4.03b, 4.03c - Adding and multiplying matrices | OCR A-Level Further Mathematics A

Add and subtract matching entries, multiply matrices by scalars, and build matrix products one row and column at a time. In $4.03b$ you will also use powers, zero matrices and identity matrices; in $4.03c$ you will see why brackets can move in a product but the matrices usually cannot. The same rules work with real or complex entries.

Addition and scalar multiples

Can two non-zero matrices multiply to give the zero matrix? Ordinary numbers cannot do this, but matrices can. We will build the arithmetic needed to produce and explain an example.

You already know

Dimensions mean rows by columns. Addition needs identical dimensions; $\mathbf{AB}$ needs the columns of $\mathbf{A}$ to match the rows of $\mathbf{B}$ . Complex entries obey the arithmetic you already know, with $i^2=-1$ .

Start with the more familiar operations. If two tables record the same quantities in the same positions, their total comes from adding corresponding entries. Thus $\begin{pmatrix}3&2\\1&4\end{pmatrix}+\begin{pmatrix}5&1\\2&0\end{pmatrix}=\begin{pmatrix}8&3\\3&4\end{pmatrix}$ . Multiplying by a single number scales every entry.

Entry-by-entry operations

For matrices $\mathbf{A}$ and $\mathbf{B}$ of the same dimensions and a scalar $k$ ,

(\mathbf{A}\pm\mathbf{B})_{ij}=a_{ij}\pm b_{ij},\qquad (k\mathbf{A})_{ij}=ka_{ij}

A scalar is a single number, real or complex, rather than a matrix. Not in the formula booklet: learn it.

Keep both matrices in the same layout while scaling; then each subtraction has a clearly matching position.

Worked example 1: a linear combination

Question. Given $\mathbf{A}=\begin{pmatrix}2&-1\\3&4\end{pmatrix}$ and $\mathbf{B}=\begin{pmatrix}1&5\\-2&0\end{pmatrix}$ , find $2\mathbf{A}-3\mathbf{B}$ .

Plan. Scale each matrix, then subtract corresponding entries.

\begin{aligned} 2\mathbf{A}-3\mathbf{B} &=\begin{pmatrix}4&-2\\6&8\end{pmatrix}-\begin{pmatrix}3&15\\-6&0\end{pmatrix} && \text{scale every entry}\\ &=\begin{pmatrix}1&-17\\12&8\end{pmatrix} && \text{subtract corresponding entries} \end{aligned}

Why this step? Writing both scaled matrices first keeps the subtraction of a negative entry visible.

Answer. $\begin{pmatrix}1&-17\\12&8\end{pmatrix}$ .

Check. Adding $3\mathbf{B}$ to the result gives $\begin{pmatrix}4&-2\\6&8\end{pmatrix}=2\mathbf{A}$ .

The second example changes only the entry type: some entries are complex. The positions and operation stay the same.

Worked example 2: a complex linear combination

Question. Find $2\mathbf{C}-3\mathbf{D}$ , where $\mathbf{C}=\begin{pmatrix}1+i&2\\-i&3\end{pmatrix}$ and $\mathbf{D}=\begin{pmatrix}i&1\\2&1-i\end{pmatrix}$ .

Plan. Scale and subtract entry by entry, collecting real and imaginary parts within each entry.

\begin{aligned} 2\mathbf{C}-3\mathbf{D} &=\begin{pmatrix}2+2i&4\\-2i&6\end{pmatrix}-\begin{pmatrix}3i&3\\6&3-3i\end{pmatrix} && \text{scale}\\ &=\begin{pmatrix}2-i&1\\-6-2i&3+3i\end{pmatrix} && \text{subtract and simplify} \end{aligned}

Why this step? Each entry is one complex number; collect its parts without mixing it with a neighbouring entry.

Answer. $\begin{pmatrix}2-i&1\\-6-2i&3+3i\end{pmatrix}$ .

