How do you find the determinant of #((2, 1, 0, 0), (1, 2, 1, 0), (0, 1, 2, 1), (0, 0, 1, 2))#?

1 Answer
Mar 21, 2016

Answer:

Transform into an upper triangular matrix then take the product of the diagnonal to find:

#abs((2, 1, 0, 0),(1, 2, 1, 0),(0, 1, 2, 1),(0, 0, 1, 2)) = 5#

Explanation:

The determinant is unchanged by adding any multiple of one row to another row, or column to another column.

Starting with:

#((2, 1, 0, 0),(1, 2, 1, 0),(0, 1, 2, 1),(0, 0, 1, 2))#

Subtract #1/2# of the first row from the second to get:

#((2, 1, 0, 0),(0, 3/2, 1, 0),(0, 1, 2, 1),(0, 0, 1, 2))#

Subtract #2/3# of the second row from the third to get:

#((2, 1, 0, 0),(0, 3/2, 1, 0),(0, 0, 4/3, 1),(0, 0, 1, 2))#

Subtract #3/4# of the third row from the fourth to get:

#((2, 1, 0, 0),(0, 3/2, 1, 0),(0, 0, 4/3, 1),(0, 0, 0, 5/4))#

The determinant of an upper triangular matrix is the product of the diagonal, so:

#abs((2, 1, 0, 0),(0, 3/2, 1, 0),(0, 0, 4/3, 1),(0, 0, 0, 5/4)) = color(red)(cancel(color(black)(2)))*color(blue)(cancel(color(black)(3)))/color(red)(cancel(color(black)(2)))*color(green)(cancel(color(black)(4)))/color(blue)(cancel(color(black)(3)))*5/color(green)(cancel(color(black)(4))) = 5#