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Given that for some system $ \mathrm{A}\mathrm{x}= \mathrm{b} $ there is a unique solutions which is given by $(\alpha, 0, 2\alpha)$, while $\alpha$ >0. What is the solution of a corresponding homogeneous system where we replace the first column of $\mathrm{A}$ by $\mathrm{b}$?

I tried to use the properties of Cramer's rule, but hadn't reach any useful conclusions. Namely, we know that $\mathrm{det}(\mathrm{A})\neq 0 $, hence the proportion of the determinants of the homogeneous system and the non-homogeneous yields $\alpha$, however it seems like wrong direction because it does not lead anywhere.

Thank you!

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$\alpha = \frac {\det(A_1)}{\det(A)}$ where $\alpha \ne 0$ $\implies \det(A_1) \ne 0$. Then $A_1$ is invertible. Meaning $$(A_1)x=0 \\ \implies x=(A_1)^{-1}0 = 0$$


Edit: OK. You asked for an alternate way so here's a method that doesn't explicitly use Cramer's rule:

We know that $$A(\alpha e_1 + 2\alpha e_3) = b$$

And we're asked about the solution to $$A_1y = 0 \\ \pmatrix{b & A_2 & A_3}y = 0 \\ \pmatrix{\alpha Ae_1 + 2\alpha Ae_3 & A_2 & A_3}y=0 \\ \pmatrix{\alpha A_1 + 2\alpha A_3 & A_2 & A_3}y=0$$

We know that $\det(A)\ne 0$ because $Ax=b$ has a unique solution.

Let's look at the determinant of $A_1$. It's $$\begin{align}\det(A_1) &= \det\pmatrix{\alpha A_1 + 2\alpha A_3 & A_2 & A_3} \\ &= \det\pmatrix{\alpha A_1 & A_2 & A_3} \\ &= \alpha\det\pmatrix{A_1 & A_2 & A_3} \\ &\ne 0\end{align}$$

Thus $A_1$ is invertible. Thus $y=0$ is the only solution to $A_1y=0$.

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  • $\begingroup$ Thank you! And I have following question. Actually, we know that every homogeneous system has at least one solution (the trivial one). Hence, in these settings, we have to prove that this is the only possible solution. Can we deduce it without using Cramer's rule? I.e., can I automatically say that $\mathrm{A_1}$ is full rank matrix (and then jump to its invertability)? $\endgroup$ – V. Vancak Nov 21 '15 at 0:47

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