# 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))#?

##### 1 Answer

#### 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:

#= 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

#= - 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

You can either continue expanding them or compute the determinant with the determinant formula for

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#