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

1 Answer
George C.
Nov 29, 2016

#|(1, -2, 3, 2), (2, -1, 0, 4), (-3, 4, 0, -2), (-1, 1, 1, 5)| = 81#

Explanation:

  • You can add or subtract multiples of any row or column to any other row or column without altering the determinant.

  • You can swap two rows or columns, resulting in inverting the sign of the determinant.

  • You can cyclically permute three rows or columns to leave the determinant unchanged (since this is equivalent to two swaps).

  • The determinant of an upper or lower triangular matrix is the product of the diagonal.

So:

#|(1, -2, 3, 2), (2, -1, 0, 4), (-3, 4, 0, -2), (-1, 1, 1, 5)|" "# subtract #3# times row #4# from row #1# row to get:

#= |(4, -5, 0, -13), (2, -1, 0, 4), (-3, 4, 0, -2), (-1, 1, 1, 5)|" "# swap columns #1# and #3# to get:

#= -|(0, -5, 4, -13), (0, -1, 2, 4), (0, 4, -3, -2), (1, 1, -1, 5)|" "# swap rows #1# and #4# to get:

#= |(1, 1, -1, 5), (0, -1, 2, 4), (0, 4, -3, -2), (0, -5, 4, -13)|" "# add #4# times row #2# to row #3# to get:

#= |(1, 1, -1, 5), (0, -1, 2, 4), (0, 0, 5, 14), (0, -5, 4, -13)|" "# subtract #5# times row #2# from row #4# to get:

#= |(1, 1, -1, 5), (0, -1, 2, 4), (0, 0, 5, 14), (0, 0, -6, -33)|" "# add #6/5# times row #3# to row #4# to get:

#= |(1, 1, -1, 5), (0, -1, 2, 4), (0, 0, 5, 14), (0, 0, 0, -81/5)|#

#=1 * (-1) * 5 * (-81/5) = 81#

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