If `A_(ij)` denotes the cofactor of the element `a_(ij)` of the determinant `|(2,-3,5),(6,0,4),(1,5,-7)|` the value of `a_11A_31+a_12A_32+a_13A_33` is:

Options

A. 0
B. 5
C. 10
D. `-5`

Answer

To solve the problem, we need to find the value of the expression \( a_{11}A_{31} + a_{12}A_{32} + a_{13}A_{33} \) where \( A_{ij} \) denotes the cofactor of the element \( a_{ij} \) of the determinant \( |(2, -3, 5), (6, 0, 4), (1, 5, -7)| \).

### Step 1: Identify the elements \( a_{ij} \)
The elements of the matrix are:
- \( a_{11} = 2 \)
- \( a_{12} = -3 \)
- \( a_{13} = 5 \)

### Step 2: Calculate the cofactors \( A_{31}, A_{32}, A_{33} \)

#### Finding \( A_{31} \)
To find \( A_{31} \), we need to calculate the determinant of the submatrix obtained by removing the third row and first column:

\[ \begin{vmatrix} -3 & 5 \\ 0 & 4 \end{vmatrix} \]

Calculating this determinant:
\[ (-3) \cdot 4 - (5) \cdot 0 = -12 \]
Since \( A_{31} = (-1)^{3+1} \cdot (-12) = 1 \cdot (-12) = -12 \).

#### Finding \( A_{32} \)
To find \( A_{32} \), we need to calculate the determinant of the submatrix obtained by removing the third row and second column:

\[ \begin{vmatrix} 2 & 5 \\ 6 & 4 \end{vmatrix} \]

Calculating this determinant:
\[ (2) \cdot (4) - (5) \cdot (6) = 8 - 30 = -22 \]
Since \( A_{32} = (-1)^{3+2} \cdot (-22) = -1 \cdot (-22) = 22 \).

#### Finding \( A_{33} \)
To find \( A_{33} \), we need to calculate the determinant of the submatrix obtained by removing the third row and third column:

\[ \begin{vmatrix} 2 & -3 \\ 6 & 0 \end{vmatrix} \]

Calculating this determinant:
\[ (2) \cdot (0) - (-3) \cdot (6) = 0 + 18 = 18 \]
Since \( A_{33} = (-1)^{3+3} \cdot 18 = 1 \cdot 18 = 18 \).

### Step 3: Substitute the values into the expression
Now we can substitute the values into the expression \( a_{11}A_{31} + a_{12}A_{32} + a_{13}A_{33} \):

\[ = 2 \cdot (-12) + (-3) \cdot 22 + 5 \cdot 18 \]

Calculating each term:
- \( 2 \cdot (-12) = -24 \)
- \( -3 \cdot 22 = -66 \)
- \( 5 \cdot 18 = 90 \)

Now, summing these values:
\[ -24 - 66 + 90 = 0 \]

### Final Answer
Thus, the value of \( a_{11}A_{31} + a_{12}A_{32} + a_{13}A_{33} \) is \( 0 \).

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