Question

The minor matrix of the matrix $\left[\begin{array}{ccc}1& -1& 2\\ 2& -1& 1\\ 1& 1& -1\end{array}\right]$ is

• A. $\left[\begin{array}{ccc}0& -3& 3\\ -1& -3& 2\\ 1& -3& 1\end{array}\right]$
• B. $\left[\begin{array}{ccc}0& -3& 1\\ 1& -3& 2\\ -1& -3& 3\end{array}\right]$
• C. $\left[\begin{array}{ccc}0& 3& 3\\ -1& -3& 2\\ 1& -3& 1\end{array}\right]$
• D. None of these

Answer 1

The correct option is A

$\left[\begin{array}{ccc}0& -3& 3\\ -1& -3& 2\\ 1& -3& 1\end{array}\right]$

We know that minor of an element is the determinant of the square matrix formed by deleting the row and column corresponding to the element from the original matrix. This is how we calculated minor in the previous question as well. Here minor matrix means that we have to calculate the minor of every element of the matrix. The matrix formed with the minors of each element is the minor matrix.

So let us use the above definition to calculate the minor of each element and finally the minor matrix. The minor for the first element of first row shall be calculated by deleting the first row and first column and then taking the determinant of the remaining matrix. So it will be $\left|\begin{array}{cc}-1& 1\\ 1& -1\end{array}\right|$ = 0. Similarly for the second entry of the first row it shall be $\left|\begin{array}{cc}2& 1\\ 1& -1\end{array}\right|$ = 2 $\times$ (-1) - (1) $\times$ (1) = -3. For the third element of the first row it shall be $\left|\begin{array}{cc}2& -1\\ 1& 1\end{array}\right|$ = 2 $\times$ 1 - (-1) $\times$ = 3.

Going the same way we can do it for the remaining rows and we can see that option (a) is the correct answer.