# Thread: Showing that a matrix is not invertible

1. ## Showing that a matrix is not invertible

The question:
Show that the following matrix is not invertible

$$M = \left[ {\begin{array}{ccc} \frac{-5}{3} &\frac{1}{3} & \frac{5}{3} \\ \frac{13}{3} &\frac{10}{3} & \frac{5}{3} \\ \frac{-7}{3} &\frac{-7}{3} & \frac{-5}{3} \\ \end{array} } \right]$$

My attempt:
I know that when the determinate is 0, a matrix is not invertible. So I did the following:

$$\left| {\begin{array}{ccc} \frac{-5}{3} &\frac{1}{3} & \frac{5}{3} \\ \frac{13}{3} &\frac{10}{3} & \frac{5}{3} \\ \frac{-7}{3} &\frac{-7}{3} & \frac{-5}{3} \\ \end{array} } \right|$$

=
$
+
$\left| {\begin{array}{cc} \frac{13}{3} &\frac{10}{3} \\ \frac{-7}{3} &\frac{-7}{3} \\ \end{array} } \right|$
$\left| {\begin{array}{cc} \frac{10}{3} &\frac{5}{3} \\ \frac{-7}{3} &\frac{-5}{3} \\ \end{array} } \right|$
-
$\left| {\begin{array}{cc} \frac{13}{3} &\frac{5}{3} \\ \frac{-7}{3} &\frac{-5}{3} \\ \end{array} } \right|$
$

Which becomes:

$(\frac{-50}{9} - (\frac{-35}{9}) - (\frac{-65}{9} - (\frac{-35}{9})) + (\frac{-91}{9} - (\frac{-70}{9}))$

=

$\frac{-15}{9} - \frac{-30}{9} + \frac{-21}{9} = \frac{-2}{3}$

But this isn't 0 like required. :/

What have I done wrong? Thanks.

2. for simplicity factor a $\frac{1}{3}$ out of the matrix and then add row 3 to both row 1 and row 2 to get

$\frac{1}{3}\begin{vmatrix} -12 & -6 & 0 \\ 6 & 3 & 0 \\ -7 & -7 & -5\end{vmatrix}$

Now expand down the third column to get

$-5(-12(3)-6(-6))=-5(0)=0$

3. Thank you, I'll try that technique. I wonder why my way didn't work.

4. The formatting on your post is a bit garbled but I don't see any coeffeints on your minors. That may be where the problem is.

5. Oh, I'm such an idiot. >_<

Thanks.