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

Mar 21, 2016

Transform into an upper triangular matrix then take the product of the diagnonal to find:

$\left\mid \begin{matrix}2 & 1 & 0 & 0 \\ 1 & 2 & 1 & 0 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 1 & 2\end{matrix} \right\mid = 5$

#### Explanation:

The determinant is unchanged by adding any multiple of one row to another row, or column to another column.

Starting with:

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

Subtract $\frac{1}{2}$ of the first row from the second to get:

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

Subtract $\frac{2}{3}$ of the second row from the third to get:

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

Subtract $\frac{3}{4}$ of the third row from the fourth to get:

$\left(\begin{matrix}2 & 1 & 0 & 0 \\ 0 & \frac{3}{2} & 1 & 0 \\ 0 & 0 & \frac{4}{3} & 1 \\ 0 & 0 & 0 & \frac{5}{4}\end{matrix}\right)$

The determinant of an upper triangular matrix is the product of the diagonal, so:

$\left\mid \begin{matrix}2 & 1 & 0 & 0 \\ 0 & \frac{3}{2} & 1 & 0 \\ 0 & 0 & \frac{4}{3} & 1 \\ 0 & 0 & 0 & \frac{5}{4}\end{matrix} \right\mid = \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \cdot \frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{3}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} \cdot \frac{\textcolor{g r e e n}{\cancel{\textcolor{b l a c k}{4}}}}{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{3}}}} \cdot \frac{5}{\textcolor{g r e e n}{\cancel{\textcolor{b l a c k}{4}}}} = 5$