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

1 Answer
Mar 27, 2016

Answer:

The determinant is #4#.

Explanation:

Since this is an upper triangular matrix, there is a theorem in linear algebra which states that its determinant is the product of the entries on the main diagonal.

We may verify this by performing co-factor expansion along any row or column of our choice.

Since row 3 contains 2 zero entries, we may use co-factor expansion along row 3 to obtain the determinant as

#Delta=2(-1)^(3+3)|(1,-3),(0,2)|=2(2)=4#.

You may also view the link below to another similar problem of this nature that I solved in full detail previously for a student:
https://socratic.org/questions/how-do-you-find-the-determinant-of-1-2-3-4-5-6-7-8-9#245541