Hi,I have:AX + X = B, all of them being matrices. I have the

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To solve the equation AX + X = B for the matrix X, the correct approach is to factor out X, leading to (A + I)X = B. Multiplying both sides by A-1 is incorrect because it does not account for the entire left side of the equation. Instead, if (A + I) is invertible, the solution can be found using X = (A + I)^{-1}B. The initial method led to the incorrect conclusion that 2X = A-1B. Understanding matrix operations is crucial to avoid errors common in matrix algebra compared to real number calculations.
Peter G.
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Hi,

I have:

AX + X = B, all of them being matrices. I have the numbers in the A and B matrices and I have to find the exact values of a,b,c,d (numbers in the X matrix)

I wanted to check if my method is correct:

I multiplied both sides by A-1.

2X = A-1B

So my values for abcd would be half of the matrix I get when I multiply A-1 by B:

Thanks,
Peter G.
 
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That's not quite right. We'd have

AX +X = (A+I)X =B,

so if A+I were invertible then X = (A+I)^{-1}B.
 


Peter G. said:
AX + X = B, all of them being matrices. I have the numbers in the A and B matrices and I have to find the exact values of a,b,c,d (numbers in the X matrix)

I wanted to check if my method is correct:

I multiplied both sides by A-1.

2X = A-1B
To elaborate on what stringy said, here is apparently what you did:
AX + X = B
A-1AX + X = A-1B
X + X = A-1B
2X = A-1B

Step 2 above is incorrect - you didn't multiply the entire left side of the equation by A-1.
 


Hey guys,

Thanks a lot for the help. Impressive how the "Matrix World" can make me commit mistakes I probably wouldn't in the "Real Number World" :redface:

Thanks,
Peter
 
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