## Problem on computing a determinant by minors and cofactors

Compute the determinant $$\text{det } \begin{pmatrix} 1 & 5 & 0 \\ 2 & 1 & 0 \\ 1 & 0 & 3 \end{pmatrix}$$ by minors and cofactors along the second column.
• ## Solution

Let $A$ be the matrix we are seeking the determinant of.
Recall that
Step 1 - We choose to expand across the second column, as directed.
Step 2 - We now form the minors for each entry in the second column.
The minor of the 1,2 entry is the determinant of the submatrix formed by removing the first row and second column: $$\begin{vmatrix} 2 & 0 \\1 & 3\end{vmatrix}.$$
The minor of the 2,2 entry is the determinant of the submatrix formed by removing the second row and the second column: $$\begin{vmatrix} 1 & 0 \\1 & 3\end{vmatrix}.$$
The minor of the 3,2 entry is the determinant of the submatrix formed by removing the third row and the second column: $$\begin{vmatrix} 1 & 0 \\2 & 0\end{vmatrix}.$$
Step 3 - We can write the minor and cofactor expansion along the second column:
\begin{align} \text{det } A = & -5\ \begin{vmatrix} 2& 0 \\ 1 & 3 \end{vmatrix} + 1\ \begin{vmatrix}1 & 0 \\ 1 & 3 \end{vmatrix} - 0\ \begin{vmatrix}1 & 0 \\ 2 & 0 \end{vmatrix} \end{align}
We double check that plus/minus signs agree with the following checkerboard pattern:
The 2,2 position is being added, while the 1,2 and 3,2 positions are being subtracted. Hence, the expansion is correct.
Hence, $$\text{det } A = -5 \cdot 6 + 1 \cdot 3 - 0 \cdot 0 = -27.$$