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

Mar 4, 2016

The determinant is $0$.

Explanation:

There are several ways to compute the determinant.

On of those is the following formula for $3 \times 3$ matrix:

For a matrix

$A = \left(\begin{matrix}a & b & c \\ d & e & f \\ g & h & i\end{matrix}\right)$,

the determinant can found with

$\det A = a \cdot e \cdot i + d \cdot h \cdot c + g \cdot b \cdot f$

$\text{ } - c \cdot e \cdot g - f \cdot h \cdot a - i \cdot b \cdot d$.

In your case, this means:

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

$= 1 \cdot 0 \cdot 3 + \left(- 2\right) \cdot 4 \cdot 1 + 1 \cdot 2 \cdot 2 - 1 \cdot 0 \cdot 1 - 2 \cdot 4 \cdot 1 - 3 \cdot 2 \cdot \left(- 2\right)$

$= 0 - 8 + 4 - 0 - 8 + 12$

$= 0$

By the way, this means that the matrix doesn't have an inverse, and that any linear equations solved with this matrix will not have a unique solution but either infinitely many solutions or none at all.