Composition, Determinants, and Inverses

Multiplication as composition

If $A$ is a transformation and $B$ is another transformation, then $AB$ is the transformation “first do $B$ , then do $A$ .” Read right-to-left, like function composition $f(g(x))$ .

(AB)\mathbf{x} \;=\; A(B\mathbf{x}).

The right-to-left order matters; it’s the same convention as function composition from Module 5. It trips people up at first, because you write $AB$ but you apply $B$ first. Own it.

The columns of AB

Given the column picture, there’s a clean way to compute $AB$ . Each column of $AB$ is $A$ applied to the corresponding column of $B$ .

AB \;=\; \bigl[\; A\,\mathbf{b}_1 \;\big|\; A\,\mathbf{b}_2 \;\big|\; \cdots \;\bigr].

If $B$ sends $\hat{\imath}$ to $\mathbf{b}_1$ , then $AB$ sends $\hat{\imath}$ to $A\mathbf{b}_1$ , because ” $A$ after $B$ ” first maps $\hat{\imath}$ to $\mathbf{b}_1$ via $B$ , then maps $\mathbf{b}_1$ to $A\mathbf{b}_1$ via $A$ . Column-picture all the way down.

The row-times-column formula $(AB)_{ij} = \sum_k A_{ik} B_{kj}$ gives the same answer. Same pattern as last lesson: column picture for meaning, rows for computation.

Multiply two matrices

Compute $\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix} \cdot \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}$ .

Give the entry at row 1, column 2 of the product.

AB ≠ BA

Matrix multiplication is not commutative. In general $AB \ne BA$ , and for most pairs of transformations the results are genuinely different.

Try “rotate 90° then shear” versus “shear then rotate 90°.” The two sequences produce visibly different final shapes. Composition cares about order because the order in which you apply transformations matters.

On the widget below, stack two transformations by eye. $R =$ 90° rotation, $S =$ shear. Do $R$ first then $S$ ; undo, do $S$ first then $R$ . Not the same.

This is the matrix $S \cdot R = \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix} \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix} = \begin{bmatrix}1 & -1 \\ 1 & 0\end{bmatrix}$ , the one from the StepCheck above.

Determinant: the signed area factor

Take any 2×2 matrix $A$ . Apply it to the unit square. The result is a parallelogram. Its signed area is called the determinant of $A$ :

\det A \;=\; ad - bc \;=\; \text{signed area of the parallelogram spanned by the columns}.

“Signed” means: positive if the transformation preserves orientation, negative if it flips (a reflection, for example). A pure rotation has determinant $+1$ . A reflection across an axis has determinant $-1$ . A matrix that stretches the plane by a factor of 3 in both directions has determinant $9$ .

Drag the columns on the widget above and watch the det readout. When the orientation flips, the color of the unit-square shape flips too. That’s the sign doing its work.

Compute a determinant

Compute $\det\begin{bmatrix}3 & 1 \\ 1 & 2\end{bmatrix}$ .

Det = 0 means collapse

When $\det A = 0$ , the parallelogram flattens. The two columns lie on the same line; their span is 1-dimensional; the transformation has squashed 2D space into a line (or, if both columns are zero, into a single point).

Four equivalent statements (memorize the equivalence, not four separate facts):

$\det A = 0$ .
The columns of $A$ are linearly dependent.
The transformation’s output span is lower-dimensional than its input.
$A$ has no inverse.

Drag your widget columns onto the same line. All four things happen at once. The determinant hits zero, the rank badge reads 1, the unit-square shape degenerates to a segment, and, as the next step will show, no inverse exists because information has been destroyed.

Inverse as undoing

The inverse of $A$ , written $A^{-1}$ , is the transformation that undoes $A$ . Apply $A$ then $A^{-1}$ and you’re back where you started.

A \cdot A^{-1} \;=\; A^{-1} \cdot A \;=\; I.

If $A$ rotates by 30°, $A^{-1}$ rotates by $-30°$ . If $A$ scales by 2, $A^{-1}$ scales by $1/2$ . Every invertible transformation has a unique “undo” operation.

But: if $A$ squashes space (say, it projects every point onto the $x$ -axis) there is no undo. Points that used to live at different $y$ -values got sent to the same place. You can’t recover information that was thrown away. That’s why $A^{-1}$ exists if and only if $\det A \ne 0$ .

The formula for 2×2 inverses (drop it in if you need to compute one):

\begin{bmatrix}a & b \\ c & d\end{bmatrix}^{-1} \;=\; \frac{1}{ad - bc}\begin{bmatrix}d & -b \\ -c & a\end{bmatrix}.

Note the $ad - bc$ in the denominator: it’s the determinant again. When it’s zero, the formula blows up. That’s the algebra telling you what the geometry already said.

Inverse of a diagonal

Given $A = \begin{bmatrix}2 & 0 \\ 0 & 3\end{bmatrix}$ , find $A^{-1}$ .

Give the entry at row 1, column 1 of the inverse. (Decimal form.)

The whole chain, in one picture

Here’s the collapse of the four ideas into one widget state. On MatrixDraggable, drag the two columns onto the same line. Simultaneously:

The unit-square polygon flattens to a line segment.
The det readout hits $0$ .
The rank badge flips from 2 to 1.
An inverse can no longer exist.

Four words for one situation. That’s linear algebra being economical.