## DLS 2 years ago Find the inverse of the following Matrix.

1. e.mccormick

2 ways to invert any invertable matrix: Augment on the identity, solve the first for the identity and the identity you adjoined becomes the inverse. 2) Find the adjoing, multiply that by the inverse of the determinant of the original.

2. DLS

$\LARGE \left[\begin{matrix}2 & 3 \\ 5 & 7\end{matrix}\right]$ Using elementary transformations only. I want to know how to think on this question or any shortcuts or such,establishing a relation between 2.3.5 and 7..

3. e.mccormick

Ah, using elementary transformations is method 1.

4. DLS

yes I know

5. e.mccormick

OK. There is only one shortcut, than that would be finding the determinant to see if it is invertable or not. Other than that, it is jsut elementary operations. The best ways to do those depend on preferences and if you like working with fractions or not.

6. e.mccormick

If you like fractions, do 5/2 times the first row and use the result to eliminate the first column from the second row. Otherwise, it is the typical back and forth.

7. DLS

$\LARGE A^{-1}=\frac{A_{adjacent}}{|A|}$ I can find the A^-1 using this but I wnt toknow the method 1

8. waterineyes

For two by two matrix, there is a shortcut method to determine the inverse of the matrix..

9. DLS

anything else than these 2? and does it involve elementary operations?it is necessary to use that,i would like to know it though :)

10. e.mccormick

$\LARGE \left[\begin{matrix}2 & 3 &1 &0\\ 5 & 7 &0 &1\end{matrix}\right]$ That is the only way to start it for if you are required to solve through elementary operations and not allowed to use the Adj.

11. waterineyes

Oh no, that is just a trick to find the inverse in a jiffy..

12. e.mccormick

This must always be written as: $\large A^{-1}=\frac{1}{det(A)}adj(A)$ You can not write this as: $\large A^{-1}=\frac{adj(A)}{det(A)}$ Because the adjoint is a matrix and there is no such thing as dividing a matrix by a number. In fact, there as no such thing as dividing a matrix by anything. You are multiplying through by a quotent.

13. DLS

$\LARGE \left[\begin{matrix}2 & 3 \\ 5 &7\end{matrix}\right]=\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]A$ I would use.. R1->R1/2 R2->R2-5R1 Then, R2-->R2/7 am i correct till here?

14. e.mccormick

Why R2/7?

15. waterineyes

7-15/2 = -1/2

16. e.mccormick

Your first steps produce: $\LARGE \left[\begin{matrix} 1 & 3/2 & 1/2 & 0\\ 0 & -1/2 & -5/2 & 1 \end{matrix}\right]$

17. e.mccormick

OK. Want to see the rest with elementary row ops from that point, or work it and have us check? Either way, I can explain the why along the way.

18. DLS

i got it ! thanks! :)

19. e.mccormick

OK. So you saw what we were talking about? I finished with: R1+3R2-->R1 $\LARGE \left[\begin{matrix} 1 & 0 & -14/2 & 3\\ 0 & -1/2 & -5/2 & 1 \end{matrix}\right]$ -2R2-->R2 $\LARGE \left[\begin{matrix} 1 & 0 & -14/2 & 3\\ 0 & 1 & 5 & -2 \end{matrix}\right]$

20. DLS

yeah!

21. e.mccormick

Which obviusly simplifies a little... hehe.. -7

22. e.mccormick

Yah, once you get the basic principal, this is easy to do. Only problem is arithmetic errors. Those still kick me in the anatomy at times...

23. DLS

is there any shortcut in this elementary method?

24. e.mccormick

Not really. It is kind of the brute force thing. However, it works pretty easy no matter the size of the matrix. The problem with th adjoint method is very large matrixes where finding the cofactors becomes a recursive loop of determinants.

25. DLS

hmm..okay! its more or less like Rubik's cube. :/

26. e.mccormick

By recursive loop of determinants I mean: Start with a $$10\times 10$$ matrix. To find the minor of the first row, first column, you need the determinant of a sub matrix that is everything remaining when you co0ver up the first row and first column. This means you need to get the determinant of a $$9\times 9$$ matrix. Now you loop into another problem because to get this, guess what you need? The determinant of an $$8\times 8$$ matrix... wheee. Where as by doing the augment with $$\mathrm{I}_n$$ and solve, it becomes the shortcut for large matrices. For $$3\times 3$$ and smaller, the adjoint method can be faster. It can even work faster for $$4\times 4$$. But for really large ones, it just becomes too much work.