Solve 2x2 Matrix in 30 Mins: An Attempt

  • Thread starter Thread starter nicknaq
  • Start date Start date
  • Tags Tags
    Matrix
AI Thread Summary
The discussion focuses on finding the inverse of a given 2x2 matrix A using elementary matrices. The user attempts to express A^-1 as a product of four elementary matrices but realizes their solution is incorrect. They detail their row operations, including swapping rows and scaling, but seek clarification on the correct elementary matrix for a specific operation. The conversation highlights the importance of accurately applying row operations to achieve the desired matrix form. The user is looking for guidance to correct their approach and finalize the solution.
nicknaq
Messages
67
Reaction score
0
I really need an answer in the next 30 min.

Homework Statement


Given the 2x2 matrix A= 0 −2
9 −5
Write A^-1 as a product of 4 elementary matrices.

The Attempt at a Solution



I got the following. It's wrong. Please find my error.

1 0 | 1/9 0 | 1 0 | 1 5/9
0 1 | 0 1 | 0 -1/2 | 0 1

Edit: Geeze. the forum won't let me show the spaces between the four matrices above. sorry.
 
Physics news on Phys.org


After swapping the two rows, dividing the first row by 9, and dividing the second row by -2, you have row-reduced A to
\begin{bmatrix}1 & -5 \\ 0 & 1\end{bmatrix}
so you have to add 5 times the second row to the first row. What elementary matrix does that?
(You seem to have kept the "9" you got rid of with the second elementary matrix.)
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top