Matrices

Order and naming. A matrix with $m$ rows and $n$ columns has order $m\times n$ (rows always stated first). A matrix that is $1\times n$ is a row matrix; one that is $m\times1$ is a column matrix. A null matrix has every entry $0$; an identity matrix $I$ is square with $1$s on the leading diagonal and $0$s everywhere else, and it acts like the number $1$ does in ordinary multiplication.

where: $A$ is a general $2\times2$ matrix and $I$ is the $2\times2$ identity, so that $AI = IA = A$ for any $2\times2$ matrix $A$.

Addition and subtraction — same order only. You may add or subtract two matrices only when their orders match exactly; then you combine corresponding entries, position by position.

where: each pair of entries in the same row-and-column position is added or subtracted independently; there is no cross-mixing between positions.

Scalar multiplication. Multiplying a matrix by a number $k$ (a scalar) scales every single entry by $k$ — nothing is compared between entries.

Matrix multiplication — a different compatibility rule entirely. $AB$ is only defined when the number of columns of $A$ equals the number of rows of $B$; the product's order is (rows of $A$) times (columns of $B$). For two $2\times2$ matrices,

where: each entry of the product is a row of the first matrix "dotted" against a column of the second — multiply matching positions and add. Multiplication is not commutative: $AB \ne BA$ in general, so order matters and you must compute the product in the sequence the question asks for. Two matrices are equal only if they share the same order and every corresponding entry is identical; this fact is often exploited to turn an "unknown entries" problem into simultaneous equations by matching entries on both sides.

Question: If $A = \begin{pmatrix}1&2\\3&-1\end{pmatrix}$ and $B = \begin{pmatrix}0&1\\2&4\end{pmatrix}$, find $AB$, state its order, and verify $AI = A$ using the $2\times2$ identity.

Step 1. Check compatibility. Both $A$ and $B$ are order $2\times2$, so columns of $A$ (2) equal rows of $B$ (2); the product $AB$ is defined and will have order $2\times2$ (rows of $A$ by columns of $B$).

Step 2. Compute each entry using row-times-column. Row $1$ of $A$, $(1\ \ 2)$, with column $1$ of $B$, $\begin{pmatrix}0\\2\end{pmatrix}$: $(1)(0)+(2)(2)=0+4=4$. Row $1$ with column $2$, $\begin{pmatrix}1\\4\end{pmatrix}$: $(1)(1)+(2)(4)=1+8=9$.

Step 3. Finish the remaining row. Row $2$ of $A$, $(3\ \ {-1})$, with column $1$ of $B$: $(3)(0)+(-1)(2)=0-2=-2$. Row $2$ with column $2$: $(3)(1)+(-1)(4)=3-4=-1$. Assembling gives $AB = \begin{pmatrix}4&9\\-2&-1\end{pmatrix}$.

Step 4. Verify with the identity matrix as a sanity check on the mechanics: for $I=\begin{pmatrix}1&0\\0&1\end{pmatrix}$, $AI$ gives $(1)(1)+(2)(0)=1$, $(1)(0)+(2)(1)=2$, $(3)(1)+(-1)(0)=3$, $(3)(0)+(-1)(1)=-1$, reproducing $A$ exactly. This confirms the row-times-column procedure was applied correctly in Steps 2–3.

$AB = \begin{pmatrix}4&9\\-2&-1\end{pmatrix}$, order $2\times2$.

Competitive stretch

For $A = \begin{pmatrix}2&-1\\3&4\end{pmatrix}$ and $B = \begin{pmatrix}5&0\\-2&1\end{pmatrix}$, find $2A-B$. First scale: $2A = \begin{pmatrix}4&-2\\6&8\end{pmatrix}$. Both matrices are order $2\times2$, so subtraction is valid entrywise: $2A-B = \begin{pmatrix}4-5&-2-0\\6-(-2)&8-1\end{pmatrix} = \begin{pmatrix}-1&-2\\8&7\end{pmatrix}$.

• Adding or subtracting matrices of different orders — this is never allowed, no matter what the entries are.
• Multiplying matrices entry-by-entry (as if it were addition) instead of applying the row-times-column rule.
• Assuming $AB=BA$ — matrix multiplication is generally not commutative, so always multiply in the order the question specifies.
• Skipping the compatibility check (columns of the first = rows of the second) before attempting a product, especially with non-square matrices.

Ask the tutor: Want to practise the two-second compatibility check on a batch of mixed-order matrices, or walk through a problem where unknown entries are found by matching equal matrices?