is neglected because it does not belong to the range [-4, -1].

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