Math Is Fun Forum

vanilla

Member

Registered: 2014-02-13

Posts: 4

Can somebody solve this matrix question

Given is the matrix equation

X-XV=U-X(U-V)

(a) Determine matrix X
(b) Compute X when

4 3 1
3 2 -1 ---> U
8 5 3

2 1 1
0 1 -2 ---> V
2 1 2

Bob

Administrator

Registered: 2010-06-20

Posts: 10,640

Re: Can somebody solve this matrix question

hi vanilla,

Welcome to the forum. 

X - XV = U -X(U-V) = U - XU + XV

X - 2XV + XU = X(I - 2V + U) = U

X = U(I - 2V + U)-¹

So compute that bracket, find its inverse, and do U times that inverse, maintaining that order.

Hope that helps, 

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

anonimnystefy

Real Member

From: Harlan's World

Registered: 2011-05-23

Posts: 16,049

Re: Can somebody solve this matrix question

Hi vanilla

If you want to check your solution, this is what I'm coming up with:

Last edited by anonimnystefy (2014-02-14 02:27:16)

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vanilla

Member

Registered: 2014-02-13

Posts: 4

Re: Can somebody solve this matrix question

Hi vanilla

If you want to check your solution, this is what I'm coming up with:

Thank you very much! It was an exercise from an exam. Indeed, I got the same result, but some colleagues with whom I spoke said that they have different determinants. Some got -3, others -6 or -7. They determined the same matrix X. That is why I was shocked because I got determinant -2 and recalculated the matrix several times since then. It is such a relief now! Thank you once again!!!

Last edited by vanilla (2014-02-14 04:43:19)

vanilla

Member

Registered: 2014-02-13

Posts: 4

Re: Can somebody solve this matrix question

hi vanilla,

Welcome to the forum. 

X - XV = U -X(U-V) = U - XU + XV

X - 2XV + XU = X(I - 2V + U) = U

X = U(I - 2V + U)-¹

So compute that bracket, find its inverse, and do U times that inverse, maintaining that order.

Hope that helps, 

Bob

Thank you very much!