Matrix `A=[(1,2,3),(1,1,5),(2,4,7)]`, then the value of `a_(31)A_(31)+a_(32)A_(32)+a_(33)A_(33)` is

Options

A. 1
B. 13
C. `-1`
D. `-13`

Explanation

To solve the problem, we need to calculate the expression \( a_{31}A_{31} + a_{32}A_{32} + a_{33}A_{33} \) for the given matrix \( A \).

The matrix \( A \) is given as:
\[ A = \begin{pmatrix} 1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7 \end{pmatrix} \]

### Step 1: Identify the elements \( a_{31}, a_{32}, a_{33} \)
- \( a_{31} \) is the element in the 3rd row and 1st column of matrix \( A \):
\[ a_{31} = 2 \]
- \( a_{32} \) is the element in the 3rd row and 2nd column of matrix \( A \):
\[ a_{32} = 4 \]
- \( a_{33} \) is the element in the 3rd row and 3rd column of matrix \( A \):
\[ a_{33} = 7 \]

### Step 2: Calculate the cofactors \( A_{31}, A_{32}, A_{33} \)
- **Cofactor \( A_{31} \)**:
\[ A_{31} = (-1)^{3+1} \cdot \text{det} \begin{pmatrix} 1 & 5 \\ 2 & 7 \end{pmatrix} \]
The determinant is calculated as:
\[ \text{det} = (1)(7) - (5)(2) = 7 - 10 = -3 \]
Thus,
\[ A_{31} = 1 \cdot (-3) = -3 \]

- **Cofactor \( A_{32} \)**:
\[ A_{32} = (-1)^{3+2} \cdot \text{det} \begin{pmatrix} 1 & 5 \\ 1 & 7 \end{pmatrix} \]
The determinant is calculated as:
\[ \text{det} = (1)(7) - (5)(1) = 7 - 5 = 2 \]
Thus,
\[ A_{32} = -1 \cdot 2 = -2 \]

- **Cofactor \( A_{33} \)**:
\[ A_{33} = (-1)^{3+3} \cdot \text{det} \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix} \]
The determinant is calculated as:
\[ \text{det} = (1)(1) - (2)(1) = 1 - 2 = -1 \]
Thus,
\[ A_{33} = 1 \cdot (-1) = -1 \]

### Step 3: Substitute values into the expression
Now we substitute the values into the expression \( a_{31}A_{31} + a_{32}A_{32} + a_{33}A_{33} \):
\[ = 2 \cdot (-3) + 4 \cdot (-2) + 7 \cdot (-1) \]
Calculating each term:
- \( 2 \cdot (-3) = -6 \)
- \( 4 \cdot (-2) = -8 \)
- \( 7 \cdot (-1) = -7 \)

Now, summing these values:
\[ -6 - 8 - 7 = -21 \]

### Final Answer
Thus, the value of \( a_{31}A_{31} + a_{32}A_{32} + a_{33}A_{33} \) is:
\[ \boxed{-21} \]