Matrices & Systems

Why rows meet columns: column $j$ of $AB$ is $A$ applied to column $j$ of $B$, and $A$ applied to a vector produces, in slot $i$, row $i$ of $A$ dotted with that vector. So the rule is not a convention; it is what "apply $B$, then $A$" forces. The size rule is the same fact: $A$ can only act on vectors with as many entries as $A$ has columns. A $2\times3$ times a $3\times4$ is $2\times4$, while a $3\times4$ times a $2\times3$ is not defined at all.

Read $A\mathbf x$ column by column: $A\begin{pmatrix}x\\y\end{pmatrix} = x\begin{pmatrix}2\\1\end{pmatrix} + y\begin{pmatrix}1\\-1\end{pmatrix}$. Solving the system means finding the weights that combine the columns into $\mathbf b$; here $3(2,1) + 1(1,-1) = (7,2)$, so $(x, y) = (3, 1)$. Had the two columns pointed along the same line, their combinations would fill only that line, most $\mathbf b$ would be out of reach, and $A$ would have no inverse — the three dialects fail together.

Test the reasoning on a near miss: $(1,2,3)$ and $(2,4,7)$ agree on the multiplier $2$ in the first two coordinates and disagree in the third ($2\cdot3 = 6 \ne 7$), so they are independent. A single mismatched coordinate is enough; two vectors are dependent only when one multiplier works in every coordinate at once.

The two pictures are the two ways of computing $A\mathbf x$. Entry by entry, slot $i$ of $A\mathbf x$ is row $i$ dotted with $\mathbf x$, so setting it equal to $b_i$ gives one equation — one line. All at once, $A\mathbf x$ is the columns weighted by the entries of $\mathbf x$. The row picture tells you where the answer is; the column picture tells you whether an answer can exist: $A\mathbf x = \mathbf b$ is solvable exactly when $\mathbf b$ lies in the span of the columns.

Column picture: the columns are $\mathbf a_1 = (1, 3)$ and $\mathbf a_2 = (2, -1)$, and the solution claims $2\mathbf a_1 + 1\mathbf a_2 = \mathbf b$. Check: $2(1,3) + (2,-1) = (4, 5)$ ✓. The row picture located the point $(2, 1)$; the column picture reads the same numbers as a recipe — two of the first column, one of the second. Because the columns are not parallel (the determinant $1\cdot(-1) - 2\cdot3 = -7$ is not $0$), every $\mathbf b$ in the plane has exactly one such recipe.

Combine slot by slot:

Check by rows: row 1 gives $1\cdot2 + 2\cdot(-1) = 0$, row 2 gives $0\cdot2 + 1\cdot(-1) = -1$, row 3 gives $3\cdot2 + (-1)(-1) = 7$ ✓. The shapes carry information too: a $3\times2$ matrix takes vectors with $2$ entries to vectors with $3$, so its outputs — combinations of just two columns — fill at most a plane inside $\mathbb R^3$. Most right-hand sides in $\mathbb R^3$ are out of its reach.

For $\mathbf b = (1, 2, 3)$: the first two coordinates force $c_1 = 1$, $c_2 = 2$, and the third must then be $c_1 + c_2 = 3$ — it is, so $\mathbf b = \mathbf a_1 + 2\mathbf a_2$ and the system $A\mathbf x = \mathbf b$ has the solution $\mathbf x = (1, 2)$. For $\mathbf b' = (1, 2, 4)$ the same weights are forced, but the third coordinate comes out $3 \ne 4$: no solution. Geometrically, every combination satisfies $z = x + y$, so the columns span that plane; $\mathbf b$ lies in it and $\mathbf b'$ sits one unit above it.

The reversal is the socks-and-shoes rule: socks go on first, so they come off last. $AB$ applies $B$ first, so undoing it must undo $A$ first: $(B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}A)B = B^{-1}B = I$. The Gauss–Jordan recipe works because each row operation is multiplication on the left by a simple invertible matrix. The operations that turn $A$ into $I$ therefore multiply to $A^{-1}$, and applying the same operations to $I$ writes that product down.

