Find the values of $x$ for which the given matrix \[\left[ {\begin{array}{*{20}{c}}
{ - x}&x&2 \\
2&x&{ - x} \\
x&{ - 2}&{ - x}
\end{array}} \right]\] will be non-singular.
A). \[ - 2 \leqslant x \leqslant 2\]
B). For all$x$other than $2$and $ - 2$
C). \[x \geqslant 2\]
D). $x \leqslant - 2$
Answer
547.2k+ views
Hint: We need to find that for which values of $x$ the given matrix \[\left[ {\begin{array}{*{20}{c}}
{ - x}&x&2 \\
2&x&{ - x} \\
x&{ - 2}&{ - x}
\end{array}} \right]\] will be non-singular. Non-singular matrix is a matrix whose determinant is non-zero. So we need to find the values of $x$ for which determinant of the above matrix will be a non-zero value.
Complete step-by-step solution:
We have to check that for which values of $x$ the given matrix \[\left[ {\begin{array}{*{20}{c}}
{ - x}&x&2 \\
2&x&{ - x} \\
x&{ - 2}&{ - x}
\end{array}} \right]\] will be non-singular. Non-singular matrix is a matrix whose determinant is non-zero. So we need to check the values of $x$ for which determinant of the above matrix will be a non-zero value.
Let us calculate the determinant of the above matrix,
\[\left| {\begin{array}{*{20}{c}}
{ - x}&x&2 \\
2&x&{ - x} \\
x&{ - 2}&{ - x}
\end{array}} \right|\]
On applying the formula for the determinant we get,
$ = - x( - {x^2} - 2x) - x( - 2x + {x^2}) + 2( - 4 - {x^2})$
On performing the multiplication of all the terms in the equation we get,
\[ = {x^3} + 2{x^2} - {x^3} + 2{x^2} - 8 - 2{x^2}\]
On performing all the operations in the above equation we get,
\[ = 2{x^2} - 8\]
Now, we need to find the condition that the matrix will be non-singular.
The condition for the matrix to be non-singular is, the determinant of the matrix should have non-zero value.
Thus,
\[2{x^2} - 8 \ne 0\]
Therefore,
\[x \ne 2{\text{ and }}x \ne - 2\]
This is a required condition for which the matrix will be non-singular.
Hence option B) For all $x$ other than $2$ and $ - 2$ is correct.
As from the calculations, required condition for which matrix will be non-singular is \[x \ne 2{\text{ and }}x \ne - 2\], hence option A) \[ - 2 \leqslant x \leqslant 2\] is incorrect.
As from the calculations, the required condition for which matrix will be non-singular is \[x \ne 2{\text{ and }}x \ne - 2\], hence option C) \[x \geqslant 2\] is incorrect.
As from the calculations, the required condition for which the matrix will be non-singular is \[x \ne 2{\text{ and }}x \ne - 2\], hence option D) $x \leqslant - 2$ is incorrect.
Note: Singular matrix is a matrix whose determinant is zero and non-singular matrix is a matrix whose determinant is non-zero. If the condition is in the square form then we need to consider both positive and negative square root values of the solution.
{ - x}&x&2 \\
2&x&{ - x} \\
x&{ - 2}&{ - x}
\end{array}} \right]\] will be non-singular. Non-singular matrix is a matrix whose determinant is non-zero. So we need to find the values of $x$ for which determinant of the above matrix will be a non-zero value.
Complete step-by-step solution:
We have to check that for which values of $x$ the given matrix \[\left[ {\begin{array}{*{20}{c}}
{ - x}&x&2 \\
2&x&{ - x} \\
x&{ - 2}&{ - x}
\end{array}} \right]\] will be non-singular. Non-singular matrix is a matrix whose determinant is non-zero. So we need to check the values of $x$ for which determinant of the above matrix will be a non-zero value.
Let us calculate the determinant of the above matrix,
\[\left| {\begin{array}{*{20}{c}}
{ - x}&x&2 \\
2&x&{ - x} \\
x&{ - 2}&{ - x}
\end{array}} \right|\]
On applying the formula for the determinant we get,
$ = - x( - {x^2} - 2x) - x( - 2x + {x^2}) + 2( - 4 - {x^2})$
On performing the multiplication of all the terms in the equation we get,
\[ = {x^3} + 2{x^2} - {x^3} + 2{x^2} - 8 - 2{x^2}\]
On performing all the operations in the above equation we get,
\[ = 2{x^2} - 8\]
Now, we need to find the condition that the matrix will be non-singular.
The condition for the matrix to be non-singular is, the determinant of the matrix should have non-zero value.
Thus,
\[2{x^2} - 8 \ne 0\]
Therefore,
\[x \ne 2{\text{ and }}x \ne - 2\]
This is a required condition for which the matrix will be non-singular.
Hence option B) For all $x$ other than $2$ and $ - 2$ is correct.
As from the calculations, required condition for which matrix will be non-singular is \[x \ne 2{\text{ and }}x \ne - 2\], hence option A) \[ - 2 \leqslant x \leqslant 2\] is incorrect.
As from the calculations, the required condition for which matrix will be non-singular is \[x \ne 2{\text{ and }}x \ne - 2\], hence option C) \[x \geqslant 2\] is incorrect.
As from the calculations, the required condition for which the matrix will be non-singular is \[x \ne 2{\text{ and }}x \ne - 2\], hence option D) $x \leqslant - 2$ is incorrect.
Note: Singular matrix is a matrix whose determinant is zero and non-singular matrix is a matrix whose determinant is non-zero. If the condition is in the square form then we need to consider both positive and negative square root values of the solution.
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