Question

What is the shortcut to finding the determinant of a $5\times 5$ matrix?

Answer 1

The determinant is used in the square matrix and is a scalar value.

Laplace expansion is used to determine the determinant of a $5\times 5$ matrix.

• First, the cofactors of every number are found in that row and column, by applying the cofactor formula ${\left(-1\right)}^{\mathrm{i}+\mathrm{j}}\left|{\mathrm{A}}_{\left(\mathrm{i},\mathrm{j}\right)}\right|$, where $\mathrm{i}$ is the row number and $\mathrm{j}$is the column number.
• After that, the cofactor is multiplied by each number in that row or column.
• Then the products are added up.

Assume a matrix $A=\left[\begin{array}{ccccc}2& 0& 0& -3& 0\\ 1& 6& 4& 6& 1\\ -2& 0& 0& 1& 4\\ 2& 0& 0& 3& -2\\ 4& 0& 0& 0& 2\end{array}\right]$

Select a row or column. Here take the second column and apply the formula.

$6{\left(-1\right)}^{2+2}\left|\begin{array}{cccc}2& 0& -3& 0\\ -2& 0& 1& 4\\ 2& 0& 3& -2\\ 4& 0& 0& 2\end{array}\right|$

$=6\left|\begin{array}{cccc}2& 0& -3& 0\\ -2& 0& 1& 4\\ 2& 0& 3& -2\\ 4& 0& 0& 2\end{array}\right|$

${\mathrm{R}}_4={\mathrm{R}}_4\times ({\mathrm{R}}_2+{\mathrm{R}}_3)$

$=6\left|\begin{array}{cccc}2& 0& -3& 0\\ -2& 0& 1& 4\\ 2& 0& 3& -2\\ 0& 0& 0& 4\end{array}\right|$

Select the fourth row, and apply the formula.

$6\left(4{\left(-1\right)}^{4+4}\left|\begin{array}{ccc}2& 0& -3\\ -2& 0& 1\\ 2& 0& 3\end{array}\right|\right)$

$=6\left(4\left|\begin{array}{ccc}2& 0& -3\\ -2& 0& 1\\ 2& 0& 3\end{array}\right|\right)$

$=24\left|\begin{array}{ccc}2& 0& -3\\ -2& 0& 1\\ 2& 0& 3\end{array}\right|$

${\mathrm{R}}_1={\mathrm{R}}_1+{\mathrm{R}}_3$

$=24\left|\begin{array}{ccc}4& 0& 0\\ -2& 0& 1\\ 2& 0& 3\end{array}\right|$

Select the first row and apply the formula.

$24\left(4{(-1)}^{1+1}\left|\begin{array}{cc}0& 1\\ 0& 3\end{array}\right|\right)$

$=24\left(4\left|\begin{array}{cc}0& 1\\ 0& 3\end{array}\right|\right)$

$=96\left|\begin{array}{cc}0& 1\\ 0& 3\end{array}\right|$

Select the first row.

$96\left({\left(-1\right)}^{1+2}\left|0\right|\right)$

$=96\left(-1\left(0\right)\right)$

$=96\left(0\right)$

$=0$

So, the determinant is $0$.