Practice: Matrices as Operators

(a) By rows. First entry: $(1)(2)+(3)(1)+(0)(3)=2+3+0=5$. Second entry: $(2)(2)+(1)(1)+(5)(3)=4+1+15=20$. So $Ax=(5,20)$.

(b) By columns. The columns are $A_1=(1,2)$, $A_2=(3,1)$, $A_3=(0,5)$, and the entries of $x$ are the weights:

Same answer, as it must be. The two readings group the same sum differently.

(c) The spaces. $x\in {\mathbb{R}}^3$ because $A$ has three columns and each needs a weight. $Ax\in {\mathbb{R}}^2$ because each column is a vector in ${\mathbb{R}}^2$ and a combination of them stays there. The action moves between spaces whenever the matrix is not square: it consumes a vector of length equal to the number of columns and produces one of length equal to the number of rows.

What the column reading adds. It shows $(5,20)$ is reachable by mixing $A$'s columns, so the system $Ay=(5,20)$ has at least one solution. The row computation gives the number without saying anything about reachability.

What went wrong. $Bv$ is not defined, and that is the answer. $B$ is $2\times 3$, so it consumes vectors of length three; $v$ has length two. Nothing rescues that, because the mismatch is the finding.

Why transposing is not a repair. $B^{\mathsf{T}}v$ is a different question with a different answer living in a different space. $Bv$ would have landed in ${\mathbb{R}}^2$; $B^{\mathsf{T}}v$ lands in ${\mathbb{R}}^3$. The student did not compute the requested quantity more cleverly. They computed something else and reported it under the original name.

The habit to build. Write the shapes before touching the entries: $(2\times 3)(3\times 1)$ works and yields $2\times 1$; $(2\times 3)(2\times 1)$ does not, and you know before any multiplication. The check is cheap precisely because it happens first.

What to answer. "$Bv$ is undefined: $B$ has three columns and $v$ has two entries." If the intent was a product that exists, say which one and why, do not silently substitute it.

First error: $AB$ and $BA$ are not the same. Matrix multiplication composes actions, $AB$ means do $B$, then $A$, and doing two things in the other order is a different thing to do. A concrete case: with and , while .

Second error: 'the same two matrices' is about inputs, not results. The same ingredients combined in a different order give a different product. For non-square shapes the point is sharper still, $A$ of size $2\times 3$ and $B$ of size $3\times 4$ make $AB$ defined and $BA$ meaningless, so they cannot be compared at all.

Third error: cancelling is division, which matrices do not support. From $AB=CB$ it does not follow that $A=C$. Take $B$ to be a matrix of zeros: then $AB=CB$ holds for every $A$ and $C$ whatsoever.

What would make the final step valid. If $B$ is square and invertible, right-multiplying both sides by $B^{-1}$ gives $AB{B}^{-1}=CB{B}^{-1}$, hence $A=C$. That is a legitimate operation, not cancellation, but multiplication by a matrix that exists only under that condition. Whether it exists is settled by the independence of $B$'s columns.

The second question first, as it is the simpler. "What does the current plan consume?" is answered by computing $Ax$, either reading gives the number, and the row reading is the natural way to execute it: each row is a resource, and dotting it with the plan gives how much of that resource the plan uses.

The first question is a column question. "Can we hit $b$ exactly?" asks whether $b$ can be written as $x_1{A}_1+\cdots +x_n{A}_n$, whether the target profile is reachable by mixing the products' requirement columns. That is precisely the column reading, and it is a question about existence rather than about arithmetic.

One column a multiple of another. Suppose $A_2=3A_1$: the second product uses exactly three times the resources of the first, in the same proportions. Then the two columns reach only a single line of profiles between them, and they add no independent flexibility, anything achievable with the pair is achievable with the first product alone, at a different quantity. In planning terms the second product is redundant as a means of hitting a resource profile, and in linear-algebra terms the columns are dependent, which is what makes some targets unreachable.

A practical consequence. If the manager's target $b$ is not on the span of the columns, no plan hits it exactly, however the quantities are adjusted. The right response is not to search harder but to recognise the target as unreachable and either relax it or change what the factory can make.

The check, first. $Ax$ is defined when the number of COLUMNS of $A$ matches the length of $x$. $A$ is $2\times 3$.

(a) $x$ has three entries, so $Ax$ is defined and lands in ${\mathbb{R}}^2$:

(b) $y$ has two entries and $A$ has three columns, so $Ay$ is not defined. The mismatch is decidable before any arithmetic, which is the point of checking first.

(c) $A^{\mathsf{T}}$ is $3\times 2$, so it takes a vector of length two. $y$ qualifies, and the result lands in ${\mathbb{R}}^3$:

Note. (b) and (c) use the same two objects and only one product exists. Transposing is not a formality for making shapes agree. It changes which vector space the result lives in.