# Linear Algebra - Matrix Inverse with an unknown element

1. Oct 13, 2011

### Jonmundsson

1. The problem statement, all variables and given/known data
Find all $r$ so that the matrix $\begin{bmatrix} 2 & 4 & 2 \\ 1 & r & 3 \\ 1 & 2 & 1\end{bmatrix}$ is inversable.

2. Relevant equations
Gauss-Jordan elimination on a matrix augmented with an identity matrix.

3. The attempt at a solution

$\begin{bmatrix} 2 & 4 & 2 & | & 1 & 0 & 0 \\ 1 & r & 3 & | & 0 & 1 & 0 \\ 1 & 2 & 1 & | & 0 & 0 & 1\end{bmatrix}$

(1/2)R1

$\begin{bmatrix} 1 & 2 & 1 & | & \frac{1}{2} & 0 & 0 \\ 1 & r & 3 & | & 0 & 1 & 0 \\ 1 & 2 & 1 & | & 0 & 0 & 1\end{bmatrix}$

R2 - R1
R3 - R1

$\begin{bmatrix} 1 & 2 & 1 & | &\frac{1}{2} & 0 & 0 \\ 0 & r - 2 & 2 & | &-\frac{1}{2} & 1 & 0 \\ 0 & 0 & 0 & | & -\frac{1}{2} & 0 & 1\end{bmatrix}$

And this is where I am stumped. Any bump in the right direction will be appreciated.

Last edited: Oct 13, 2011
2. Oct 13, 2011

### Staff: Mentor

That would be "invertible."
You're sort of doing things at random. Instead, add -1 times row 1 to row 2 to eliminate the leading entry in row 2. Add -1 times row 1 to row 3 to eliminate the leading entry in row 3.

IOW, R2 <-- -R1 + R2, and R3 <-- -R1 + R3.

Next, use the leading entry in row 2 to eliminate the entries above and below it.

Finally, use the remaining entry in row 3 to eliminate the two entries above it.

3. Oct 13, 2011

### Jonmundsson

Sorry. I posted my matrixes as A and I before I thought of a way to augment A and it got all messed up. I think I've fixed all the errors now.

4. Oct 13, 2011

### Staff: Mentor

Are you sure you have the right matrix? I just noticed that the first and third rows are the same. That means that your matrix is row-equivalent to one with zeroes in the bottom row, which means that its determinant is zero, which means that it is not invertible.

5. Oct 13, 2011

### Jonmundsson

It's the right matrix. I guess the answer is that it isn't invertible.

6. Oct 13, 2011

### Staff: Mentor

Trick question...