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I am having a bad time with this matrix and vector situation, and I think the solution is kind a trick in some part of the computation, but I don't know how to find this:

Find 3 vectors (different than Zero), $\vec{x},\vec{y},\vec{b} \in \mathbb{R}^3 $, where $A\vec{x}=\vec{b}$ and $B\vec{y}=\vec{b}$ $$A =\begin{bmatrix} 0 && 1 && 2\\ 1 && 1 && 1\\ 2 && 1 && 0 \end{bmatrix}$$ $$B =\begin{bmatrix} 1 && -1 && 0\\ 0 && 0 && 1\\ 1 && 1 && 0 \end{bmatrix}$$

Basically, what I have done, is multiplying the matrices by a vector $\vec{X} = (X_1, X_2, X_3)$ and the other matrix by a vector $\vec{Y} = (Y_1, Y_2, Y_3)$.

Then, as I am multiplying a matrix $(3 \times 3)$ with a vector $(3 \times 1)$, my result is a new vector $(3 \times 1)$.

But then, I don't know what to do next, can someone help me with a kind advice on this?

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    $\begingroup$ This is easy. You can take $x=y=b=(0,0,0)$. $\endgroup$ – Dietrich Burde Apr 23 at 15:13
  • $\begingroup$ Thank you @DietrichBurde, I forgot to mention that should be different than Zero. $\endgroup$ – jguzmaje Apr 23 at 15:38
  • $\begingroup$ The only thing that I can think of is to let $$\vec{x} = \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}$$ Do the same for $\vec{y}$. Since $A\vec{x} = \vec{b}$ and $B\vec{y} = \vec{b}$, we have that $A\vec{x} = B\vec{y}$. Doing the expansion will yield three equations with six unknowns. Not solvable. The only other thing I could think of would be to find $A^{-1}$ and $B^{-1}$ because you have equations $\vec{x} = A^{-1}B\vec{y}$ and $\vec{y} = B^{-1}A\vec{x}$. This may just give you the same three equations... $\endgroup$ – JacobCheverie Apr 23 at 15:41
  • $\begingroup$ No “tricks” necessary, but a solid understanding of the properties of systems of linear equations helps make this much easier. $\endgroup$ – amd Apr 23 at 19:19
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I would start by looking for candidates for $\vec b$. For the equation $A\vec x=\vec b$ to have any solution at all, $\vec b$ must lie in $A$’s column space, and likewise it must also lie in $B$’s column space. $B$’s column space is clearly all of $\mathbb R^3$, but the columns of $A$ are linearly dependent (the sum of the outer columns is twice the central column). So, pick any convenient-looking multiple of any column of $A$ (or a convenient-looking linear combination of them) as $\vec b$ and then solve the two equations for $\vec x$ and $\vec y$. For any particular choice of $\vec b$, there’s a degree of freedom in the choice of a corresponding $\vec x$, but if you construct $\vec b$ as I suggest you should be able to write down a value for $\vec x$ without doing any more work. (Hint: interpret the product $A\vec x$ as a linear combination of the columns of $A$.)

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  • $\begingroup$ Thank you @amd . I didn't notice the linearly dependency, that was key for the evaluation. Again, thank you. $\endgroup$ – jguzmaje Apr 24 at 12:28

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