Matrix equation

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Randyyy

Junior Member
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May 14, 2020
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hey, I am given the following:
matrix.png
And I am asked to solve the equations: [MATH]XA=B+2X[/MATH] and [MATH]CYA=D[/MATH]. The second one is not that bad, I only need to multiply by the inverse of C and A and I get: [MATH]Y=A^{-1}C^{-1}D[/MATH] and I just find my inverse for A and C and plug in. My issue is with the first equation, how do I Isolate X?
My attempt: [MATH]XA-2X=B \iff X(A-2)=B[/MATH] I can´t divide by a matrix so I try multiplying by the inverse of A and stick it into the parenthesis: [MATH]XA^{-1}(A-2)=BA^{-1} \iff X(E-2A^{-1})=BA^{-1}[/MATH] where E is the identity matrix. But Here I get stuck in an endless loop, my issue is that I can´t divide, only multiply by the inverse but how do I then handle (A-2)? Can I multiply by [MATH](A-2)^{-1}[/MATH]? But that doesn´t feel like a valid operation I am allowed to do but it is the only option I can think of to isolate X.
 
A - 2 is meaningless. You need to calclulate A - 2E. You will need to multiply by the inverse of (A -2E). Why do you think that is not valid?
Also, with the second one, you'll need to rethink the order in which you multiply by the inverse matrices.
 
Ah so for the first one I need to calculate [MATH]X(A-2E)=B \iff X(A-2E)(A-2E)^{-1}=B(A-2E)^{-1}[/MATH] which yields [MATH]XE=B(A-2E)^{-1}[/MATH] and for the second one I need to do it in the order of: [MATH]CYA=D \iff YA=C^{-1}D \iff Y=C^{-1}DA^{-1}[/MATH].
Regarding your question, I had only ever seen from our lectures where we multiplied by the inverse of one Matrix, not a parenthesis so I just assumed it was an invalid operation. I also assume that [MATH]AX-2X = X(A-2E)[/MATH] and not [MATH]X(A-2)[/MATH] like I had originally written.
 
Yes you are on the right track now.

(A-2E) IS one matrix when you do the calculation, so you can find the inverse of it.
 
In mathematics it is sometimes useful to think of A+B, where A and B are two matrices (or two real numbers) as just one matrix (or one real number). After all, when you add two matrices you do get back one matrix (one real number)!

As a student I was told that a(b+c) = ab+ac, (b+c)a = ab+ac and that this was called the distributive law. When I saw (a+b)(c+d) no one had to tell me what this equaled to. I just thought of (a+b) as one number initially and then as the sum of two numbers. Here is what I got.
(a+b)(c+d) =(a+b)c + (a+b)d = ac+bc + ad+bd. Is this clear?
 
Yes everythings crystsal clear Jomo, thank you both! I am pretty confident I know how to solve the question, I'll post a solution hopefully later today when I am not attending lectures.
 
Hi, it took longer than expected to finally solve the task, but I finally got about doing it!

solution.jpg
 
Randyyy said:
Correction, I realized I had messed up the order, It should be [MATH]B(A-2E)[/MATH] and so [MATH]x=[/MATH]{{1,-1,-1},{5,4,2},{2,2,1}}
Click to expand...
Actually, you have that mistake in the first line of your response. You may want to be careful about that - may find yourself collecting a bad grade due to some oversight.

However, it is excellent that you caught your oversite and corrected it.
 
Subhotosh Khan said:
Actually, you have that mistake in the first line of your response. You may want to be careful about that - may find yourself collecting a bad grade due to some oversight.

However, it is excellent that you caught your oversite and corrected it.
Click to expand...

Agreed, if I do that mistake on a final, it´ll cost.
 
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