The complete solution

One answer, plus a whole line of freedom. Watch a line of inputs slide out, and every point on it solve the same equation.

One answer, then all of them

Take the matrix from The null space, the one that crushes the line of $(-1,-1,1)$, and ask it to produce $\mathbf{b}=(2,3,5)$. One answer is easy to check:

Now add any multiple of the crushed direction. The matrix sends that part to zero, so the output does not change:

Every point of the line through $(2,3,0)$ in the direction $(-1,-1,1)$ is a solution. That is the shifted line in the picture.

Are there any others? Suppose $\mathbf{x}_1$ and $\mathbf{x}_2$ both solve $\mathbf{A}\mathbf{x}=\mathbf{b}$. Then $\mathbf{A}(\mathbf{x}_1-\mathbf{x}_2)=\mathbf{b}-\mathbf{b}=\mathbf{0}$, so any two solutions differ by a vector of the null space. So once you have one solution, you have them all:

Here $\mathbf{x}_\text{part}$ is any one particular solution, and $\mathbf{x}_\text{null}$ runs over the whole null space. This is the complete solution. When the null space is only $\mathbf{0}$, the solution is unique; otherwise there are infinitely many.

Finding a particular solution

Run elimination on $\mathbf{A}$ and $\mathbf{b}$ together, written side by side as one wide table $[\,\mathbf{A}\mid\mathbf{b}\,]$, and go on to the reduced form of the null space page. For the matrix above:

The simplest particular solution sets every free variable to 0. Here $x_3$ is free, so $x_3=0$, and the pivot rows give $x_1=2$ and $x_2=3$: $\mathbf{x}_\text{part}=(2,3,0)$, the point the line slid out to.

When there is no solution

The last row of the reduced table says $0=0$. That came from $5-2-3$: the third entry of $\mathbf{b}$ minus the first two. For a general $\mathbf{b}$ the same steps leave

in the last row. If $b_3\ne b_1+b_2$, that row says 0 equals something that is not 0, and no $\mathbf{x}$ can fix it. In the picture, drag the third entry of $\mathbf{b}$ off 5 and the line of solutions fades away.

This is the column space test of Vector spaces and subspaces, now worked out by elimination: every zero row of the reduced matrix gives one condition that $\mathbf{b}$ must meet to be in $C(\mathbf{A})$.

Rank and the four cases

The rank $r$ is the number of pivots. An $m\times n$ matrix has $m$ rows and $n$ columns, and each pivot needs its own row and its own column, so $r\le m$ and $r\le n$. Comparing $r$ with $m$ and $n$ sorts every system into one of four cases.

• Zero rows come from $r<m$. There are $m-r$ of them, and each puts a condition on $\mathbf{b}$.
• Free variables come from $r<n$. There are $n-r$ of them, and each adds a direction of freedom to the solution.

Worked example

1. Two equations, four unknowns. Reduce the table for $\mathbf{A}=\begin{bmatrix}1&2&1&3\\2&4&3&8\end{bmatrix}$ and $\mathbf{b}=(2,5)$:

With the free variables $x_2=x_4=0$, the pivot rows give $x_1=1$ and $x_3=1$: $\mathbf{x}_\text{part}=(1,0,1,0)$. Check: $\mathbf{A}\mathbf{x}_\text{part}=(1+1,\;2+3)=(2,5)$.

Add the null space from the null space page, with weights $c$ and $d$:

With $c=d=1$ this is $(-2,1,-1,1)$, and $\mathbf{A}\mathbf{x}=(-2+2-1+3,\;-4+4-3+8)=(2,5)$, as it should be.

2. One equation, two unknowns. All solutions of $x+y=3$: the particular solution $(3,0)$ plus the null line of $(-1,1)$, so $(3,0)+c\,(-1,1)$.

3. A condition on b. $\begin{bmatrix}1&1\\2&2\end{bmatrix}\mathbf{x}=\mathbf{b}$ has a solution only when $b_2=2b_1$.

4. Counting. A $3\times5$ matrix of rank 3 has $r=m$, so every $\mathbf{b}$ works, with $5-3=2$ free variables. A $3\times2$ matrix of rank 2 is the case $r=n<m$.

Predict before you drag: what happens to the line of solutions if the third entry of $\mathbf{b}$ becomes 6? It vanishes, since $2+3\ne6$.

Where you’ll meet this

Natural and forced responses. A circuit or filter driven by an input settles into a forced response, but the way it starts depends on its natural response, the part that solves the equation with no input. Matching the starting values picks one point on the line of solutions.

Too few sensors. A structure with five unknown forces and three strain gauges gives three equations in five unknowns. If the rank is 3, every reading has a solution, but a whole two-dimensional family of forces fits it. The measurements alone can’t pick one; extra knowledge, such as a preference for the smallest forces, must.

Reference card