Question

If \[{\rm{X}} = \left[ {\begin{array}{*{20}{c}}5&{ - 2}\\3&1\end{array}} \right]\], then find the inverse of the matrix \[X\].
A. \[\dfrac{1}{{11}}\left[ {\begin{array}{*{20}{c}}1&2\\{ - 3}&5\end{array}} \right]\]
B. \[\left[ {\begin{array}{*{20}{c}}1&2\\{ - 3}&5\end{array}} \right]\]
C. \[\dfrac{1}{{13}}\left[ {\begin{array}{*{20}{c}}{ - 2}&5\\1&3\end{array}} \right]\]
D. \[\left[ {\begin{array}{*{20}{c}}1&3\\{ - 2}&5\end{array}} \right]\]

Answer 1

Hint: In the given question, we need to find inverse of the matrix \[X\]. For this, we need to find the cofactor matrix of a given matrix using the following formula. After that, we will take the transpose of that matrix to get the adjacent matrix of \[{\rm{X}}\]. By multiplying the determinant of the matrix \[X\] with the adjacent matrix of \[X\], we get the inverse of the matrix \[X\].

Formula used: The adjoint of matrix \[{\rm{X}}\] is the transpose of co-factor matrix of given matrix.
The co-factor matrix of \[{\rm{X}} = \left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]\] can be calculated as \[{C_{ij}} = {\left( { - 1} \right)^{i + j}}{M_{ij}}\]
Here, \[{M_{ij}}\] is the minor that is determinant of sub matrix formed by deleting row \[i\] and column \[j\] from the given matrix \[{\rm{X}}\].
The inverse of matrix \[{\rm{X}}\] is \[{{\rm{X}}^{ - 1}} = \dfrac{1}{{\left| X \right|}}adj\left( X \right)\]

Complete step by step solution:
To check whether the inverse exists or not calculate the determinant of the matrix.
Now, the determinant of the given matrix is given by
\[\left| X \right| = \left| {\begin{array}{*{20}{c}}5&{ - 2}\\3&1\end{array}} \right|\]
Thus, we get
\[\begin{array}{c}\left| X \right| = 5\left( 1 \right) - \left[ { - 2\left( 3 \right)} \right]\\ = 5 + 6\\ = 11\end{array}\]
Since the determinant is a non-zero number.
Therefore, the inverse for the matrix exists.
We know that \[{\rm{X}} = \left[ {\begin{array}{*{20}{c}}5&{ - 2}\\3&1\end{array}} \right]\]
First, we will find the co-factor matrix of the above matrix.
\[{\rm{X}} = \left[ {\begin{array}{*{20}{c}}5&{ - 2}\\3&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}\\{{a_{21}}}&{{a_{22}}}\end{array}} \right]\]
Let the co-factor matrix be \[\left[ {\begin{array}{*{20}{c}}{{C_{11}}}&{{C_{12}}}\\{{C_{21}}}&{{C_{22}}}\end{array}} \right]\]
Thus, we get
\[\begin{array}{c}{C_{11}} = {( - 1)^{1 + 1}}\left( {{M_{11}}} \right)\\ = {( - 1)^2}\left( 1 \right)\\ = 1\end{array}\]
\[\begin{array}{c}{C_{12}} = {( - 1)^{1 + 2}}\left( {{M_{12}}} \right)\\ = {( - 1)^3}\left( 3 \right)\\ = - 3\end{array}\]
\[\begin{array}{c}{C_{21}} = {( - 1)^{2 + 1}}\left( {{M_{21}}} \right)\\ = {( - 1)^3}\left( { - 2} \right)\\ = 2\end{array}\]
\[\begin{array}{c}{C_{22}} = {( - 1)^{2 + 2}}\left( {{M_{22}}} \right)\\ = {( - 1)^4}\left( 5 \right)\\ = 5\end{array}\]
Hence, we get the following co-factor matrix of the given matrix \[{\rm{X}}\]. is \[\left[ {\begin{array}{*{20}{c}}{{C_{11}}}&{{C_{12}}}\\{{C_{21}}}&{{C_{22}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ - 3}\\2&5\end{array}} \right]\]

Now, the transpose of co-factor matrix is \[\left[ {\begin{array}{*{20}{c}}1&2\\{ - 3}&5\end{array}} \right]\]
Therefore, \[adj\left( X \right) = \left[ {\begin{array}{*{20}{c}}1&2\\{ - 3}&5\end{array}} \right]\]
Thus, the inverse of matrix \[{\rm{X}}\]. Is given by
\[{{\rm{X}}^{ - 1}} = \dfrac{1}{{\left| X \right|}}adj\left( X \right)\]
\[{{\rm{X}}^{ - 1}} = \dfrac{1}{{11}}\left[ {\begin{array}{*{20}{c}}1&2\\{ - 3}&5\end{array}} \right]\]
Hence, the inverse of the matrix \[{\rm{X}}\] is \[\dfrac{1}{{11}}\left[ {\begin{array}{*{20}{c}}1&2\\{ - 3}&5\end{array}} \right]\].
Therefore, the correct option is (A).

Additional information: The adjoint of the matrix is defined as the transpose of a cofactor matrix of the square matrix. The adjoint of the matrix \[A\] is denoted by \[adj\left( A \right)\]. It is essential to determine the adjoint of a matrix to calculate the inverse of a matrix. This can be applied to only square matrices.

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. Students may make mistakes while finding the co-factor matrix from the given matrix. Even a small sign mistake can make a bigger difference in the required result. So, we need to be careful while calculating the minors, co-factors, and determinants.