# Linear algebra -- compute the following without finding invA

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1. Mar 6, 2015

### SchrodingersMu

1. The problem statement, all variables and given/known data

2. Relevant equations
A=LU, U^-1 * L^-1= A^-1 , U^-1 * L^-1 * U^-1 * L^-1 = A^-2,

3. The attempt at a solution
I used MATLAB and the relations:
U^-1 * L^-1= A^-1 , U^-1 * L^-1 * U^-1 * L^-1 = A^-2,
to find a solution

I found U^-1*L^-1 , let =B

Then, found B^2 and took the inverse of B to get A^-2.

B=A^-1

C=inv(B^2)

so we have B*x+C*y= [2;5;10]

My question is: is there an easier way to do this? We are supposed to do this problem by hand. This would have taken a good amount fo effort to find so many inverse matrices.

Any input is appreciated!

Thank you!

2. Mar 6, 2015

### Staff: Mentor

L and U are triangular matrices, so I don't think it would have been all that hard to compute their inverses by hand, especially since they're only 3 x 3. Also, once you have A-1, just multiply it by itself to get (A-1)2. You don't have to do the product of four matrices you show.

3. Mar 6, 2015

### SammyS

Staff Emeritus
When you found matrices, B and C, that violated your instruction which was that you should not form A-1 and A-2 explicitly.

Simply find L-1 and U-1 and use those. They are easy to find by hand.

4. Mar 6, 2015

### HallsofIvy

As you say, since A= LU, $A^{-1}= L^{-1}U^{-1}$ and $A^{-2}= L^{-1}U^{-1}L^{-1}U^{-1}$ so the first thing I would do is find $L^{-1}$ and $U^{-1}$. Since L and U are triangular, that should be easy to do "by hand".

(Well, that was three answer in quick succession- and we are all saying the same thing.)

5. Mar 7, 2015

### Ray Vickson

You can do it by solving equations with $A$ on the left, which is doable easily from the LU decomposition. You want to compute $v+w$, where $v = A^{-1} x$ and $w = A^{-2} y$. We can get $v$ as the solution of the system $A v = x$ and we can get $w$ as the solution of $A^2 w = y$, and this, in turn, can be obtained from the systems $A w_0 = y$, $A w = w_0$.

I don't know if this is what the questioner really wants.