Let \(A = \begin{bmatrix} 1 & 3 & -2 \\ 0 & 2 & 1 \\ 4 & -1 & 3 \end{bmatrix}\). Let \(M_{ij}\) and \(C_{ij}\) denote the minor and the cofactor of the element \(a_{ij}\). Which of the following statements are correct?
(A) \(|A| = 35\)
(B) \(M_{12} = 4\)
(C) \(C_{12} = -4\)
(D) \(a_{21}C_{11} + a_{22}C_{12} + a_{23}C_{13} = 0\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (D) only
The minors of the first-row elements, and the cofactors \(C_{ij} = (-1)^{i + j}M_{ij}\):
\( \displaystyle M_{11} = \begin{vmatrix} 2 & 1 \\ -1 & 3 \end{vmatrix} = 6 + 1 = 7, \quad C_{11} = 7 \)
\( \displaystyle M_{12} = \begin{vmatrix} 0 & 1 \\ 4 & 3 \end{vmatrix} = 0 - 4 = -4, \quad C_{12} = 4 \)
\( \displaystyle M_{13} = \begin{vmatrix} 0 & 2 \\ 4 & -1 \end{vmatrix} = 0 - 8 = -8, \quad C_{13} = -8 \)
Expanding along the first row:
\(|A| = 1(7) + 3(4) + (-2)(-8) = 7 + 12 + 16 = 35\)
The elements of one row multiplied by the cofactors of a different row always add up to 0:
\(a_{21}C_{11} + a_{22}C_{12} + a_{23}C_{13} = 0(7) + 2(4) + 1(-8) = 0\)
So |A| = 35, \(M_{12} = -4\), \(C_{12} = 4\), and the mixed sum of row 2 with the row 1 cofactors is 0.
Checking the statements
(A) \(|A| = 35\) — Correct. \(7 + 12 + 16 = 35\).
(B) \(M_{12} = 4\) — Incorrect. \(M_{12} = 0 \times 3 - 1 \times 4 = -4\).
(C) \(C_{12} = -4\) — Incorrect. \(C_{12} = (-1)^{1 + 2} \times (-4) = 4\).
(D) \(a_{21}C_{11} + a_{22}C_{12} + a_{23}C_{13} = 0\) — Correct. \(0 + 8 - 8 = 0\).