Vectors, Linear Combinations and Systems

2 MIN READ · PRACTISE AT YOUR OWN PACE · FREE

A vector in $\mathbb R^n$ is an ordered list of $n$ real numbers. Add vectors componentwise and multiply each component by a scalar. In an engineering model, components may record currents, displacements or concentrations; their order and units must be specified.

A linear combination $c_1\mathbf v_1+\cdots+c_k\mathbf v_k$ uses scalar weights to combine vectors. If the vectors are the columns of a matrix $A$ , the same expression is $A\mathbf c$ . Thus solving $A\mathbf x=\mathbf b$ asks which column weights produce the target vector.

Visual guide

VISUAL GUIDEAdd displacements head to tail

Translated v

Blue u = (2, 1), followed by orange v = (−1, 2), ends at u + v = (1, 3). The dashed segment shows the second vector translated without changing its direction or length.

Worked example: two views of one system

Let $A=\begin{pmatrix}1&2\\2&1\end{pmatrix}$ and $\mathbf b=(5,4)^T$ . The equation is equivalent to $x+2y=5$ and $2x+y=4$ . Solving gives $x=1$ , $y=2$ .

In the column view, $1(1,2)^T+2(2,1)^T=(5,4)^T$ . In the row view, each row tests one scalar equation. Both views encode the same arithmetic but emphasize different geometry.

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

If A has columns a₁ and a₂, what does Ax for x=(2,−1) mean?

2a₁−a₂

a₁+2a₂

2a₁a₂

Hint 1 · Find a starting point

The entries of x weight the columns of A.

Hint 2 · Take the next step

Multiply the first column by 2 and the second by −1.

Show the reasoning

Answer: 2a₁−a₂

Matrix–vector multiplication forms the column combination 2a₁−a₂.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Worked example: an unreachable target

The columns $(1,2)^T$ and $(2,4)^T$ lie on the same line. Every combination is $(s,2s)^T$ , so no combination produces $(1,3)^T$ . The corresponding system is inconsistent. The target $(1,2)^T$ , in contrast, is reached by infinitely many weights satisfying $x+2y=1$ .

An $m\times n$ matrix takes an $n$ -component input to an $m$ -component output. The number of unknowns is the number of columns, and the number of scalar equations is the number of rows. A square shape alone does not guarantee a unique solution.

Practice

Compute $2(1,-1)^T+3(0,2)^T$ .
Write $x+3y=2$ , $2x-y=5$ as $A\mathbf x=\mathbf b$ .
Can $(2,5)^T$ be a multiple of $(1,2)^T$ ?

Show worked solutions

$(2,4)^T$ .
$A=\begin{pmatrix}1&3\\2&-1\end{pmatrix}$ , $\mathbf x=(x,y)^T$ , $\mathbf b=(2,5)^T$ .
No. The first component requires multiplier $2$ , but then the second would be $4$ .

MAKE IT YOURS

Pause before the next idea.

Can you explain how the visualization connects to this lesson’s goal? If a step still feels uncertain, put this lesson on your review list and try the check again another day.

Optional marks, not a grade. Saved in this browser only. Open notebook →