Question

Show that this matrix is un-invertible regardless of what values the variables are:

Answer 1

\[\left[\begin{matrix}0 & a & 0 & 0 \\ b & 0 & c & 0 \\ 0 & c & 0 & d \\ 0 & 0 & e & 0\end{matrix}\right]\]

Answer 2

First I'll try to construct an augmented matrix with it:
\[\left[\begin{matrix}
0 & a & 0 & 0 & | & 1 & 0 & 0 & 0\\
b & 0 & c & 0 & | & 0 & 1 & 0 & 0\\
0 & c & 0 & d & | & 0 & 0 & 1 & 0\\
0 & 0 & e & 0 & | & 0 & 0 & 0 & 1
\end{matrix}\right]\]

Multiply 2nd row by 1/b and add it to 1st row:

\[\left[\begin{matrix}
1 & a & \frac{c}{b} & 0 & | & 1 & \frac{1}{b} & 0 & 0\\
b & 0 & c & 0 & | & 0 & 1 & 0 & 0\\
0 & c & 0 & d & | & 0 & 0 & 1 & 0\\
0 & 0 & e & 0 & | & 0 & 0 & 0 & 1
\end{matrix}\right]\]

Multiply 1st row by -b and add it to 2nd row

\[\left[\begin{matrix}
1 & a & \frac{c}{b} & 0 & | & 1 & \frac{1}{b} & 0 & 0\\
0 & -\frac{a}{b} & 0 & 0 & | & -b & 0 & 0 & 0\\
0 & c & 0 & d & | & 0 & 0 & 1 & 0\\
0 & 0 & e & 0 & | & 0 & 0 & 0 & 1
\end{matrix}\right]\]

Multiply 2nd row by -b/a

\[\left[\begin{matrix}
1 & a & \frac{c}{b} & 0 & | & 1 & \frac{1}{b} & 0 & 0\\
0 & 1 & 0 & 0 & | & \frac{b^{2}}{a} & 0 & 0 & 0\\
0 & c & 0 & d & | & 0 & 0 & 1 & 0\\
0 & 0 & e & 0 & | & 0 & 0 & 0 & 1
\end{matrix}\right]\]

Switch 3rd and 4th row:

\[\left[\begin{matrix}
1 & a & \frac{c}{b} & 0 & | & 1 & \frac{1}{b} & 0 & 0\\
0 & 1 & 0 & 0 & | & \frac{b^{2}}{a} & 0 & 0 & 0\\
0 & 0 & e & 0 & | & 0 & 0 & 0 & 1 \\
0 & c & 0 & d & | & 0 & 0 & 1 & 0
\end{matrix}\right]\]

Multiply 3rd row by 1/e

\[\left[\begin{matrix}
1 & a & \frac{c}{b} & 0 & | & 1 & \frac{1}{b} & 0 & 0\\
0 & 1 & 0 & 0 & | & \frac{b^{2}}{a} & 0 & 0 & 0\\
0 & 0 & 1 & 0 & | & 0 & 0 & 0 & \frac{1}{e} \\
0 & c & 0 & d & | & 0 & 0 & 1 & 0
\end{matrix}\right]\]

Multiply 2nd row by -c and add it to 4th row

\[\left[\begin{matrix}
1 & a & \frac{c}{b} & 0 & | & 1 & \frac{1}{b} & 0 & 0\\
0 & 1 & 0 & 0 & | & \frac{b^{2}}{a} & 0 & 0 & 0\\
0 & 0 & 1 & 0 & | & 0 & 0 & 0 & \frac{1}{e} \\
0 & 0 & 0 & d & | & \frac{-cb^{2}}{a} & 0 & 1 & 0
\end{matrix}\right]\]

Multiply 4th row by 1/d

\[\left[\begin{matrix}
1 & a & \frac{c}{b} & 0 & | & 1 & \frac{1}{b} & 0 & 0\\
0 & 1 & 0 & 0 & | & \frac{b^{2}}{a} & 0 & 0 & 0\\
0 & 0 & 1 & 0 & | & 0 & 0 & 0 & \frac{1}{e} \\
0 & 0 & 0 & 1 & | & \frac{-cb^{2}}{ad} & 0 & \frac{1}{d} & 0
\end{matrix}\right]\]

I can easily finish this augmented matrix can get my inverted matrix on the right side, so how is this not invertible?
Did I do something wrong?

Answer 3

No