Question

If \( A = \begin{bmatrix} 1 & -2 & 2 \\ -2 & -6 & 5 \\ 0 & 0 & 4 \end{bmatrix} \) then Adj A =

• A. \( \begin{bmatrix} -24 & 8 & 2 \\ 17 & -6 & -1 \\ 30 & -10 & 2 \end{bmatrix} \)
• B. \( \begin{bmatrix} -24 & 8 & 2 \\ 17 & -6 & 1 \\ -30 & 10 & -2 \end{bmatrix} \)
• C. \( \begin{bmatrix} -24 & 8 & 2 \\ 8 & 4 & -9 \\ 0 & 0 & -10 \end{bmatrix} \)
• D. \( \begin{bmatrix} -24 & 8 & 2 \\ 17 & -6 & -1 \\ 30 & -10 & 2 \end{bmatrix} \)

Hint:

The cofactor \(C_{ij}\) of an element \(a_{ij}\) is \( (-1)^{i+j} M_{ij} \), where \(M_{ij}\) is the determinant of the submatrix obtained by deleting row i and column j.
The Adjugate (or Adjoint) matrix, Adj(A), is the transpose of the cofactor matrix C.
Be careful with signs when calculating cofactors.

Answer 1

The Correct Option is C

Given \( A = \begin{bmatrix} 1 & -2 & 2
-2 & -6 & 5
0 & 0 & 4 \end{bmatrix} \). The adjugate (or adjoint) of A, Adj(A), is the transpose of the cofactor matrix of A. Let \(C_{ij}\) be the cofactor of the element \(a_{ij}\). \(C_{11} = (-1)^{1+1} \begin{vmatrix} -6 & 5
0 & 4 \end{vmatrix} = (-6)(4) - (5)(0) = -24\). \(C_{12} = (-1)^{1+2} \begin{vmatrix} -2 & 5
0 & 4 \end{vmatrix} = -((-2)(4) - (5)(0)) = -(-8) = 8\). \(C_{13} = (-1)^{1+3} \begin{vmatrix} -2 & -6
0 & 0 \end{vmatrix} = ((-2)(0) - (-6)(0)) = 0\). \(C_{21} = (-1)^{2+1} \begin{vmatrix} -2 & 2
0 & 4 \end{vmatrix} = -((-2)(4) - (2)(0)) = -(-8) = 8\). \(C_{22} = (-1)^{2+2} \begin{vmatrix} 1 & 2
0 & 4 \end{vmatrix} = (1)(4) - (2)(0) = 4\). \(C_{23} = (-1)^{2+3} \begin{vmatrix} 1 & -2
0 & 0 \end{vmatrix} = -((1)(0) - (-2)(0)) = 0\). \(C_{31} = (-1)^{3+1} \begin{vmatrix} -2 & 2
-6 & 5 \end{vmatrix} = ((-2)(5) - (2)(-6)) = -10 - (-12) = -10+12 = 2\). \(C_{32} = (-1)^{3+2} \begin{vmatrix} 1 & 2
-2 & 5 \end{vmatrix} = -((1)(5) - (2)(-2)) = -(5 - (-4)) = -(5+4) = -9\). \(C_{33} = (-1)^{3+3} \begin{vmatrix} 1 & -2
-2 & -6 \end{vmatrix} = ((1)(-6) - (-2)(-2)) = -6 - 4 = -10\). Cofactor matrix \( C = \begin{bmatrix} -24 & 8 & 0
8 & 4 & 0
2 & -9 & -10 \end{bmatrix} \). Adjugate matrix Adj(A) = \(C^T\). \[ \text{Adj}(A) = \begin{bmatrix} -24 & 8 & 2
8 & 4 & -9
0 & 0 & -10 \end{bmatrix} \] This matches option (c) from the re-transcribed options based on the image. \[ \boxed{\begin{pmatrix} -24 & 8 & 2
8 & 4 & -9
0 & 0 & -10 \end{pmatrix}} \]