# How do you find the inverse of A=((-2, -2, -2), (-3, 2, -3), (1, -2, 1))?

Apr 12, 2017

This matrix is not invertible

#### Explanation:

For a matrix $A$ to be invertible, the determinant $\det A \ne 0$

Let's calculate

$\det A = | \left(- 2 , - 2 , - 2\right) , \left(- 3 , 2 , - 3\right) , \left(1 , - 2 , 1\right) |$

$= - 2 \cdot | \left(2 , - 3\right) , \left(- 2 , 1\right) | + 2 \cdot | \left(- 3 , - 3\right) , \left(1 , 1\right) | - 2 \cdot | \left(- 3 , 2\right) , \left(1 , - 2\right) |$

$= - 2 \cdot \left(2 - 6\right) + 2 \cdot \left(- 3 + 3\right) - 2 \cdot \left(6 - 2\right)$

$= 8 + 0 - 8$

$= 0$

As $\det A = 0$, matrix $A$ is not invertible

Apr 12, 2017

The first and third column vectors are the same, ie linearly dependent. So by inspection the matrix is non invertible. In simple algebra terms, you have three unknowns and 2 equations.

If the column vectors are linearly dependent, it follows that the row vectors are too.

Here, you will find that the second row vector minus twice the first row vector gives you the third row vector.

These and that fact of the zero determinant and non-invertibility are all equivalent statements.