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

  • Thread starter Thread starter arkturus
  • Start date Start date
  • Tags Tags
    Inverse Matrix
AI Thread Summary
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.
arkturus
Messages
27
Reaction score
0

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?
 
Physics news on Phys.org
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.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

Similar threads

Back
Top