If \(c_{ij}\) denotes the cofactor of element \(a_{ij}\) of the matrix \(A = \begin{bmatrix}1&2&-1\\0&-3&2\\4&2&3\end{bmatrix}\), then the value of \(c_{21}.c_{33}\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 24
The cofactor is \(c_{ij} = (-1)^{i+j}M_{ij}\), where the minor \(M_{ij}\) is the determinant left after deleting row i and column j.
For \(c_{21}\), delete row 2 and column 1:
\( M_{21} = \begin{vmatrix}2&-1\\2&3\end{vmatrix} = 6 + 2 = 8 \)
\( c_{21} = (-1)^{2+1}(8) = -8 \)
For \(c_{33}\), delete row 3 and column 3:
\( M_{33} = \begin{vmatrix}1&2\\0&-3\end{vmatrix} = -3 - 0 = -3 \)
\( c_{33} = (-1)^{3+3}(-3) = -3 \)
Hence
\( c_{21} \cdot c_{33} = (-8)(-3) = 24 \)
Why the other options are incorrect
Option (1) −8 — Incorrect. This is \(c_{21}\) alone.
Option (2) −3 — Incorrect. This is \(c_{33}\) alone.
Option (3) −24 — Incorrect. This is \(M_{21} \cdot M_{33} = 8 \times (-3)\), the product of the minors. The cofactor \(c_{21}\) carries the sign \((-1)^{2+1} = -1\).