26. Determinants: examples

Example:

0.00 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).

0.58 To clear the first column, we do $R_{2}\mapsto R_{2}+2R_{1}$ and $R_{3}\mapsto R_{3}-3R_{1}$ which gives: $\begin{pmatrix}1&4&-4\\ 0&6&-12\\ 0&-15&15\end{pmatrix}$ 2.36 To clear the $-15$ below the diagonal, we do $R_{3}\mapsto R_{3}+\frac{5}{2}R_{2}$ , giving: $\begin{pmatrix}1&4&-4\\ 0&6&-12\\ 0&0&-15\end{pmatrix}.$ 3.41 This is in echelon form. Therefore $\det(A)=1\times 6\times(-15)=-90$ .

Example:

4.37 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: $\begin{pmatrix}2&-3&-1&4\\ 0&0&3&0\\ 0&2&-3&-7\\ 0&0&5&-2\end{pmatrix}$ 6.56 Now let's swap $R_{2}\leftrightarrow R_{3}$ : $\begin{pmatrix}2&-3&-1&4\\ 0&2&-3&-7\\ 0&0&3&0\\ 0&0&5&-2\end{pmatrix}.$ 7.44 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.

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