# How do you find the determinant of ((1, 7, -4, -5), (1, 5, 7, -4), (3, 21, -12, -15), (5, 25, -8, 0))?

Feb 16, 2016

This is a long recursive method. Details are given below.

#### Explanation:

To find the determinant of a 4x4 Matrix, one has to go through a
recursive process working way down from the 4x4 determinant to 3x3 determinants and then to 2x2 determinants.

The determinant $\left(\begin{matrix}1 & 7 & - 4 & - 5 \\ 1 & 5 & 7 & - 4 \\ 3 & 21 & - 12 & - 15 \\ 5 & 25 & - 8 & 0\end{matrix}\right)$ can be written as

$1 \left(\begin{matrix}5 & 7 & - 4 \\ 21 & - 12 & - 15 \\ 25 & - 8 & 0\end{matrix}\right) - 7 \left(\begin{matrix}1 & 7 & - 4 \\ 3 & - 12 & - 15 \\ 5 & - 8 & 0\end{matrix}\right) + \left(- 4\right) \left(\begin{matrix}1 & 5 & - 4 \\ 3 & 21 & - 15 \\ 5 & - 8 & 0\end{matrix}\right) - \left(- 5\right) \left(\begin{matrix}1 & 5 & 7 \\ 3 & 21 & - 12 \\ 5 & 25 & - 8\end{matrix}\right)$ (A)

Here we have chosen first row as multipliers to each $3 X 3$ matrix and latter is formed by eliminating rows and columns of that multiplier. Also note that starting from each multiplier as positive, alternate multipliers are of opposite sign.

To find the value of each $3 X 3$ matrix, say $\left(\begin{matrix}5 & 7 & - 4 \\ 21 & - 12 & - 15 \\ 25 & - 8 & 0\end{matrix}\right)$, let us say one chooses again first row of this $3 X 3$ determinant and it is equal to

$5 \left(\begin{matrix}- 12 & - 15 \\ - 8 & 0\end{matrix}\right) - 7 \left(\begin{matrix}21 & - 15 \\ 25 & 0\end{matrix}\right) + \left(- 4\right) \left(\begin{matrix}21 & - 12 \\ 25 & - 8\end{matrix}\right)$

This way expand each $3 X 3$ determinant into $2 X 2$ determinants.

Value of a $2 X 2$ determinant $\left(\begin{matrix}a & b \\ c & d\end{matrix}\right)$ is given by $\left(a d - b c\right)$