Let \(A = \begin{bmatrix}2 & 1 & 3\\ 0 & -1 & 4\\ 1 & 2 & 0\end{bmatrix}\), and let \(M_{ij}\) and \(C_{ij}\) denote the minor and the cofactor of the element \(a_{ij}\) of A. Consider the following statements:
(A) \(M_{12} = -4\)
(B) \(C_{12} = -4\)
(C) \(|A| = -9\)
(D) \(2C_{21} + C_{22} + 3C_{23} = 0\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (C) and (D) only
The minor \(M_{ij}\) is the determinant left after deleting row i and column j, and the cofactor is \(C_{ij} = (-1)^{i+j}M_{ij}\).
For the first row:
\(M_{11} = (-1)(0) - (4)(2) = -8,\quad C_{11} = -8\)
\(M_{12} = (0)(0) - (4)(1) = -4,\quad C_{12} = 4\)
\(M_{13} = (0)(2) - (-1)(1) = 1,\quad C_{13} = 1\)
Expanding along the first row:
\(|A| = 2(-8) + 1(4) + 3(1)\)
\(|A| = -9\)
For the second row:
\(M_{21} = (1)(0) - (3)(2) = -6,\quad C_{21} = 6\)
\(M_{22} = (2)(0) - (3)(1) = -3,\quad C_{22} = -3\)
\(M_{23} = (2)(2) - (1)(1) = 3,\quad C_{23} = -3\)
Then
\(2C_{21} + C_{22} + 3C_{23} = 12 - 3 - 9 = 0\)
This had to be 0: the elements of one row (2, 1, 3 of the first row) multiplied by the cofactors of another row always add up to 0.
So \(M_{12} = -4\), \(C_{12} = 4\), \(|A| = -9\) and \(2C_{21} + C_{22} + 3C_{23} = 0\).
Checking the statements
(A) \(M_{12} = -4\) — Correct. \((0)(0) - (4)(1) = -4\).
(B) \(C_{12} = -4\) — Incorrect. \(C_{12} = (-1)^{1+2}M_{12} = +4\).
(C) \(|A| = -9\) — Correct. \(2(-8) + 1(4) + 3(1) = -9\).
(D) \(2C_{21} + C_{22} + 3C_{23} = 0\) — Correct. \(2(6) + (-3) + 3(-3) = 0\).