# How do you find the determinant of ((1, 2, 0, 0, 4), (0, 3, 0, 0, 5), (0, 1, 0, 2, 0), (3, 0, 0, 1, -1), (0, -2, -1, 0, -3))?

Mar 2, 2016

$\det A = 13$

#### Explanation:

For a matrix with that many zeroes, the Laplace expansion method would be a good way to compute the determinant.

First of all, you should pick a row or a column with the most amount of zeroes. This one looks like a good pick to me:

( (1, 2, color(red)(0), 0, 4), (0, 3, color(red)(0), 0, 5), (0, 1, color(red)(0), 2, 0), (3, 0, color(red)(0), 1, -1), (0, -2, color(red)(-1), 0, -3))

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Let me remind you very shortly how to use the Laplace expansion here:

• You need to take every element of the chosen column/row one by one with alternating signs.
• The signs can be determined by ${\left(- 1\right)}^{i + j}$ where $i$ is the row number and $j$ is the column number, so in our case the sign of e.g. the first zero in the first row, third column is ${\left(- 1\right)}^{1 + 3} = {\left(- 1\right)}^{4} = + 1$, so a "$+$". The signs after
that will alternate.
• Each element ${a}_{i j}$ needs to be multiplied with the determinant of the matrix that remains if you delete the $i$-th row and the $j$-th column.
• Your determinante is the sum of the products ${a}_{i j}$ times determinants of submatrices.

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So, let's expand along that column:

det ( (1, 2, color(red)(0)^(color(blue)(+)), 0, 4), (0, 3, color(red)(0)^(color(blue)(-)), 0, 5), (0, 1, color(red)(0)^(color(blue)(+)), 2, 0), (3, 0, color(red)(0)^(color(blue)(-)), 1, -1), (0, -2, color(red)(-1)^(color(blue)(+)), 0, -3))

= 0 * det (...) - 0 * det (...) + 0 * det (...) - 0 * det (...) + (-1) * det ((1, 2, 0, 4), (0, 3, 0, 5), (0, 1, 2, 0), (3, 0, 1, -1))

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

It's easier already. Now, let's take this $4 \times 4$ matrix and again, choose a row or column with a maximum amount of zeroes. This time, there are several good possibilities. Let me pick one randomly:

$= - \det \left(\begin{matrix}1 & 2 & 0 & 4 \\ {\textcolor{red}{0}}^{\textcolor{b l u e}{-}} & {\textcolor{red}{3}}^{\textcolor{b l u e}{+}} & {\textcolor{red}{0}}^{\textcolor{b l u e}{-}} & {\textcolor{red}{5}}^{\textcolor{b l u e}{+}} \\ 0 & 1 & 2 & 0 \\ 3 & 0 & 1 & - 1\end{matrix}\right)$

$= - \left[- 0 \cdot \det \left(\ldots\right) + 3 \cdot \det \left(\begin{matrix}1 & 0 & 4 \\ 0 & 2 & 0 \\ 3 & 1 & - 1\end{matrix}\right) - 0 \cdot \det \left(\ldots\right) + 5 \cdot \det \left(\begin{matrix}1 & 2 & 0 \\ 0 & 1 & 2 \\ 3 & 0 & 1\end{matrix}\right)\right]$

$= - 3 \cdot \det \left(\begin{matrix}1 & 0 & 4 \\ 0 & 2 & 0 \\ 3 & 1 & - 1\end{matrix}\right) - 5 \cdot \det \left(\begin{matrix}1 & 2 & 0 \\ 0 & 1 & 2 \\ 3 & 0 & 1\end{matrix}\right)$

Now, the only thing left to do is compute the determinants of the two $3 \times 3$ matrices.
You can either continue expanding them or compute the determinant with the determinant formula for $3 \times 3$ matrices.

As the first matrix has a row with two zeroes, I would like to expand it one more time:

$\det \left(\begin{matrix}1 & 0 & 4 \\ {\textcolor{red}{0}}^{\textcolor{b l u e}{-}} & {\textcolor{red}{2}}^{\textcolor{b l u e}{+}} & {\textcolor{red}{0}}^{\textcolor{b l u e}{-}} \\ 3 & 1 & - 1\end{matrix}\right) = - 0 \cdot \det \left(\ldots\right) + 2 \cdot \det \left(\begin{matrix}1 & 4 \\ 3 & - 1\end{matrix}\right) - 0 \cdot \det \left(\ldots\right)$

$= 2 \cdot \left[1 \cdot \left(- 1\right) - 3 \cdot 4\right] = 2 \cdot \left(- 13\right) = - 26$

For the second matrix, it doesn't matter that much, so let me use the formula for a change:

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

$= 1 + 12 = 13$

So, let's complete the total computation:

$\ldots = - 3 \cdot \det \left(\begin{matrix}1 & 0 & 4 \\ 0 & 2 & 0 \\ 3 & 1 & - 1\end{matrix}\right) - 5 \cdot \det \left(\begin{matrix}1 & 2 & 0 \\ 0 & 1 & 2 \\ 3 & 0 & 1\end{matrix}\right)$

$= - 3 \cdot \left(- 26\right) - 5 \cdot 13$

$= 13$