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))?

Mar 28, 2016

Multiply the diagonal to get: $- 6$

Explanation:

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

In this case $1 \times 2 \times - 3 \times 1 = - 6$

Mar 28, 2016

The determinant is $- 6$

Explanation:

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.

$\therefore \Delta = 1 {\left(- 1\right)}^{1 + 1} | \left(2 , 0 , 0\right) , \left(0 , - 3 , 0\right) , \left(\sqrt{37} , \pi , 1\right) | = 2 \left(- 3 \times 0\right) = - 6$

For a full example of how to work out a matrix with more non-zero entries, see this link to a similar question:
https://socratic.org/questions/how-do-you-find-the-determinant-of-1-2-3-4-5-6-7-8-9#245541