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

Nov 29, 2016

$| \left(1 , - 2 , 3 , 2\right) , \left(2 , - 1 , 0 , 4\right) , \left(- 3 , 4 , 0 , - 2\right) , \left(- 1 , 1 , 1 , 5\right) | = 81$

#### Explanation:

• You can add or subtract multiples of any row or column to any other row or column without altering the determinant.

• You can swap two rows or columns, resulting in inverting the sign of the determinant.

• You can cyclically permute three rows or columns to leave the determinant unchanged (since this is equivalent to two swaps).

• The determinant of an upper or lower triangular matrix is the product of the diagonal.

So:

$| \left(1 , - 2 , 3 , 2\right) , \left(2 , - 1 , 0 , 4\right) , \left(- 3 , 4 , 0 , - 2\right) , \left(- 1 , 1 , 1 , 5\right) | \text{ }$ subtract $3$ times row $4$ from row $1$ row to get:

$= | \left(4 , - 5 , 0 , - 13\right) , \left(2 , - 1 , 0 , 4\right) , \left(- 3 , 4 , 0 , - 2\right) , \left(- 1 , 1 , 1 , 5\right) | \text{ }$ swap columns $1$ and $3$ to get:

$= - | \left(0 , - 5 , 4 , - 13\right) , \left(0 , - 1 , 2 , 4\right) , \left(0 , 4 , - 3 , - 2\right) , \left(1 , 1 , - 1 , 5\right) | \text{ }$ swap rows $1$ and $4$ to get:

$= | \left(1 , 1 , - 1 , 5\right) , \left(0 , - 1 , 2 , 4\right) , \left(0 , 4 , - 3 , - 2\right) , \left(0 , - 5 , 4 , - 13\right) | \text{ }$ add $4$ times row $2$ to row $3$ to get:

$= | \left(1 , 1 , - 1 , 5\right) , \left(0 , - 1 , 2 , 4\right) , \left(0 , 0 , 5 , 14\right) , \left(0 , - 5 , 4 , - 13\right) | \text{ }$ subtract $5$ times row $2$ from row $4$ to get:

$= | \left(1 , 1 , - 1 , 5\right) , \left(0 , - 1 , 2 , 4\right) , \left(0 , 0 , 5 , 14\right) , \left(0 , 0 , - 6 , - 33\right) | \text{ }$ add $\frac{6}{5}$ times row $3$ to row $4$ to get:

$= | \left(1 , 1 , - 1 , 5\right) , \left(0 , - 1 , 2 , 4\right) , \left(0 , 0 , 5 , 14\right) , \left(0 , 0 , 0 , - \frac{81}{5}\right) |$

$= 1 \cdot \left(- 1\right) \cdot 5 \cdot \left(- \frac{81}{5}\right) = 81$