Question

The matrix \(A = \left[ {\begin{array}{*{20}{c}} { - 4}&{ - 3}&{ - 3}\\ 1&0&1\\ 4&4&x \end{array}} \right]\) is its own adjoint. The value of x will be
1. 5
2. 3
3. -3
4. -5

Answer 1

Correct Answer - Option 2 : 3

Concept:

For the given matrix A:

Cofactor matrix = \(\left[ {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_{12}}}&{{A_{13}}}\\ {{A_{21}}}&{{A_{22}}}&{{A_{23}}}\\ {{A_{31}}}&{{A_{32}}}&{{A_{33}}} \end{array}} \right]\)

Cofactor of A = \(\left[ {\begin{array}{*{20}{c}} { - 4}&{4 - x}&4\\ {3x - 12}&{12 - 4x}&4\\ -3&1&3 \end{array}} \right]\)

Adjoint of A = adj (A) = [cofactor matrix]^T

adj (A) = \(\left[ {\begin{array}{*{20}{c}} { - 4}&{3x - 12}&{ - 3}\\ {4 - x}&{12 - 4x}&1\\ 4&4&3 \end{array}} \right]\)

Given: adj (A) = A

i.e. \(\left[ {\begin{array}{*{20}{c}} { - 4}&{3x - 12}&{ - 3}\\ {4 - x}&{12 - 4x}&1\\ 4&4&3 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 4}&{ - 3}&{ - 3}\\ 1&0&1\\ 4&4&x \end{array}} \right]\)

∴ x = 3