Let \(A = [a_{ij}]\) is given by \(A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{bmatrix}\). Then the matrix \(B = [b_{ij}]\), where \(b_{ij}\) = Minor of \(a_{ij}\), is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\begin{bmatrix} 7 & 19 & -11 \\ -1 & -1 & 1 \\ -3 & -11 & 7 \end{bmatrix}\)
The minor of \(a_{ij}\) is the determinant left after deleting row i and column j (no sign attached, unlike the cofactor).
Row 1:
\(b_{11} = (4)(3) - (-5)(-1) = 12 - 5 = 7\)
\(b_{12} = (3)(3) - (-5)(2) = 9 + 10 = 19\)
\(b_{13} = (3)(-1) - (4)(2) = -3 - 8 = -11\)
Row 2:
\(b_{21} = (-1)(3) - (2)(-1) = -3 + 2 = -1\)
\(b_{22} = (1)(3) - (2)(2) = 3 - 4 = -1\)
\(b_{23} = (1)(-1) - (-1)(2) = -1 + 2 = 1\)
Row 3:
\(b_{31} = (-1)(-5) - (2)(4) = 5 - 8 = -3\)
\(b_{32} = (1)(-5) - (2)(3) = -5 - 6 = -11\)
\(b_{33} = (1)(4) - (-1)(3) = 4 + 3 = 7\)
So:
\( B = \begin{bmatrix} 7 & 19 & -11 \\ -1 & -1 & 1 \\ -3 & -11 & 7 \end{bmatrix} \)