The expansion of the determinant along the second row is:

Steps to Solve

1. Identify the determinant and its entries

The determinant to be expanded is given as:

$$ D = \begin{vmatrix} 4 & 5 & 6 \ -1 & 7 & 3 \ 2 & 3 & 8 \end{vmatrix} $$

2. Use the second row for expansion

For the expansion along the second row, we will use the formula:

$$ D = a_{21} \cdot D_{21} + a_{22} \cdot D_{22} + a_{23} \cdot D_{23} $$

where $a_{ij}$ are the elements of the second row and $D_{ij}$ are the corresponding minors.

3. Calculate the determinants of minors

• First Element: $a_{21} = -1$

The minor $D_{21}$ is:

$$ D_{21} = \begin{vmatrix} 5 & 6 \ 3 & 8 \end{vmatrix} = (5 \cdot 8) - (6 \cdot 3) = 40 - 18 = 22 $$

• Second Element: $a_{22} = 7$

The minor $D_{22}$ is:

$$ D_{22} = \begin{vmatrix} 4 & 6 \ 2 & 8 \end{vmatrix} = (4 \cdot 8) - (6 \cdot 2) = 32 - 12 = 20 $$

• Third Element: $a_{23} = 3$

The minor $D_{23}$ is:

$$ D_{23} = \begin{vmatrix} 4 & 5 \ 2 & 3 \end{vmatrix} = (4 \cdot 3) - (5 \cdot 2) = 12 - 10 = 2 $$

4. Putting it all together

Now plug in the values:

$$ D = (-1) \cdot 22 + 7 \cdot 20 + 3 \cdot 2 $$

Calculating gives:

$$ D = -22 + 140 + 6 = 124 $$