Linear algebra - matrix equation

[AncientOne455]

I tried to solve this matrix equasion but i don't even know where to start solving.
We are given a matrix equasion (picture).
Questions are folowing:
a) For what parameter a has matrix equasion solution?
b) Define matrix X if parameter a=1

Solutions for this problems are:
a) a ≠ 7/2
b) X = 1/5 [2X2] a11= 7, a12=-1, a21=-3, a22=-1
I don't understand this equasion because i can't get X on only one side, pls help, thanks very much.

 

[Steven G]

Did you try letting X= be an arbitrary 2x2 matrix. Do you see why it needs to be 2x2?. Then try to solve for X.

 

[AncientOne455]

So you can only solve this equasion by putting random 2x2 matrix which seems right?

 

[AncientOne455]

You can't make it arbitrary on one side thats the problem.

 

[topsquark]

You can't make it arbitrary on one side thats the problem.

What do you mean by "on one side"?

Let \(\displaystyle X = \left [ \begin{matrix} b & c \\ d & e \end{matrix} \right ] \)

Now plug that into your equation.

-Dan

 

[Otis]

... i can't get X on only one side ...

Hello AncientOne455. We cannot solve for X that way because X is a matrix. In other words, we cannot use beginning-algebra steps to isolate X. We need a linear-algebra approach.

When Jomo suggested using an arbitrary matrix for X, he wasn't talking about a random matrix. He meant a symbolic matrix (like the matrix shown by topsquark, in post #5). Matrix X contains four unknown elements, and you're asked to find them, so you need to pick four different symbols to represent those unknowns.

When I substitute the symbolic matrix X from post #5 into the given equation, followed by completing the two matrix multiplications, I get a system of four linear equations containing four variables (b,c,d,e) and one parameter (the constant a).

An augmented-coefficient matrix for that system is:

\(\displaystyle \left[ \begin{matrix} 2-a & -1 & 1 & 0 & 1 \\ 1 & 1 & 0 & 1 & 1 \\ 1 & 0 & 2-a & -1 & 1 \\ 0 & 1 & 1 & 1 & -1 \end{matrix} \right]\)

Putting that matrix into reduced row echelon form gives the solutions for b,c,d,e in terms of a.

For example: b = (2a - 9)/(2a - 7)

If you would like more help, please show the work you've done so far. Cheers

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[Steven G]

I suspect that when the OP said i can't get X on only one side the OP meant that there was no way to factor out X.
The OP wanted something like AX +BX=b =>(A+B)X =b so X = (A+B)^-1b (if of course A+B was invertible)

 

[pka]

I suspect that when the OP said i can't get X on only one side the OP meant that there was no way to factor out X.
The OP wanted something like AX +BX=b =>(A+B)X =b so X = (A+B)^-1b (if of course A+B was invertible)

On the other hand, I taught a course in linear algebra ever third semester for of twenty years, but I have absolutely no idea what this post is about.

 

[Otis]

... I have absolutely no idea what [post #7] is about.

I think Jomo is speculating that the OP initially wanted to isolate X by factoring the form AX-XB as (A-B)X but then realized that it can't be done (matrix multiplication is not commutative) and hence the OP posted, "I can't get X on only one side".

Of course, I'm just speculating.

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[AncientOne455]

Yes, i tried to isolate X to only one side of the equasion, but it can't be done that way, you need other trick to solve matrix X.