Question

Find the inverse matrix of the matrix \[\left[ {\begin{array}{*{20}{c}}0&1&2\\1&2&3\\3&1&1\end{array}} \right]\] .
A. \[\left[ {\begin{array}{*{20}{c}}{\dfrac{1}{2}}&{ - \dfrac{1}{2}}&{\dfrac{1}{2}}\\{ - 4}&3&{ - 1}\\{\dfrac{5}{2}}&{\dfrac{{ - 3}}{2}}&{\dfrac{1}{2}}\end{array}} \right]\]
B. \[\left[ {\begin{array}{*{20}{c}}{\dfrac{1}{2}}&{ - 4}&{\dfrac{5}{2}}\\1&{ - 6}&3\\1&2&{ - 1}\end{array}} \right]\]
C. \[\dfrac{1}{2}\left[ {\begin{array}{*{20}{c}}1&2&3\\3&2&1\\4&2&3\end{array}} \right]\]
D. \[\dfrac{1}{2}\left[ {\begin{array}{*{20}{c}}1&{ - 1}&{ - 1}\\{ - 8}&6&{ - 2}\\5&{ - 3}&1\end{array}} \right]\]

Answer 1

Hint: First, calculate the determinant of the given matrix to check whether the inverse matrix is defined or not. If the determinant is non-zero, then calculate the adjoint matrix by calculating the co-factors of the matrix. In the end, substitute the values in the formula of the inverse matrix \[{A^{ - 1}} = \dfrac{1}{{\left| A \right|}}\left( {adj A} \right)\] and get the required answer.

Formula used:
The inverse matrix of a non-singular matrix \[A\] is: \[{A^{ - 1}} = \dfrac{1}{{\left| A \right|}}\left( {adj A} \right)\]
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: \[\left| A \right| = {a_{11}}\left( {{a_{22}}{a_{33}} - {a_{32}}{a_{23}}} \right) - {a_{12}}\left( {{a_{21}}{a_{33}} - {a_{31}}{a_{23}}} \right) + {a_{13}}\left( {{a_{21}}{a_{32}} - {a_{31}}{a_{22}}} \right)\]

Complete step by step solution:
The given matrix is \[\left[ {\begin{array}{*{20}{c}}0&1&2\\1&2&3\\3&1&1\end{array}} \right]\].

Let consider,
\[A = \left[ {\begin{array}{*{20}{c}}0&1&2\\1&2&3\\3&1&1\end{array}} \right]\]
To check whether the inverse exists or not calculate the determinant of the matrix.
Apply the formula for the determinant of the \[3 \times 3\] matrix.
\[\left| A \right| = 0\left( {2 \times 1 - 1 \times 3} \right) - 1\left( {1 \times 1 - 3 \times 3} \right) + 2\left( {1 \times 1 - 3 \times 2} \right)\]
\[ \Rightarrow \left| A \right| = 0\left( {2 - 3} \right) - 1\left( {1 - 9} \right) + 2\left( {1 - 6} \right)\]
\[ \Rightarrow \left| A \right| = 0 - 1\left( { - 8} \right) + 2\left( { - 5} \right)\]
\[ \Rightarrow \left| A \right| = 8 - 10\]
\[ \Rightarrow \left| A \right| = - 2\]
Since the determinant is a non-zero number.
Therefore, the inverse for the matrix exists.

Now calculate the adjoint matrix of the given matrix \[A\].
Let’s calculate the co-factors of the matrix.
\[{A_{11}} = {\left( { - 1} \right)^{1 + 1}}\left( {2 \times 1 - 1 \times 3} \right) = - 1\]
\[{A_{12}} = {\left( { - 1} \right)^{1 + 2}}\left( {1 \times 1 - 3 \times 3} \right) = 8\]
\[{A_{13}} = {\left( { - 1} \right)^{1 + 3}}\left( {1 \times 1 - 3 \times 2} \right) = - 5\]
\[{A_{21}} = {\left( { - 1} \right)^{2 + 1}}\left( {1 \times 1 - 1 \times 2} \right) = 1\]
\[{A_{22}} = {\left( { - 1} \right)^{2 + 2}}\left( {0 \times 1 - 3 \times 2} \right) = - 6\]
\[{A_{23}} = {\left( { - 1} \right)^{2 + 3}}\left( {0 \times 1 - 3 \times 1} \right) = 3\]
\[{A_{31}} = {\left( { - 1} \right)^{3 + 1}}\left( {1 \times 3 - 2 \times 2} \right) = - 1\]
\[{A_{32}} = {\left( { - 1} \right)^{3 + 2}}\left( {0 \times 3 - 1 \times 2} \right) = 2\]
\[{A_{33}} = {\left( { - 1} \right)^{3 + 3}}\left( {0 \times 2 - 1 \times 1} \right) = - 1\]
So, the co-factor matrix of the given matrix is \[\left[ {\begin{array}{*{20}{c}}{ - 1}&8&{ - 5}\\1&{ - 6}&3\\{ - 1}&2&{ - 1}\end{array}} \right]\]

We know that the cofactor matrix is the transpose of the adjoint matrix.
So, the adjoint matrix of the given matrix is,
\[adj A = \left[ {\begin{array}{*{20}{c}}{ - 1}&1&{ - 1}\\8&{ - 6}&2\\{ - 5}&3&{ - 1}\end{array}} \right]\]

Now substitute the values of determinant and adjoint matrix in the formula of the inverse matrix \[{A^{ - 1}} = \dfrac{1}{{\left| A \right|}}\left( {adj A} \right)\].
We get,
\[{A^{ - 1}} = \dfrac{1}{{ - 2}}\left[ {\begin{array}{*{20}{c}}{ - 1}&1&{ - 1}\\8&{ - 6}&2\\{ - 5}&3&{ - 1}\end{array}} \right]\]
\[ \Rightarrow {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}{\dfrac{1}{2}}&{ - \dfrac{1}{2}}&{\dfrac{1}{2}}\\{ - 4}&3&{ - 1}\\{\dfrac{5}{2}}&{ - \dfrac{3}{2}}&{\dfrac{1}{2}}\end{array}} \right]\]
Hence the correct option is A.

Note: Students should keep in mind that the inverse matrix of any matrix exists if the determinant of that matrix is non-zero. So, first, check whether the determinant is nonzero or not. And the product of the inverse matrix and the original matrix is an identity matrix.