If `A=[(3,2,4),(1,2,1),(3,2,6)]` and `A_(ij)` are the cofactors of `a_(ij)`, then `a_(11)A_(11)+a_(12)A_(12)+a_(13)A_(13)` is equal to

Options

A. 8
B. 6
C. 4
D. 0

Explanation

To solve the problem, we need to compute the expression \( a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13} \) where \( A \) is the matrix given by

\[ A = \begin{pmatrix} 3 & 2 & 4 \\ 1 & 2 & 1 \\ 3 & 2 & 6 \end{pmatrix} \]

### Step 1: Identify the elements \( a_{11}, a_{12}, a_{13} \)

From the matrix \( A \):
- \( a_{11} = 3 \)
- \( a_{12} = 2 \)
- \( a_{13} = 4 \)

### Step 2: Calculate the cofactors \( A_{11}, A_{12}, A_{13} \)

#### Finding \( A_{11} \)

The cofactor \( A_{11} \) is given by:

\[ A_{11} = (-1)^{1+1} \cdot \text{det} \begin{pmatrix} 2 & 1 \\ 2 & 6 \end{pmatrix} \]

Calculating the determinant:

\[ \text{det} \begin{pmatrix} 2 & 1 \\ 2 & 6 \end{pmatrix} = (2 \cdot 6) - (1 \cdot 2) = 12 - 2 = 10 \]

Thus,

\[ A_{11} = 10 \]

#### Finding \( A_{12} \)

The cofactor \( A_{12} \) is given by:

\[ A_{12} = (-1)^{1+2} \cdot \text{det} \begin{pmatrix} 1 & 1 \\ 3 & 6 \end{pmatrix} \]

Calculating the determinant:

\[ \text{det} \begin{pmatrix} 1 & 1 \\ 3 & 6 \end{pmatrix} = (1 \cdot 6) - (1 \cdot 3) = 6 - 3 = 3 \]

Thus,

\[ A_{12} = -3 \]

#### Finding \( A_{13} \)

The cofactor \( A_{13} \) is given by:

\[ A_{13} = (-1)^{1+3} \cdot \text{det} \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix} \]

Calculating the determinant:

\[ \text{det} \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix} = (1 \cdot 2) - (2 \cdot 3) = 2 - 6 = -4 \]

Thus,

\[ A_{13} = -4 \]

### Step 3: Substitute the values into the expression

Now we substitute \( a_{11}, a_{12}, a_{13} \) and their corresponding cofactors into the expression:

\[ a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13} = 3 \cdot 10 + 2 \cdot (-3) + 4 \cdot (-4) \]

Calculating each term:

- \( 3 \cdot 10 = 30 \)
- \( 2 \cdot (-3) = -6 \)
- \( 4 \cdot (-4) = -16 \)

Adding these together:

\[ 30 - 6 - 16 = 30 - 22 = 8 \]

### Final Answer

Thus, the value of \( a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13} \) is

\[ \boxed{8} \]