Using Cofactors of elements of second row, evaluate `Delta=|(5, 3, 8),( 2, 0, 1),( 1, 2 ,3)|`

Options

A. `7`
B. `8`
C. `9`
D. `10`

Explanation

To evaluate the determinant \( \Delta = \begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix} \) using the cofactors of the elements of the second row, we will follow these steps:

### Step 1: Calculate the Minors of the Elements in the Second Row
The second row consists of the elements \( 2, 0, 1 \). We will calculate the minors for each of these elements.

1. **Minor \( M_{21} \)** (for element 2):
- Delete the second row and the first column:
\[ M_{21} = \begin{vmatrix} 3 & 8 \\ 2 & 3 \end{vmatrix} = (3)(3) - (8)(2) = 9 - 16 = -7 \]

2. **Minor \( M_{22} \)** (for element 0):
- Delete the second row and the second column:
\[ M_{22} = \begin{vmatrix} 5 & 8 \\ 1 & 3 \end{vmatrix} = (5)(3) - (8)(1) = 15 - 8 = 7 \]

3. **Minor \( M_{23} \)** (for element 1):
- Delete the second row and the third column:
\[ M_{23} = \begin{vmatrix} 5 & 3 \\ 1 & 2 \end{vmatrix} = (5)(2) - (3)(1) = 10 - 3 = 7 \]

### Step 2: Calculate the Cofactors
Now we will calculate the cofactors using the minors.

1. **Cofactor \( A_{21} \)**:
\[ A_{21} = (-1)^{2+1} M_{21} = -(-7) = 7 \]

2. **Cofactor \( A_{22} \)**:
\[ A_{22} = (-1)^{2+2} M_{22} = 1 \cdot 7 = 7 \]

3. **Cofactor \( A_{23} \)**:
\[ A_{23} = (-1)^{2+3} M_{23} = -7 \]

### Step 3: Apply the Cofactor Expansion
Now we will use the cofactors and the elements of the second row to evaluate the determinant.

\[ \Delta = a_{21} A_{21} + a_{22} A_{22} + a_{23} A_{23} \]
Where \( a_{21} = 2, a_{22} = 0, a_{23} = 1 \).

Substituting the values:
\[ \Delta = 2 \cdot 7 + 0 \cdot 7 + 1 \cdot (-7) = 14 + 0 - 7 = 7 \]

### Final Result
Thus, the value of the determinant is:
\[ \Delta = 7 \]