# Solving Vector Equation?

$2$. a: For what vectors $$v = \begin{pmatrix}a \\ b \\ c\end{pmatrix}$$ does the equation $Ax = v$ have a solution if $$A = \left(\begin{array}{crc} 1 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 3 \end{array}\right),\quad \text{ and }\quad x = \begin{pmatrix}x_1 \\ x_2 \\ x_3\end{pmatrix}.$$

I'm confused about this question. I would know how to solve it if the components of $x$ were constants, but I don't know how it works when there are two unknown vectors like this.

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You mean, "if the components of $v$ were constants", rather than the components of $x$, right? – Arturo Magidin Mar 7 '12 at 6:04
The way I'm thinking of the question, I meant what I said. I don't know if I'm trying to do it the right way though. – Kyle V. Mar 7 '12 at 6:24
That would not correspond to solving an equation; it would correspond to plugging in values and finding the answer, which is a very different thing. Did you cover "solving systems of linear equations" at some point in your linear algebra course? $x_1,x_2,x_3$ are the unknowns. If $a,b,c$ were specific numbers, then the equation $Ax=v$ corresponds to a system of 3 linear equations in three unknowns, and it has "a solution" if the system has a solution. They are asking you to find all triples of values of $a,b,c$ for which the corresponding system of linear equations has a solution. – Arturo Magidin Mar 7 '12 at 6:28

Added. Okay, let's clarify this, since you express confusion. If we multiply out $Ax$ and set it equal to $v$, we get: $$\left(\begin{array}{c} x_1 + x_3 \\ -x_2 + 2x_3\\ x_1 - x_2 + 3x_3 \end{array}\right) = \left(\begin{array}{c}a\\b\\c\end{array}\right),$$ which leads to the system of linear equations in three unknowns: $$\begin{array}{rcccccl} x_1 & & & + & x_3 & = & a\\ & - & x_2 & + & 2x_3 & = & b\\ x_1 & - & x_2 & + & 3x_3 & = & c \end{array}$$

The question is: for what vectors $v$ does this system have a solution $x$?

For example, if $$v =\left(\begin{array}{c}0\\0\\0\end{array}\right)$$ then the system has a solution (namely, $x_1=x_2=x_3=0$). If $$v = \left(\begin{array}{c}1\\2\\3\end{array}\right)$$ then the system has a solution (for example, $x_1 = 1$, $x_2=-2$, $x_3=0$; or $x_1=0$, $x_2=0$, $x_3=3$; or any of a number of others, but at least one). On the other hand, if $$v=\left(\begin{array}{c}0\\1\\0\end{array}\right)$$ then the system has no solutions: from $x_1+x_3=0$ we get $x_1=-x_3$. Substituting into the third equation, we would get $-x_2 + 2x_3 = 0$, so $x_2 = 2x_3$; but if we substitute that into the second equation, we get $-2x_3+2x_3 = 1$, which is impossible. So for this $v$, there are no solutions.

So... for some choices of $v$, the system has a solution. For other choices of $v$, the system has no solutions.

The question is: what are the choices of $v$ that will give you a system with a solution?

You are being asked to find the values of $a$, $b$, and $c$ for which the system $$\begin{array}{rcccccl} x_1 & & & + & x_3 & = & a\\ & - & x_2 & + & 2x_3 & = & b\\ x_1 & - & x_2 & + & 3x_3 & = & c \end{array}$$ has a solution. I hope you are familiar with solving systems of equations by Gaussian elimination, because that is what I'm going to do next. The augmented matrix of this system is just: $$\left(\begin{array}{crr|c} 1 & 0 & 1 & a\\ 0 & -1 & 2 & b\\ 1 & -1 & 3 & c \end{array}\right)$$ (that is, it is nothing more than $(A|v)$). We need to know when this system is consistent. If $a$, $b$, and $c$ were numbers, then Gaussian elimination is a recipe that tells us how to figure that out.

So then let's pretend they are specific numbers, and see where that leads us! We subtract the first row from the third row, and we obtain $$\left(\begin{array}{rrr|c} 1 & 0 & 1 & a\\ 0 & -1 & 2 & b\\ 0 & -1 & 2 & c-a \end{array}\right).$$ And now, subtracting the second row from the third row, we get $$\left(\begin{array}{rrr|c} 1 & 0 & 1 & a\\ 0 & -1 & 2 & b\\ 0 & 0 & 0 & c-a-b \end{array}\right).$$

Now... under what conditions would you know this system has no solutions? Under what conditions will it certainly have solutions?

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I'm confused, where does the vector x fit into all of this? – Kyle V. Mar 7 '12 at 6:23
It is not there, don't you know this technique when $a$, $b$ and $c$ are numbers? If you don't, perhaps your confusion comes from there.. – Patrick Da Silva Mar 7 '12 at 6:24
If you multiply out $Ax=v$ and write it out as equations, you get $$\begin{array}{rcccccl} 1x_1 & + & 0x_2 & + & 1x_3 &=&a\\ 0x_1 & - & 1x_2 & + & 2x_3 & = & b\\1x_1 & - & 1x_2 & + & 3x_3 & = & c\end{array}$$ If you try to solve the system by Gaussian elimination, you will simply get back to the matrix I start with. – Arturo Magidin Mar 7 '12 at 6:26
I'm still not clear on where the x1,x2,x3's go... – Kyle V. Mar 7 '12 at 6:37
@StickFigs: I've added stuff at the beginning; hopefully this helps. – Arturo Magidin Mar 7 '12 at 6:42