# GAUSS elimination ¶

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To find the determinant, you need to reduce a matrix to an upper diagonal matrix, then use the property "$det(A)$=product of the values on the diagonal" to find the determinant.

If you swapped lines, then multiply the result by

• $det(A) = det(A) * (-1)^s$
• with $s$ the number of swaps

## Example 1 ¶

What's the determinant of A?

$A = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix}$

$\begin{split}\begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix} \Leftrightarrow^{L1 \leftrightarrow L2} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix} \Leftrightarrow^{L3 \leftrightarrow L4} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\end{split}$

The determinant is $det(A) = 1*1*1*1 = 1$ but since we swapped lines 2 times, then $det(A) = 1 * (-1)^2 = 1$.

## Example 2 ¶

What's the determinant of A?

$A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{pmatrix}$

$\begin{split}\begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{pmatrix} \Leftrightarrow^{L_2 \leftarrow L_2 + -2*L_1 } \begin{pmatrix} 1 & 2 & 3 \\ 0 & -1 & -2 \\ 3 & 4 & 5 \end{pmatrix} \Leftrightarrow^{L_3 \leftarrow L_3 + -3*L_1 } \\ \begin{pmatrix} 1 & 2 & 3 \\ 0 & -1 & -2 \\ 0 & -2 & -4 \end{pmatrix} \Leftrightarrow^{L_3 \leftarrow L_3 + 2*L_2 } \begin{pmatrix} 1 & 2 & 3 \\ 0 & -1 & -2 \\ 0 & 0 & 0 \end{pmatrix}\end{split}$

The determinant is $det(A) = 1*-1*0 = 0$.