Let \(M_{ij}\) and \(A_{ij}\) denote respectively minors and co-factors of the element in the \(i\)th row and \(j\)th column of the matrix \(\begin{bmatrix}1&2&-1\\3&2&3\\4&-1&0\end{bmatrix}\). Then
(A) \(M_{32} = 6\)
(B) \(M_{23} = 9\)
(C) \(A_{32}=-6\)
(D) \(A_{23}=-9\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (C) only
The minor \(M_{ij}\) is the determinant left after deleting row i and column j. The cofactor is \(A_{ij} = (-1)^{i+j} M_{ij}\).
Deleting row 3 and column 2:
\( \displaystyle M_{32} = \begin{vmatrix}1&-1\\3&3\end{vmatrix} = 3 + 3 = 6 \)
Deleting row 2 and column 3:
\( \displaystyle M_{23} = \begin{vmatrix}1&2\\4&-1\end{vmatrix} = -1 - 8 = -9 \)
For both cofactors \(i + j = 5\) is odd, so the sign changes:
\( \displaystyle A_{32} = -M_{32} = -6 \)
\( \displaystyle A_{23} = -M_{23} = 9 \)
So \(M_{32} = 6\), \(M_{23} = -9\), \(A_{32} = -6\) and \(A_{23} = 9\).
Checking the statements
(A) \(M_{32} = 6\) — Correct. \(M_{32}\) is 6.
(B) \(M_{23} = 9\) — Incorrect. \(M_{23}\) is \(-9\).
(C) \(A_{32}=-6\) — Correct. \(A_{32} = (-1)^{3+2} M_{32} = -6\).
(D) \(A_{23}=-9\) — Incorrect. \(A_{23}\) is 9; −9 is the minor \(M_{23}\), not the cofactor.