Consider a system AX = B, where B ≠ 0. Let X and Y satisfy this system. Find all constants a and b such that aX + bY also satisfies this system.

Question

I would like to know if the following approach is correct:

Given X and Y satisfy AX = B.

Hence, AX = B, AY = B.

X = Y.

Therefore, A(aX+bY) = A(aX+bX) = B. As AX = B, a + b must equal 1.

This approach seems very wrong to me at the X = Y step. Consider a system of equations:

3a + 2b + c = 1

a + 2b + c = 1

Where A = (a b c), X = (3 2 1)$^T$, Y = (1 2 3)$^T$, B = 1. Obviously, X ≠ Y.

However, the conclusion is correct (a + b = 1). Is there something wrong with this approach - it just accidentally leads to the correct answer - or my intuition is incorrect?

Answer 1

The approach is wrong. There is no reason to assume that $X=Y$. You made no justification of why the equation $X=Y$ should be true.

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A better approach is this. Let $X,Y$ both satisfy the system, that is, let $AX=AY=B$. Also, let $aX+bY$ also satisfy the system. Now, because it satisfies the system, we know that $A(aX+bY)=B$. Further, we know that $A(aX+bY)=aAx+bAY$. Can you continue from here?