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

1 Answer
Mar 4, 2016

Answer:

The determinant is #0#.

Explanation:

There are several ways to compute the determinant.

On of those is the following formula for #3 times 3# matrix:

For a matrix

#A = ((a, b, c), (d, e, f), (g, h, i))#,

the determinant can found with

#det A = a * e * i + d * h * c + g * b * f#

#" "- c * e * g - f * h * a - i * b * d#.

In your case, this means:

#det ((1, 2, 1), (-2, 0, 2), (1, 4, 3))#

# = 1 * 0 * 3 + (-2) * 4 * 1 + 1 * 2 * 2 - 1 * 0 * 1 - 2 * 4 * 1 - 3 * 2 * (-2)#

# = 0 - 8 + 4 - 0 - 8 + 12 #

# = 0 #

By the way, this means that the matrix doesn't have an inverse, and that any linear equations solved with this matrix will not have a unique solution but either infinitely many solutions or none at all.