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
Mar 2, 2016

Answer:

#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))#

====================

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.

====================

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#