Find Inverse of Matrix: x1, -x, -x1, 1-x

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The discussion focuses on finding the inverse of a specific 4x4 matrix using the augmented matrix method. Participants suggest that row reduction is the most effective approach to transform the matrix into the identity matrix, allowing the inverse to be extracted from the augmented section. An alternative method involving determinants and minors is mentioned, but row reduction is emphasized as the preferred technique. The conversation includes practical steps, such as swapping rows and performing row operations to facilitate the reduction process. Ultimately, row reduction is highlighted as the simplest and best method for this problem.
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Homework Statement


Determine the inverse of the matrix:

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


Homework Equations


Augmented matrix method


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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arkturus said:

Homework Statement


Determine the inverse of the matrix:

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


Homework Equations


Augmented matrix method


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.
 
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 a_{ij} as b_{ji} 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
\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}
Now add x times the first row to the second row, etc.
 
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