Question

If \[A = \left[ {\begin{array}{*{20}{c}}0&0&1\\0&1&0\\1&0&0\end{array}} \right]\], then find the value of \[{A^{ - 1}}\].
A. \[I\]
B. \[ - I\]
C. \[ - A\]
D. \[A\]

Answer 1

Hint: First, calculate the determinant of the given matrix. Then calculate the adjoint matrix of the given matrix. In the end, substitute the values in the formula for the inverse of the matrix and get the required answer.

Formula used:
The determinant of a \[3 \times 3\] matrix \[A = \left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{array}} \right]\] is: \[\left| A \right| = {a_{11}}\left( {{a_{22}}{a_{33}} - {a_{32}}{a_{23}}} \right) - {a_{12}}\left( {{a_{21}}{a_{33}} - {a_{31}}{a_{23}}} \right) + {a_{13}}\left( {{a_{21}}{a_{32}} - {a_{31}}{a_{22}}} \right)\]
The inverse matrix of a non-singular matrix \[A\] is: \[{A^{ - 1}} = \dfrac{1}{{\left| A \right|}}\left( {adj A} \right)\]

Complete step by step solution:
The given matrix is \[A = \left[ {\begin{array}{*{20}{c}}0&0&1\\0&1&0\\1&0&0\end{array}} \right]\].

Let’s calculate the determinant of the given matrix.
Apply the formula of the determinant of a \[3 \times 3\] matrix.
We get,
\[\left| A \right| = 0\left( {0 - 0} \right) - 0\left( {0 - 0} \right) + 1\left( {0 - 1} \right)\]
\[ \Rightarrow \left| A \right| = - 1\] \[.....\left( 1 \right)\]

Now calculate the adjoint matrix of the given matrix by using the co-factor method.
Let’s calculate the co-factors of the matrix.
\[{A_{11}} ={\left( { - 1} \right)^{1 + 1}}\left[ {\begin{array}{*{20}{c}}1&0\\0&0\end{array}} \right]
= {\left( { - 1} \right)^{1 + 1}}\left( {1 \times 0 - 0 \times 0} \right) = 0\]
\[{A_{12}} =\left[ {\begin{array}{*{20}{c}}0&0\\1&0\end{array}} \right]
= {\left( { - 1} \right)^{1 + 3}}{\left( { - 1} \right)^{1 + 2}}\left( {0 \times 0 - 1 \times 0} \right) = 0\]
\[{A_{13}} ={\left( { - 1} \right)^{1 + 3}}\left[ {\begin{array}{*{20}{c}}0&1\\1&0\end{array}} \right]
= {\left( { - 1} \right)^{1 + 3}}\left( {0 \times 0 - 1 \times 1} \right) = - 1\]
\[{A_{21}} ={\left( { - 1} \right)^{2 + 1}}\left[ {\begin{array}{*{20}{c}}0&1\\0&0\end{array}} \right]
= {\left( { - 1} \right)^{2 + 1}}\left( {0 \times 0 - 0 \times 1} \right) = 0\]
\[{A_{22}} ={\left( { - 1} \right)^{2 + 2}} \left[ {\begin{array}{*{20}{c}}0&1\\1&0\end{array}} \right]
={\left( { - 1} \right)^{2 + 2}}\left( {0 \times 0 - 1 \times 1} \right) = - 1\]
\[{A_{23}} ={\left( { - 1} \right)^{2 + 3}}\left[ {\begin{array}{*{20}{c}}0&0\\1&0\end{array}} \right]
= {\left( { - 1} \right)^{2 + 3}}\left( {0 \times 0 - 1 \times 0} \right) = 0\]
\[{A_{31}} ={\left( { - 1} \right)^{3 + 1}}\left[ {\begin{array}{*{20}{c}}0&1\\1&0\end{array}} \right]
= {\left( { - 1} \right)^{3 + 1}}\left( {0 \times 0 - 1 \times 1} \right) = - 1\]
\[{A_{32}} ={\left( { - 1} \right)^{3 + 2}} \left[ {\begin{array}{*{20}{c}}0&1\\0&0\end{array}} \right] ={\left( { - 1} \right)^{3 + 2}}\left( {0 \times 0 - 1 \times 0} \right) = 0\]
\[{A_{33}} = {\left( { - 1} \right)^{3 + 3}}\left[ {\begin{array}{*{20}{c}}0&0\\0&1\end{array}} \right]= {\left( { - 1} \right)^{3 + 3}}\left( {0 \times 1 - 0 \times 0} \right) = 0\]
So, the co-factor matrix of the given matrix is \[\left[ {\begin{array}{*{20}{c}}0&0&{ - 1}\\0&{ - 1}&0\\{ - 1}&0&0\end{array}} \right]\].
We know that the cofactor matrix is the transpose of the adjoint matrix.
So, the adjoint matrix of the given matrix is,
\[adj A = \left[ {\begin{array}{*{20}{c}}0&0&{ - 1}\\0&{ - 1}&0\\{ - 1}&0&0\end{array}} \right]\] \[.....\left( 2 \right)\]
Substitute the equations \[\left( 1 \right)\] and \[\left( 2 \right)\] in the formula of the inverse matrix \[{A^{ - 1}} = \dfrac{1}{{\left| A \right|}}\left( {adj A} \right)\].
Then,
\[{A^{ - 1}} = \dfrac{1}{{ - 1}}\left[ {\begin{array}{*{20}{c}}0&0&{ - 1}\\0&{ - 1}&0\\{ - 1}&0&0\end{array}} \right]\]
\[ \Rightarrow {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}0&0&1\\0&1&0\\1&0&0\end{array}} \right]\]
\[ \Rightarrow {A^{ - 1}} = A\]
Hence the correct option is D.

Note: If the determinant of a matrix is 0, then \[\dfrac{1}{{det A}}\] is undefined. So, the matrix with a 0 determinant has no inverse.
While calculating the inverse matrix, first check whether the determinant is nonzero or not.