Row times column: entry $(1,1)$ is $2\cdot3 + 1\cdot(-5) = 1$; entry $(1,2)$ is $2\cdot(-1) + 1\cdot2 = 0$; entry $(2,1)$ is $5\cdot3 + 3\cdot(-5) = 0$; entry $(2,2)$ is $5\cdot(-1) + 3\cdot2 = 1$. So $AB = I$ ✓. The system is $A\mathbf x = (4, 11)$; multiplying both sides on the left by $A^{-1} = B$ isolates $\mathbf x$: $\mathbf x = B\begin{pmatrix}4\\11\end{pmatrix} = \begin{pmatrix}12 - 11\\-20 + 22\end{pmatrix} = \begin{pmatrix}1\\2\end{pmatrix}$. Check in the original equations: $2 + 2 = 4$ and $5 + 6 = 11$ ✓.

The second pivot is already $1$, so clear above it: $R_1 \gets R_1 - 2R_2$ gives $\left[\begin{array}{cc|cc}1&0&7&-2\\0&1&-3&1\end{array}\right]$. The left half is $I$, so the right half is the inverse: $A^{-1} = \begin{pmatrix}7&-2\\-3&1\end{pmatrix}$. Check:

The same march works at any size. Had a row of zeros appeared in the left half, there would be no pivot to scale to $1$, and $A$ would have no inverse.

The shear happened last, so it must be undone first:

Check:

The other order, $A^{-1}B^{-1} = \begin{pmatrix}\tfrac12&-1\\0&1\end{pmatrix}$, fails: $AB$ sends $\mathbf e_2$ to $(1, 1)$, and $A^{-1}B^{-1}$ sends $(1,1)$ to $(-\tfrac12, 1)$ instead of back to $\mathbf e_2$.

But a linear map is determined by where it sends basis vectors, so the matrix whose columns are $A(B\mathbf e_j)$ Is the matrix of $T_A\circ T_B$. The row-times-column rule is just this computed entry by entry — and composition of maps is famously order-sensitive, which is why $AB \ne BA$. $\blacksquare$

Any other input is a combination $\mathbf x = x_1\mathbf e_1 + x_2\mathbf e_2$, and linearity forces $A\mathbf x = x_1(\text{col}_1) + x_2(\text{col}_2)$ — matrix-times-vector is a weighted sum of columns. Example: $A = \begin{pmatrix}2&1\\0&3\end{pmatrix}$ sends the unit square spanned by $\mathbf e_1, \mathbf e_2$ to the parallelogram spanned by $(2,0)$ and $(1,3)$: the plane is stretched, sheared, and every area is multiplied by $|\det A| = 6$. This picture explains the multiplication rule too: the columns of $AB$ are $A$ applied to the columns of $B$ — do $B$ first, then $A$ — which is why $AB$ and $BA$ genuinely differ: shear-then-rotate is not rotate-then-shear.

Entry $(2,2)$: $3\cdot1 + 4\cdot2 = 11$. So $AB = \begin{pmatrix}10&5\\20&11\end{pmatrix}$. Now the other order: $BA = \begin{pmatrix}0\cdot1+1\cdot3 & 0\cdot2+1\cdot4\\ 5\cdot1+2\cdot3 & 5\cdot2+2\cdot4\end{pmatrix} = \begin{pmatrix}3&4\\11&18\end{pmatrix}$ — completely different, as composition order predicts. One determinant check seals both: $\det A\cdot\det B = (-2)(-5) = 10$, and $\det(AB) = 10\cdot11 - 5\cdot20 = 10$ ✓.

Way 1 — elimination: add the equations to kill $y$: $3x = 9$, so $x = 3$, then $y = 7 - 6 = 1$.

Way 2 — inverse matrix: $A = \begin{pmatrix}2&1\\1&-1\end{pmatrix}$ has $\det A = -2 - 1 = -3$, so $A^{-1} = \tfrac{1}{-3}\begin{pmatrix}-1&-1\\-1&2\end{pmatrix}$, and $\mathbf x = A^{-1}\begin{pmatrix}7\\2\end{pmatrix} = \tfrac{1}{-3}\begin{pmatrix}-9\\-3\end{pmatrix} = \begin{pmatrix}3\\1\end{pmatrix}$.

Way 3 — column picture: find weights making $x\begin{pmatrix}2\\1\end{pmatrix} + y\begin{pmatrix}1\\-1\end{pmatrix} = \begin{pmatrix}7\\2\end{pmatrix}$; the weights $3$ and $1$ work. Three languages, one answer $(3, 1)$ — and fluency means moving between them at will.