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

Mar 26, 2016

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 ${\left(- 1\right)}^{i + j}$, 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 $3 \times 3$ in dimension, it will result in three $2 \times 2$ determinants which we can find from definition
$| \left(a , b\right) , \left(c , d\right) | = a d - b c$

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

$\Delta = 1 {\left(- 1\right)}^{1 + 1} | \left(5 , 6\right) , \left(8 , 9\right) | + 2 {\left(- 1\right)}^{1 + 2} | \left(4 , 6\right) , \left(7 , 9\right) | + 3 {\left(- 1\right)}^{1 + 3} | \left(4 , 5\right) , \left(7 , 8\right) |$

$= 1 \left(45 - 48\right) + \left(- 2\right) \left(36 - 42\right) + 3 \left(32 - 35\right)$

$= - 3 + 12 - 9$

$= 0$