Question

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))`?

Answer 1

`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 `(-1)^(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 `(-1)^(1+3) = (-1)^4 = + 1`, so a "`+`". The signs after
  that will alternate.
• Each element `a_(ij)` 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_(ij)` 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 ((1, 2, 0, 4), (0, 3, 0, 5), (0, 1, 2, 0), (3, 0, 1, -1))`

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

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

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

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((1, 0, 4), (color(red)(0)^(color(blue)(-)), color(red)(2)^(color(blue)(+)), color(red)(0)^(color(blue)(-))), (3, 1, -1)) = - 0 * det (...) + 2 * det ((1, 4), (3, -1)) - 0 * det (...)`

`= 2 * [1 * (-1) - 3 * 4] = 2 * (-13) = -26`

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

`det((1, 2, 0), (0, 1, 2), (3, 0, 1)) = 1 * 1 * 1 + 0 * 0 * 0 + 3 * 2 * 2 - 0 * 1 * 3 - 2 * 0 * 1 - 1 * 2 * 0`

`= 1 + 12 = 13`

So, let's complete the total computation:

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

`= - 3 * (-26) - 5 * 13`

`= 13`