Find the cofactor matrix for the matrix \(A = \left[ {\begin{array}{*{20}{c}} { - 1}&2&3\\ { - 2}&3&5\\ 4&{ - 2}&1 \end{array}} \right]\) ?

CONCEPT:

In order to find the minor of an element of a determinant, we need to delete the row and column passing through the element aij , thus obtained is called the minor of aij and is usually denoted by Mij .

The co-factor of an element aij is given by (-1)i + j ⋅ Mij, where Mij is the minor of element aij and it is denoted by Cij.

Thus \({C_{ij}} = \;\left\{ {\begin{array}{*{20}{c}} {{M_{ij}},\;when\;i + j\;is\;even}\\ { - \;{M_{ij}},\;when\;i + j\;is\;odd} \end{array}} \right.\)

CALCULATION:

Given: \(A = \left[ {\begin{array}{*{20}{c}} { - 1}&2&3\\ { - 2}&3&5\\ 4&{ - 2}&1 \end{array}} \right]\)

Here, we have to find the cofactor matrix for the given matrix A

As we know that, the cofactor of an element aij is given by: \({C_{ij}} = \;\left\{ {\begin{array}{*{20}{c}} {{M_{ij}},\;when\;i + j\;is\;even}\\ { - \;{M_{ij}},\;when\;i + j\;is\;odd} \end{array}} \right.\)

⇒ \({C_{11}} = {\left( { - 1} \right)^2} \times \left| {\begin{array}{*{20}{c}} 3&5\\ { - 2}&1 \end{array}} \right| = 13\)

⇒ \({C_{12}} = {\left( { - 1} \right)^3} \times \left| {\begin{array}{*{20}{c}} { - 2}&5\\ 4&1 \end{array}} \right| = \;22\)

⇒ \({C_{13}} = {\left( { - 1} \right)^4} \times \left| {\begin{array}{*{20}{c}} { - 2}&3\\ 4&{ - 2} \end{array}} \right| = \; - 8\)

Similarly, we can say that C₂₁ = - 8, C₂₂ = - 13 and C₂₃ = 6

Similarly, we can also say that, C₃₁ = 1, C₃₂ = - 1 and C₃₃ = 1

So, the required cofactor matrix is \(\left[ {\begin{array}{*{20}{c}} {13}&{22}&{ - 8}\\ { - 8}&{ - 13}&6\\ 1&{ - 1}&1 \end{array}} \right]\)

Hence, option B is the correct answer.