Inverse of a matrix

  • Thread starter arkturus
  • Start date
1. The problem statement, all variables and given/known data
Determine the inverse of the matrix:

-x 1 0 0
1 -x 0 0
0 0 -x 1
0 0 1 -x


2. Relevant equations
Augmented matrix method


3. The attempt at a solution
The augmented matrix would be the matrix above with the identity matrix alongside it. I'm unsure how to manipulate the given matrix in order to make it the identity matrix.

Is there another way of finding the inverse?
 
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1. The problem statement, all variables and given/known data
Determine the inverse of the matrix:

-x 1 0 0
1 -x 0 0
0 0 -x 1
0 0 1 -x


2. Relevant equations
Augmented matrix method


3. The attempt at a solution
The augmented matrix would be the matrix above with the identity matrix alongside it. I'm unsure how to manipulate the given matrix in order to make it the identity matrix.

Is there another way of finding the inverse?
Nope. Form your augmented 4 x 8 matrix and use row reduction to reduce the left half to the identity matrix. When you're done, you'll have the inverse in the right half.
 

HallsofIvy

Science Advisor
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821
Well, yes, there is another way of finding the inverse. Many ways, in fact.

While I consider row-reduction as simplest- and best for this problem, you can also find the determinant, then take the matrix formed by the "minors" of [itex]a_{ij}[/itex] as [itex]b_{ji}[/itex] and divide by the determinant to get the inverse matrix.

But, as I said and Mark44 implied, row reduction is still best. Start, say, by swapping the first and second rows. That will give
[tex]\begin{bmatrix}1 & -x & 0 & 0 \\ -x & 1 & 0 & 0 \\ 0 & 0 & -x & 1\\ 0 & 0 & 1 & -x\end{bmatrix}\begin{bmatrix}0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{bmatrix}[/tex]
Now add x times the first row to the second row, etc.
 

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