Question

If $A=\begin{vmatrix}1 & -2 & 4 \\2 & 0 & 5 \\3 & 4 & -7\end{vmatrix}$, then the value of $a_{11}A_{31}-a_{12}A_{32}+a_{13}A_{33}$ is:

• A. 2
• B. -2
• C. 12
• D. 0

Answer 1

Answer: (C)

12

Answer 2

For the matrix

$$A = \begin{pmatrix} 1 & -2 & 4 \\ 2 & 0 & 5 \\ 3 & 4 & -7 \end{pmatrix},$$

we are given the expression

$$a_{11}A_{31} - a_{12}A_{32} + a_{13}A_{33}.$$

Recall that the cofactor $A_{ij} = (-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor obtained by eliminating the $i$th row and $j$th column.

For the third row cofactors:

• $A_{31} = (-1)^{3+1} M_{31} = M_{31}$.
• $A_{32} = (-1)^{3+2} M_{32} = -M_{32}$.
• $A_{33} = (-1)^{3+3} M_{33} = M_{33}$.

Thus, the given expression becomes:

$$a_{11}M_{31} - a_{12}(-M_{32}) + a_{13}M_{33} = a_{11}M_{31} + a_{12}M_{32} + a_{13}M_{33}.$$

Now, compute the minors:

1. Minor $M_{31}$ (remove row 3 and column 1):

   $$M_{31} = \begin{vmatrix} -2 & 4 \\ 0 & 5 \end{vmatrix} = (-2)(5) - (4)(0) = -10.$$

2. Minor $M_{32}$ (remove row 3 and column 2):

   $$M_{32} = \begin{vmatrix} 1 & 4 \\ 2 & 5 \end{vmatrix} = (1)(5) - (4)(2) = 5 - 8 = -3.$$

3. Minor $M_{33}$ (remove row 3 and column 3):

   $$M_{33} = \begin{vmatrix} 1 & -2 \\ 2 & 0 \end{vmatrix} = (1)(0) - (-2)(2) = 0 + 4 = 4.$$

With the elements:

• $a_{11} = 1$
• $a_{12} = -2$
• $a_{13} = 4$

Substitute the values:

$$\begin{align*} \text{Expression} &= 1(-10) + (-2)(-3) + 4(4) \\ &= -10 + 6 + 16 \\ &= 12. \end{align*}$$

So, the value is 12.