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

Mar 11, 2016

$\det = 4$

#### Explanation:

$D = 1 | \left(4 , 3\right) , \left(2 , 2\right) | - 2 | \left(0 , 3\right) , \left(1 , 2\right) | + 1 | \left(0 , 4\right) , \left(1 , 2\right) |$

$= 1 \left(8 - 6\right) - 2 \left(0 - 3\right) + 1 \left(0 - 4\right)$-use Cramer's rule $| \left(r , s\right) , \left(t , u\right) | = r u - s t$

$= 2 + 6 - 4$
$\det = 4$

Mar 11, 2016

Use alternative method to find determinant is $4$

#### Explanation:

Use two properties of determinants:

1. The determinant of a matrix is unchanged by adding any multiple of a row to another row.
2. The determinant of an upper triangular matrix is the product of the diagonal.

Given:

$\left(\begin{matrix}1 & 2 & 1 \\ 0 & 4 & 3 \\ 1 & 2 & 2\end{matrix}\right)$

Subtract the first row from the third (i.e. add $- 1$ times the first row to the third) to get:

$\left(\begin{matrix}1 & 2 & 1 \\ 0 & 4 & 3 \\ 0 & 0 & 1\end{matrix}\right)$

The determinant of this upper triangular matrix is the product of the diagonal:

$1 \times 4 \times 1 = 4$