# 26. Determinants: examples

## 26. Determinants: examples

Example:

Let $A=\begin{pmatrix}1&4&-4\\ -2&-2&-4\\ 3&-3&3\end{pmatrix}$ . Rather than applying our formula (sum over permutations), we're going to apply type I row operations to put this matrix into echelon form; this won't change the determinant, so we can take the determinant afterwards (at which point it will be the product of diagonal entries).

To clear the first column, we do $R_{2}\mapsto R_{2}+2R_{1}$ and $R_{3}\mapsto R_{3}-3R_{1}$ which gives: $\begin{pmatrix}1&4&-4\\ 0&6&-12\\ 0&-15&15\end{pmatrix}$ To clear the $-15$ below the diagonal, we do $R_{3}\mapsto R_{3}+\frac{5}{2}R_{2}$ , giving: $\begin{pmatrix}1&4&-4\\ 0&6&-12\\ 0&0&-15\end{pmatrix}.$ This is in echelon form. Therefore $\det(A)=1\times 6\times(-15)=-90$ .

Example:

Let $B=\begin{pmatrix}2&-3&-1&4\\ 2&-3&2&4\\ 2&-1&-4&-3\\ 2&-3&4&2\end{pmatrix}$ . Let's clear the first column with $R_{2}\mapsto R_{2}-R_{1}$ , $R_{3}\mapsto R_{3}-R_{1}$ , $R_{4}\mapsto R_{4}-R_{1}$ , yielding: $\begin{pmatrix}2&-3&-1&4\\ 0&0&3&0\\ 0&2&-3&-7\\ 0&0&5&-2\end{pmatrix}$ Now let's swap $R_{2}\leftrightarrow R_{3}$ : $\begin{pmatrix}2&-3&-1&4\\ 0&2&-3&-7\\ 0&0&3&0\\ 0&0&5&-2\end{pmatrix}.$ The determinant just changed sign! We'll need to remember to stick an extra minus sign in at the end. To clear the final $5$ below the diagonal, we do $R_{4}\mapsto R_{4}-\frac{5}{3}R_{3}$ : $\begin{pmatrix}2&-3&-1&4\\ 0&2&-3&-7\\ 0&0&3&0\\ 0&0&0&-2\end{pmatrix}$ which is now in echelon form.

The determinant of $B$ is therefore $-(2\times 2\times 3\times-2)=24$ .