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: Write the determinant and identify the second row
We have the determinant:
\[ \Delta = \begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix} \]
The second row is \((2, 0, 1)\).

### Step 2: Expand the determinant using the second row
Using the second row, we can expand the determinant as follows:
\[ \Delta = 2 \cdot C_{21} + 0 \cdot C_{22} + 1 \cdot C_{23} \]
where \(C_{ij}\) is the cofactor of the element in the \(i^{th}\) row and \(j^{th}\) column.

### Step 3: Calculate the cofactors
#### Cofactor \(C_{21}\)
The cofactor \(C_{21}\) is calculated as:
\[ C_{21} = (-1)^{2+1} \cdot M_{21} \]
where \(M_{21}\) is the minor of the element at position (2,1). To find \(M_{21}\), we remove the second row and first column:
\[ M_{21} = \begin{vmatrix} 3 & 8 \\ 2 & 3 \end{vmatrix} = (3 \cdot 3) - (8 \cdot 2) = 9 - 16 = -7 \]
Thus,
\[ C_{21} = -1 \cdot (-7) = 7 \]

#### Cofactor \(C_{22}\)
Since the element is multiplied by 0, we do not need to calculate \(C_{22}\).

#### Cofactor \(C_{23}\)
The cofactor \(C_{23}\) is calculated as:
\[ C_{23} = (-1)^{2+3} \cdot M_{23} \]
where \(M_{23}\) is the minor of the element at position (2,3). To find \(M_{23}\), we remove the second row and third column:
\[ M_{23} = \begin{vmatrix} 5 & 3 \\ 1 & 2 \end{vmatrix} = (5 \cdot 2) - (3 \cdot 1) = 10 - 3 = 7 \]
Thus,
\[ C_{23} = 1 \cdot 7 = 7 \]

### Step 4: Substitute the cofactors back into the determinant expansion
Now substituting the cofactors back into the determinant expansion:
\[ \Delta = 2 \cdot 7 + 0 \cdot C_{22} + 1 \cdot 7 = 14 + 0 + 7 = 21 \]

### Final Result
Thus, the value of the determinant \(\Delta\) is:
\[ \Delta = 21 \]