Are equations and vector combinations the same problem?

We have used vectors to build reachable targets and equations to describe shared constraints. Here is a problem that can be read both ways:

$$2x_1+x_2+x_3=6,\qquad x_2-x_3=2.$$

The subscripts label three different unknowns. They are not powers. Think of them as three adjustable inputs producing two measured outputs.

Keep a record of your investigation

Use first matrix entry 2 and input $(2,2,0)$ as your reference. Write the predicted two outputs before entering the input. For the coefficient experiment, keep that input, the target, and every other matrix entry fixed. Change only the first entry. Record the outputs before and after, checking them both by rows and by weighted columns; explain which calculation accounts for the difference.

One input, two measurements

Before using the explorer, test the input $(2,2,0)$ on paper. What should each equation produce?

Enter those values in Input x1, Input x2, and Input x3. Compare the equation rows with the vector path to the target.

Then change First matrix entry from 2 to 3 while keeping the input fixed. Predict which output changes, and by how much.

One input, four descriptionsReset

Three input coordinates; two output measurements. Target b = (6, 2).

Input x1

Input x2

Input x3

First matrix entry

Augmented matrix [A | b]; A has 2 rows and 3 columns

x₁	x₂	x₃	equals
2	1	1	6
0	1	-1	2

Equations: 2x₁ + x₂ + x₃ = 6; x₂ − x₃ = 2.

Column combination: (0)(2, 0) + (0)(1, 1) + (0)(1, -1) = (0, 0).

Row measurements: Row 1: (0) + (0) + (0) = 0. Row 2: (0) + (0) + (0) = 0.

Matrix equation: Ax = (0, 0). Ax ≠ b: this input does not satisfy both constraints.

Compare all the readouts

Initially, $2(2)+2+0=6$ and $2-0=2$, so $(2,2,0)$ satisfies both constraints. The weighted columns reach the target $(6,2)$.

Changing the first coefficient to 3 changes the first output to $3(2)+2+0=8$. The second remains 2. The output becomes $(8,2)$, so the same input no longer solves the changed problem.

Four descriptions of the starting problem

The two equations can be recorded as an augmented matrix:

$$\left[\begin{array}{ccc|c}2&1&1&6\\0&1&-1&2\end{array}\right].$$

The divider separates coefficients from required outputs. Let $A$ contain only the coefficients, let $\mathbf x$ contain the unknown inputs, and let $\mathbf b$ contain the target:

$$A=\begin{bmatrix}2&1&1\\0&1&-1\end{bmatrix},\quad \mathbf x=\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix},\quad \mathbf b=\begin{bmatrix}6\\2\end{bmatrix}.$$

Each input scales one column. Adding those scaled columns gives a vector equation:

$$x_1\begin{bmatrix}2\\0\end{bmatrix} +x_2\begin{bmatrix}1\\1\end{bmatrix} +x_3\begin{bmatrix}1\\-1\end{bmatrix} =\begin{bmatrix}6\\2\end{bmatrix}.$$

We define $A\mathbf x$ to mean that column combination. The compact matrix equation is therefore

$A\mathbf x=\mathbf b.$

Equations, augmented matrix, vector equation, and matrix equation express the same requirements. Switching notation does not change which inputs solve them.

Rows and columns answer different questions

Read by columns: how does each input contribute to the output vector? For $\mathbf x=(2,2,0)$, the contributions are $(4,0)$, $(2,2)$, and $(0,0)$, giving $(6,2)$.

Read by rows: how is each output measured from the whole input? Each output is a dot product:

$$A\mathbf x= \begin{bmatrix} (2,1,1)\mathbin{\cdot}(x_1,x_2,x_3)\\ (0,1,-1)\mathbin{\cdot}(x_1,x_2,x_3) \end{bmatrix}.$$

The two calculations agree because they add the same contributions to each output coordinate. The explorer’s graph is the two-dimensional output space; the input has three coordinates and is shown by the controls.

In general, an $m\times n$ matrix has $m$ rows and $n$ columns. It takes an $n$-component input and returns an $m$-component output. The dimensions need not match.

Reachability becomes consistency

The equation $A\mathbf x=\mathbf b$ is solvable exactly when $\mathbf b$ is in the span of the columns of $A$: that is what finding the input coefficients means.

A small counterexample helps. If the columns are $(1,2)$ and $(2,4)$, every output has second coordinate twice its first. The target $(3,5)$ is unreachable, so the corresponding system is inconsistent. Having coefficients to adjust does not guarantee that every target is possible.

The previous lesson also fits this notation. If $A\mathbf p=\mathbf b$, every solution is $\mathbf p+\mathbf z$ where $A\mathbf z=\mathbf0$. The set of these zero-output changes is called the null space of $A$. In our starting example, $\mathbf p=(2,2,0)$ and changes $t(-1,1,1)$ give all solutions.

Finish with a new matrix

Take $B$ with rows $(1,0,-1)$ and $(2,1,0)$. Predict the output for $(3,-2,1)$. Calculate it once by rows and once by columns.

Compare both calculations

The rows give $3-1=2$ and $2(3)-2=4$. The columns give $3(1,2)-2(0,1)+( -1,0)=(2,4)$. The input has three components; the output has two.

Design your own comparison

Choose your own three-component input within the explorer’s range. With first entry 2, record both outputs. Predict the effect of changing just that entry to 3, then test it while keeping the input fixed. Explain the output difference using both a row and a column.

Design a second input for which the same matrix-entry change produces no output change. Test your proposal and explain why it works.

Write a short conclusion: what you predicted, what you observed, and how the evidence changed or supported your explanation.

Check the reasoning in your investigation

For input $(r,s,t)$, increasing the first entry from 2 to 3 changes the output by $(r,0)$. An input with $r=0$ makes the change invisible. The affected row adds one extra copy of $r$; the affected column changes by $(1,0)$ and is weighted by $r$.

Question 1 of 30 checked · 0 correct

A matrix has 2 rows and 4 columns. What sizes does its matrix-vector product use?

Two input coordinates and four output coordinates.Four input coordinates and two output coordinates.Two input coordinates and two output coordinates.

Check answer

Start over

Explain what you can now do

Write a short answer without the explorer: Why can three inputs produce two outputs, and why might several different inputs produce the same output? Use the starting matrix to give two such inputs.

Compare your explanation

Each row produces one measurement of the three inputs. The input difference $(-1,1,1)$ contributes zero to both measurements, so $(2,2,0)$ and $(1,3,1)$ both produce $(6,2)$. More generally, adding any multiple of this difference leaves the output unchanged.

You have completed the opening route through vectors and systems. Return to any investigation where the prediction still feels uncertain. These same ideas will let us study matrices as transformations, and ask exactly when their actions can be undone.