The question:
Show that the following matrix is not invertible
$\displaystyle \[
M =
\left[ {\begin{array}{ccc}
\frac{5}{3} &\frac{1}{3} & \frac{5}{3} \\
\frac{13}{3} &\frac{10}{3} & \frac{5}{3} \\
\frac{7}{3} &\frac{7}{3} & \frac{5}{3} \\
\end{array} } \right]
\]$
My attempt:
I know that when the determinate is 0, a matrix is not invertible. So I did the following:
$\displaystyle \[
\left {\begin{array}{ccc}
\frac{5}{3} &\frac{1}{3} & \frac{5}{3} \\
\frac{13}{3} &\frac{10}{3} & \frac{5}{3} \\
\frac{7}{3} &\frac{7}{3} & \frac{5}{3} \\
\end{array} } \right
\]$
=
$\displaystyle
+
\[
\left {\begin{array}{cc}
\frac{13}{3} &\frac{10}{3} \\
\frac{7}{3} &\frac{7}{3} \\
\end{array} } \right
\]
\[
\left {\begin{array}{cc}
\frac{10}{3} &\frac{5}{3} \\
\frac{7}{3} &\frac{5}{3} \\
\end{array} } \right
\]

\[
\left {\begin{array}{cc}
\frac{13}{3} &\frac{5}{3} \\
\frac{7}{3} &\frac{5}{3} \\
\end{array} } \right
\]
$
Which becomes:
$\displaystyle (\frac{50}{9}  (\frac{35}{9})  (\frac{65}{9}  (\frac{35}{9})) + (\frac{91}{9}  (\frac{70}{9}))$
=
$\displaystyle \frac{15}{9}  \frac{30}{9} + \frac{21}{9} = \frac{2}{3}$
But this isn't 0 like required. :/
What have I done wrong? Thanks.