Question

If \[{\rm{X}} = \left[ {\begin{array}{*{20}{c}}{ - x}&{ - y}\\z&t\end{array}} \right]\], then find the value of \[adj\left( X \right)\].
A. \[\left[ {\begin{array}{*{20}{c}}t&z\\{ - y}&{ - x}\end{array}} \right]\]
B. \[\left[ {\begin{array}{*{20}{c}}t&y\\{ - z}&{ - x}\end{array}} \right]\]
C. \[\left[ {\begin{array}{*{20}{c}}t&{ - z}\\y&{ - x}\end{array}} \right]\]
D. None of these

Answer 1

Hint: In the given question, we need to find the transpose of \[adj\left( X \right)\]. For this, we will 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 desired result which is an adjacent matrix of \[{\rm{X}}\].

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 the determinant of sub matrix formed by deleting row \[i\] and column \[j\] from the given matrix \[{\rm{X}}\].


Complete step by step solution:
We know that \[{\rm{X}} = \left[ {\begin{array}{*{20}{c}}{ - x}&{ - y}\\z&t\end{array}} \right]\]
First, we will find the co-factor matrix of the above matrix.
\[{\rm{X}} = \left[ {\begin{array}{*{20}{c}}{ - x}&{ - y}\\z&t\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}\\{{a_{21}}}&{{a_{22}}}\end{array}} \right]\]
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
\[\begin{array}{c}{C_{11}} = {( - 1)^{1 + 1}}\left( {{M_{11}}} \right)\\ = {( - 1)^2}\left( t \right)\\ = t\end{array}\]
\[\begin{array}{c}{C_{12}} = {( - 1)^{1 + 2}}\left( {{M_{12}}} \right)\\ = {( - 1)^3}\left( z \right)\\ = - z\end{array}\]
\[\begin{array}{c}{C_{21}} = {( - 1)^{2 + 1}}\left( {{M_{21}}} \right)\\ = {( - 1)^3}\left( { - y} \right)\\ = y\end{array}\]
\[\begin{array}{c}{C_{22}} = {( - 1)^{2 + 2}}\left( {{M_{22}}} \right)\\ = {( - 1)^4}\left( { - x} \right)\\ = - x\end{array}\]
Hence, we get the following co-factor matrix of given matrix \[{\rm{X}}\]. is \[\left[ {\begin{array}{*{20}{c}}{{C_{11}}}&{{C_{12}}}\\{{C_{21}}}&{{C_{22}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}t&{ - z}\\y&{ - x}\end{array}} \right]\]

Now, the transpose of co-factor matrix is \[\left[ {\begin{array}{*{20}{c}}t&y\\{ - z}&{ - x}\end{array}} \right]\]
Therefore, \[adj\left( X \right) = \left[ {\begin{array}{*{20}{c}}t&y\\{ - z}&{ - x}\end{array}} \right]\]
Therefore, the correct option is (B).

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 applicable only to square matrices.


Note: Students may make mistakes while finding the co-factor matrix from the given matrix. Specifically, they may get wrong the sign of elements of the co-factor matrix. Even a small sign mistake can make bigger difference in the overall result. So, we need to be careful while calculating the minors as well as co-factors.