# How do you find the determinant of ((9, -4, 3, 5), (2, 7, 6, -5), (4, 1, -2, 0), (7, 3, 4, 10))?

Jun 25, 2017

The answer is $= - 4025$

#### Explanation:

We reduce the matrix to row echelon form

$A = \left(\begin{matrix}9 & - 4 & 3 & 5 \\ 2 & 7 & 6 & - 5 \\ 4 & 1 & - 2 & 0 \\ 7 & 3 & 4 & 10\end{matrix}\right)$

We perform

${R}_{2} \leftarrow {R}_{2} - \frac{2}{9} {R}_{1}$

$= \left(\begin{matrix}9 & - 4 & 3 & 5 \\ 0 & \frac{71}{9} & \frac{16}{3} & - \frac{55}{9} \\ 4 & 1 & - 2 & 0 \\ 7 & 3 & 4 & 10\end{matrix}\right)$

${R}_{3} \leftarrow {R}_{3} - \frac{4}{9} {R}_{1}$

$= \left(\begin{matrix}9 & - 4 & 3 & 5 \\ 0 & \frac{71}{9} & \frac{16}{3} & - \frac{55}{9} \\ 0 & \frac{25}{9} & - \frac{10}{3} & - \frac{2}{90} \\ 7 & 3 & 4 & 10\end{matrix}\right)$

${R}_{4} \leftarrow {R}_{4} - \frac{7}{9} {r}_{1}$

$= \left(\begin{matrix}9 & - 4 & 3 & 5 \\ 0 & \frac{71}{9} & \frac{16}{3} & - \frac{55}{9} \\ 0 & \frac{25}{9} & - \frac{10}{3} & - \frac{2}{90} \\ 0 & \frac{55}{9} & \frac{5}{3} & \frac{55}{9}\end{matrix}\right)$

${R}_{3} \leftarrow {R}_{3} - \frac{25}{71} R - 2$

$= \left(\begin{matrix}9 & - 4 & 3 & 5 \\ 0 & \frac{71}{9} & \frac{16}{3} & - \frac{55}{9} \\ 0 & 0 & - \frac{370}{71} & - \frac{5}{71} \\ 0 & \frac{55}{9} & \frac{5}{3} & \frac{55}{9}\end{matrix}\right)$

${R}_{4} \leftarrow {R}_{4} - \frac{55}{71} {R}_{2}$

$= \left(\begin{matrix}9 & - 4 & 3 & 5 \\ 0 & \frac{71}{9} & \frac{16}{3} & - \frac{55}{9} \\ 0 & 0 & - \frac{370}{71} & - \frac{5}{71} \\ 0 & 0 & - \frac{175}{71} & \frac{770}{71}\end{matrix}\right)$

${R}_{4} \leftarrow {R}_{4} - \frac{35}{74} {R}_{3}$

$= \left(\begin{matrix}9 & - 4 & 3 & 5 \\ 0 & \frac{71}{9} & \frac{16}{3} & - \frac{55}{9} \\ 0 & 0 & - \frac{370}{71} & - \frac{5}{71} \\ 0 & 0 & 0 & \frac{805}{74}\end{matrix}\right)$

The determinant of the matrix is equal to the diagonal product of the matrix

$\det \left(A\right) = 9 \cdot \frac{71}{9} \cdot \left(- \frac{370}{71}\right) \cdot \frac{805}{74} = - 5 \cdot 805 = - 4025$