How do you find the determinant of #((1, 0, 0, 0), (sqrt3, 2, 0, 0), (-0.003, 0, -3, 0), (1.432, sqrt37, pi, 1))#?

2 Answers
Mar 28, 2016

Multiply the diagonal to get: #-6#


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

In this case #1 xx 2 xx -3 xx 1 = -6#

Mar 28, 2016

The determinant is #-6#


Since this is a lower triangular matrix, there is a theorem in linear algebra which says that is determinant is the product of entries on the main diagonal

To verify this, use the method of co-factor expansion along row 1 since it contains the most zeroes.

#therefore Delta=1(-1)^(1+1)|(2,0,0),(0,-3,0),(sqrt37,pi,1)|=2(-3xx0)=-6#

For a full example of how to work out a matrix with more non-zero entries, see this link to a similar question: