Let `A = (a_(ij))_(n xx n)` and adj A = `(alpha_(ij))`
If A = `[{:(1,2,3),(4,5,4),(2,3,-1):}]`, what is the value of `alpha_(23)` ?

Options

A. 1
B. -1
C. 8
D. -8

Explanation

To find the value of \( \alpha_{23} \) (the element in the 2nd row and 3rd column of the adjoint of matrix \( A \)), we first need to understand that the adjoint of a matrix is related to the cofactors of the matrix.

Given the matrix:
\[ A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 4 \\ 2 & 3 & -1 \end{pmatrix} \]

### Step 1: Identify the element \( A_{32} \)
The element \( A_{32} \) is located in the 3rd row and 2nd column of matrix \( A \).
\[ A_{32} = 3 \]

**Hint:** To find \( \alpha_{23} \), we need to calculate the cofactor of \( A_{32} \).

### Step 2: Calculate the cofactor \( C_{32} \)
The cofactor \( C_{32} \) is given by:
\[ C_{32} = (-1)^{3+2} \cdot \text{det}(M_{32}) \]
where \( M_{32} \) is the minor matrix obtained by deleting the 3rd row and 2nd column of \( A \).

**Hint:** Remember that the sign of the cofactor depends on the position of the element in the matrix.

### Step 3: Form the minor matrix \( M_{32} \)
To form \( M_{32} \), we remove the 3rd row and 2nd column from \( A \):
\[ M_{32} = \begin{pmatrix} 1 & 3 \\ 4 & 4 \end{pmatrix} \]

**Hint:** The minor matrix is formed by excluding the row and column of the element for which we are calculating the cofactor.

### Step 4: Calculate the determinant of \( M_{32} \)
Now we calculate the determinant of \( M_{32} \):
\[ \text{det}(M_{32}) = (1)(4) - (3)(4) = 4 - 12 = -8 \]

**Hint:** Use the formula for the determinant of a 2x2 matrix, which is \( ad - bc \).

### Step 5: Calculate the cofactor \( C_{32} \)
Now substituting back into the cofactor formula:
\[ C_{32} = (-1)^{5} \cdot (-8) = -(-8) = 8 \]

**Hint:** The exponent \( (-1)^{m+n} \) gives the sign for the cofactor, where \( m \) and \( n \) are the row and column indices of the element.

### Step 6: Find \( \alpha_{23} \)
Since \( \alpha_{23} \) is equal to the cofactor \( C_{32} \):
\[ \alpha_{23} = C_{32} = 8 \]

Thus, the value of \( \alpha_{23} \) is:
\[ \boxed{8} \]