How do you find the determinant of #((1, 2, 3), (4, 5, 6), (7, 8, 9))#?

1 Answer
Mar 26, 2016

Answer:

The determinant is #0#.

Explanation:

Since column 3 of the given matrix consists of numbers which are all constant multiples of each other, there is a theorem which states that the determinant of this matrix is #0#.
However, I will prove it from calculation just to verify and show how you would calculate such determinants should you not be aware of the theorems of linear matrix algebra.

Use co-factor expansion along any row or column of your choice.

This involves adding the products of the entries in each row with their co-factors #(-1)^(i+j)#, for an entry in row I and column j, multiplied by the minor of the entry, formed by evaluating the resulting determinant when you delete row I and column j.
In this case, since the given matrix is #3xx3# in dimension, it will result in three #2xx2# determinants which we can find from definition
#|(a,b),(c,d)|=ad-bc#

I will take co-factor expansion along row 1 as an example to find the determinant of the given matrix as :

#Delta=1(-1)^(1+1)|(5,6),(8,9)|+2(-1)^(1+2)|(4,6),(7,9)|+3(-1)^(1+3)|(4,5),(7,8)|#

#=1(45-48)+(-2)(36-42)+3(32-35)#

#=-3+12-9#

#=0#