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Linear Algebra - Matrix Inverse with an unknown element

  1. Oct 13, 2011 #1
    1. The problem statement, all variables and given/known data
    Find all [itex]r[/itex] so that the matrix [itex]\begin{bmatrix} 2 & 4 & 2 \\ 1 & r & 3 \\ 1 & 2 & 1\end{bmatrix}[/itex] is inversable.


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


    3. The attempt at a solution

    [itex]\begin{bmatrix} 2 & 4 & 2 & | & 1 & 0 & 0 \\ 1 & r & 3 & | & 0 & 1 & 0 \\ 1 & 2 & 1 & | & 0 & 0 & 1\end{bmatrix}[/itex]


    (1/2)R1

    [itex]\begin{bmatrix} 1 & 2 & 1 & | & \frac{1}{2} & 0 & 0 \\ 1 & r & 3 & | & 0 & 1 & 0 \\ 1 & 2 & 1 & | & 0 & 0 & 1\end{bmatrix}[/itex]

    R2 - R1
    R3 - R1

    [itex]\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}[/itex]

    And this is where I am stumped. Any bump in the right direction will be appreciated.
     
    Last edited: Oct 13, 2011
  2. jcsd
  3. Oct 13, 2011 #2

    Mark44

    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.
     
  4. Oct 13, 2011 #3
    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.
     
  5. Oct 13, 2011 #4

    Mark44

    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.
     
  6. Oct 13, 2011 #5
    It's the right matrix. I guess the answer is that it isn't invertible.
     
  7. Oct 13, 2011 #6

    Mark44

    Staff: Mentor

    Trick question...
     
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