# Finding the vector form of the general solution Ax = 0

Suppose that $x_1 = -1$, $x_2 = 2$, $x_3 = 4$, $x_4 = -3$ is a solution of a non-homogeneous linear system $A\mathbf{x} = \mathbf{b}$ and that the solution set of the homogeneous system $A\mathbf{x} =\mathbf{0}$ is given by the formulas: \begin{align*} x_1 &= -3r + 4s,\\ x_2 &= r - s,\\ x_3 &= r,\\ x_4 &= s. \end{align*} Find the vector form of the general solutions of Ax = 0 and Ax = b

I ended up with something like:

( -3 4)
( 1 -1)
( 1 0)
( 0 1)


where I separated the $r$ and $s$ values, I haven't tried to actually solve though because I'm kinda confused about what I'm suppose to do with this.

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## 3 Answers

You answer of the solutions to $A\vec{x}=0$ being equal to the span of $\begin{bmatrix} -3\\ 1\\ 1\\ 0 \end{bmatrix}$ and $\begin{bmatrix} 4\\ -1\\ 0\\ 1 \end{bmatrix}$ is correct; now let's find the solutions to $A\vec{x}=\vec{b}$. If the kernel of a matrix is not equal to only the zero vector (as in our case, it actually has a dimension of two), then the matrix is non-invertible, and has infinitely many solutions for $\vec{b}$. By the property of linear transformations, if $A\vec{x}=0$, then $A(\vec{p}+\vec{x})=\vec{b}$ (where $\vec{p}$ is a vector such that $A\vec{p}=\vec{b}$) because $A(\vec{p}+\vec{x})=A\vec{p}+A\vec{x}=A\vec{p}+\vec{0}=\vec{b}+\vec{0}=\vec{b}$. Since you know that $\begin{bmatrix} -1\\ 2\\ 4\\ -3 \end{bmatrix}$ is a solution to $A\vec{x}=\vec{b}$, the set of solutions to this equations will consist of $\vec{p}$+any solution to $A\vec{x}=0$, which you found to be the span of $\begin{bmatrix} -3\\ 1\\ 1\\ 0 \end{bmatrix}$ and $\begin{bmatrix} 4\\ -1\\ 0\\ 1 \end{bmatrix}$. Or, in other words, the set of $\left \{\vec{p}+\vec{x}| A\vec{x}=0 \right \}$ is a set of solutions to $A\vec{x}=\vec{b}$. Or, represented numerically, solutions to $A\vec{x}=\vec{b}$ are of form $\underset{\text{A given solution to } A\vec{x}=\vec{b}}{\underbrace{\begin{bmatrix} -1\\ 2\\ 4\\ -3 \end{bmatrix}}}$ $+$ $\underset{\text{span of Kernel } A} {\underbrace{r\begin{bmatrix} -3\\ 1\\ 1\\ 0 \end{bmatrix} + s \begin{bmatrix} 4\\ -1\\ 0\\ 1 \end{bmatrix}}}$, as Arturo pointed out.

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So for the second part I'm just proving that the span of Ax = 0 is equal to vector b? –  user8485 Mar 20 '11 at 4:12
@Justin: careful with the wording -- span of the kernel of $A$, or such $\vec{x}$ that satisfy $A\vec{x}=0$. They are not equal. However, any element in kernel of $A$ plus a given solution for $A\vec{x}=\vec{b}$ will still be a solution for $\vec{b}$, because $A(\vec{p}+\vec{x})=A\vec{p}+A\vec{x}=A\vec{p}+\vec{0}=\vec{b}+\vec{0}=\vec{b}$ where $\vec{x}$ is a solution to $A\vec{x}=\vec{0}$ and $\vec{p}$ is a solution to $A\vec{x}=\vec{b}$. –  InterestedGuest Mar 20 '11 at 4:17
@Justin: I edited the last part of the answer a bit, hopefully that will make things more obvious. –  InterestedGuest Mar 20 '11 at 4:27
So if I do it this way I'll get something like ( 0 2 5 -2)? –  user8485 Mar 20 '11 at 4:51
@Justin: if you let r and s be equal to 1, yes. But they can be equal to anything, let r=2, s=5, then $\begin{bmatrix} 13\\ -1\\ 6\\ 2 \end{bmatrix}$ will be a solution, too. That is exactly the point -- there are infinitely many solutions, since the matrix is not invertible. –  InterestedGuest Mar 20 '11 at 4:57

The key is to find a single solution to the non-homogeneous, which can be done by Gaussian elimination if such a solution exists. The general solution is then this solution plus any linear combination of solutions to the homogeneous equation.

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You are close, but you aren't writing it right.

First, let's look at the solution to the homogeneous system. Writing out $r$ and $s$ explicitly in all the equation, we get: \begin{align*} x_1 & = -3r + 4s\ x_2 & = 1r -1 s\ x_3 & = 1r + 0s\ x_4 &= 0r + 1s. \end{align*} To write it as a sum of vectors, the vectors should have four coordinates; they are column vectors, not row vectors:\left(\begin{array}{c}x_1\\x_2\\x_3\\x_4\end{array}\right) = \left(\begin{array}{r}-3\\1\\1\\0\end{array}\right)r + \left(\begin{array} {r}4\\-1\\0\\1\end{array}\right)s.$$So you see, the vectors go down, not across. Now, the particular solution you have can be written analogously:$$\left(\begin{array}{c}x_1\\x_2\\x_3\\x_4\end{array}\right) = \left(\begin{array}{r}-1\\2\\4\\-3\end{array}\right).$$The solutions to the system A\mathbf{x}=\mathbf{b} are then equal to the particular solution plus the general solution to the homogeneous system. So putting it all together, you get:$$\left(\begin{array}{c}x_1\\x_2\\x_3\\x_4\end{array}\right) = \left(\begin{array}{r}-1\\2\\4\\-3\end{array}\right) + \left(\begin{array}{r}-3\\1\\1\\0\end{array}\right)r + \left(\begin{array}{r}4\\-1\\0\\1\end{array}\right)s.

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I know they are column vectors I just didn't know how to format it on here. –  user8485 Mar 20 '11 at 4:19
@Justin: Well, sorry; you wrote them as four row 2x1 vectors... –  Arturo Magidin Mar 20 '11 at 4:21