Question

If the cofactors of the elements 3, 7, and 6 of the matrix \( \begin{bmatrix} 1 & 2 & 3 \\ 4 & -1 & 7 \\ 2 & 4 & 6 \end{bmatrix} \) are \( a, b, c \) respectively, then evaluate: \[ \left[ a \ b \ c \right] \begin{bmatrix} 1 \\ 4 \\ 2 \end{bmatrix} + \left[ a \ b \ c \right] \begin{bmatrix} 3 \\ 7 \\ 6 \end{bmatrix} \]

• A. \( -1 \)
• B. \( 1 \)
• C. \( 0 \)
• D. \( 3 \)

Hint:

When summing dot products, it’s efficient to combine vectors first, then apply cofactors. Pay attention to signs in cofactor expansion.

Answer 1

The Correct Option is C

We are given matrix: \[ A = \begin{bmatrix} 1 & 2 & 3
4 & -1 & 7
2 & 4 & 6 \end{bmatrix} \] and the cofactors of entries 3, 7, and 6, which are the elements \( A_{13}, A_{23}, A_{33} \) respectively. We define: \[ [a\ b\ c] = \text{cofactors of } (3, 7, 6) \] We need to evaluate: \[ [a\ b\ c] \cdot \left( \begin{bmatrix} 1
4
2 \end{bmatrix} + \begin{bmatrix} 3
7
6 \end{bmatrix} \right) = [a\ b\ c] \cdot \begin{bmatrix} 4
11
8 \end{bmatrix} \] Now calculate each cofactor: - \( a = \text{Cofactor of } 3 = (-1)^{1+3} \cdot \begin{vmatrix} 4 & -1
2 & 4 \end{vmatrix} = 1 \cdot (16 + 2) = 18 \) - \( b = \text{Cofactor of } 7 = (-1)^{2+3} \cdot \begin{vmatrix} 1 & 2
2 & 4 \end{vmatrix} = -1 \cdot (4 - 4) = 0 \) - \( c = \text{Cofactor of } 6 = (-1)^{3+3} \cdot \begin{vmatrix} 1 & 2
4 & -1 \end{vmatrix} = 1 \cdot (-1 - 8) = -9 \) So, \( [a\ b\ c] = [18\ 0\ -9] \) Now compute: \[ 18 \cdot 4 + 0 \cdot 11 + (-9) \cdot 8 = 72 + 0 - 72 = 0 \] % Final Answer \[ \boxed{0} \]