# Linear Algebra - Matrix Inverse with an unknown element

## Homework Statement

Find all $r$ so that the matrix $\begin{bmatrix} 2 & 4 & 2 \\ 1 & r & 3 \\ 1 & 2 & 1\end{bmatrix}$ is inversable.

## Homework Equations

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

## 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:

Mark44
Mentor

## Homework Statement

Find all $r$ so that the matrix $\begin{bmatrix} 2 & 4 & 2 \\ 1 & r & 3 \\ 1 & 2 & 1\end{bmatrix}$ is inversable.
That would be "invertible."

## Homework Equations

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

## 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

$\begin{bmatrix} 1 & 2 & 1 & | & \frac{1}{2} & 0 & 0 \\ 1 & r & 3 & | & 0 & 1 & 0 \\ 1 & 2 & 1 & | & 0 & 0 & 1\end{bmatrix}$
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.
R2 - R1
R3 - R1

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

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

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.

Mark44
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.

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

Mark44
Mentor
Trick question...