Question

For what values of x, the matrix A is singular?
$A=\left[ \begin{align}
  & 3-x\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2 \\
 & 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,4-x\,\,\,\,\,\,\,\,\,\,\,\,1 \\
 & -2\,\,\,\,\,\,\,\,\,\,\,\,-4\,\,\,\,\,\,\,\,\,\,\,-1-x \\
\end{align} \right]\,$
(a) $x=0,2$
(b) $x=1,2$
(c) $x=2,3$
(d) $x=0,3$

Answer 1

Hint: In this question, we will use the condition of the square matrix to be singular, i.e. determinant of the matrix is zero, and solve that equation of condition to find value for x.

Complete step-by-step answer:
In the given question, we are asked to find the values of x such that, for that values A will be singular matrix.
Now, a square matrix is singular matrix when its, inverse matrix does not exist. And, inverse matrix of any square matrix does not exist only when the value of determinant of that square matrix is zero.
Therefore, for given square matrix A to be singular, the value of its determinant must be zero.
Let us first find the determinant of given matrix A. determinant of matrix A is given as,
$|A|=\left[ \begin{align}
  & 3-x\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2 \\
 & 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,4-x\,\,\,\,\,\,\,\,\,\,\,\,1 \\
 & -2\,\,\,\,\,\,\,\,\,\,\,\,-4\,\,\,\,\,\,\,\,\,\,\,-1-x \\
\end{align} \right]\,$
Writing determinant along first row, we get:
$\left| A \right|=\left( 3-x \right)\left| \begin{matrix}
   4-x & 1 \\
   -4 & -1-x \\
\end{matrix} \right|-2\left| \begin{matrix}
   2 & 1 \\
   -2 & -1-x \\
\end{matrix} \right|+2\left| \begin{matrix}
   2 & 4-x \\
   -2 & -4 \\
\end{matrix} \right|$
Writing values of all $2\times 2$ determinants we get,
$\begin{align}
  & |A|=\left( 3-x \right)\left( \left( 4-x \right)\left( -1-x \right)-\left( -4 \right)\left( 1 \right) \right) \\
 & \,\,\,\,\,\,\,\,\,\,\,\,-2\left( \left( 2 \right)\left( -1-x \right)-\left( -2 \right)\left( 1 \right) \right) \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+2\left( \left( 2 \right)\left( -4 \right)-\left( -2 \right)\left( 4-x \right) \right) \\
\end{align}$
\[\Rightarrow |A|=\left( 3-x \right)\left( -4-4x+x+{{x}^{2}}+4 \right)-2\left( -2-2x+2 \right)+2\left( -8+8-2x \right)\]
\[\Rightarrow |A|=\left( 3-x \right)\left( {{x}^{2}}-3x \right)\left( -2\left( -2x \right)+2\left( -2x \right) \right)\]
\[\Rightarrow |A|=\left( 3-x \right)\left( {{x}^{2}}-3x \right)+4x-4x\]
\[\Rightarrow |A|=\left( 3-x \right)\left( {{x}^{2}}-3x \right)\]
\[\Rightarrow |A|=3{{x}^{2}}-9x-{{x}^{3}}+3{{x}^{2}}\]
\[\Rightarrow |A|=-{{x}^{3}}+6{{x}^{2}}-9x\]
Taking -x common, we get,
\[|A|=-x\left( {{x}^{2}}-6x+9 \right)\]
Splitting the middle term $-6x\,\,to\,-3x-3x$, to factorize the expression, we get,
\[|A|=-x\left( {{x}^{2}}-3x-3x+9 \right)\]
Taking common terms out, we get,
$\begin{align}
  & |A|=-x\left( x\left( x-3 \right)-3\left( x-3 \right) \right) \\
 & \,\,\,\,\,\,\,\,=-x\left( x-3 \right)\left( x-3 \right) \\
 & \,\,\,\,\,\,\,\,\,\,\,=-x{{\left( x-3 \right)}^{2}} \\
\end{align}$
Now, for A to be singular, condition is,
$|A|=0$
Putting value of |A|, we get,
$-x\left( x-3 \right)=0$
Therefore, either $-x=0\,\,or\,\,\,x-3=0$ .
 $\Rightarrow x=0\,\,\,\,or\,\,x=3$
Hence, for x = 0, 3, A is singular.
Therefore, the correct answer is option (d).


Note: In this type of quotations, while finding determinant, it can be opened along any row or relevant column, the value will remain the same.