Check. The bottom-right entry plus $3-3i$ is $6$ , the corresponding entry of $2\mathbf{C}$ .

Subtracting a matrix means subtracting each complete entry, including its sign.

Watch out

Wrong: $6-(-6)=0$ . Subtracting the negative bottom-left entry in the first example gives $6+6=12$ . Keep the brackets until the sign is settled.

The next calculation varies the scalar as well as the entries.

Row-by-column multiplication

Matrix multiplication combines a whole row with a whole column; it does not multiply corresponding entries.

For a concrete case, suppose the row $\begin{pmatrix}2&3&1\end{pmatrix}$ records quantities and the column $\begin{pmatrix}4\\5\\6\end{pmatrix}$ records prices. Their product is the one-entry matrix $\begin{pmatrix}2\times4+3\times5+1\times6\end{pmatrix}=\begin{pmatrix}29\end{pmatrix}$ . There are three products because the inner dimensions are both $3$ .

Matrix product

If $\mathbf{A}$ is $m$ by $n$ and $\mathbf{B}$ is $n$ by $p$ , then $\mathbf{AB}$ is $m$ by $p$ , with

(\mathbf{AB})_{ij}=a_{i1}b_{1j}+a_{i2}b_{2j}+\cdots+a_{in}b_{nj}

Use row $i$ of the first matrix and column $j$ of the second. Not in the formula booklet: learn it.

The output has one entry for each possible row-column pairing. For a $2$ by $3$ matrix times a $3$ by $4$ matrix, there are eight pairings, and each entry is a sum of three products. Numerical calculations may be done by calculator unless detailed reasoning is required, but the order and dimensions are still your responsibility.

Worked example 3: four row-column pairings

Question. Find $\mathbf{AB}$ for $\mathbf{A}=\begin{pmatrix}1&2\\3&-1\end{pmatrix}$ , $\mathbf{B}=\begin{pmatrix}4&0\\-2&5\end{pmatrix}$ .

Plan. The product is $2$ by $2$ ; calculate each entry from its row and column.

\begin{aligned} \mathbf{AB}&=\begin{pmatrix}1(4)+2(-2)&1(0)+2(5)\\3(4)+(-1)(-2)&3(0)+(-1)(5)\end{pmatrix} && \text{row by column}\\ &=\begin{pmatrix}0&10\\14&-5\end{pmatrix} && \text{simplify} \end{aligned}

Why this step? The top-right entry needs the first row and second column, not the two top-right entries.

Answer. $\mathbf{AB}=\begin{pmatrix}0&10\\14&-5\end{pmatrix}$ .

Check. The second column is $\mathbf{A}\begin{pmatrix}0\\5\end{pmatrix}=5\begin{pmatrix}2\\-1\end{pmatrix}=\begin{pmatrix}10\\-5\end{pmatrix}$ .

The second example changes only the entries to complex numbers. The row-column pattern is unchanged.

Worked example 4: a complex matrix product

Question. Find $\mathbf{CD}$ , where $\mathbf{C}=\begin{pmatrix}1&i\\2&1-i\end{pmatrix}$ , $\mathbf{D}=\begin{pmatrix}i&1\\1&-i\end{pmatrix}$ .

Plan. Form each row-column sum, then replace $i^2$ by $-1$ .

\begin{aligned} \mathbf{CD}&=\begin{pmatrix}i+i&1-i^2\\2i+1-i&2+(1-i)(-i)\end{pmatrix} && \text{row by column}\\ &=\begin{pmatrix}2i&2\\1+i&1-i\end{pmatrix} && i^2=-1 \end{aligned}

Why this step? Complex arithmetic happens inside each row-column sum; it changes no matrix rule.

Answer. $\mathbf{CD}=\begin{pmatrix}2i&2\\1+i&1-i\end{pmatrix}$ .

Check. The bottom-right entry is $2-i+i^2=1-i$ , agreeing with the result.

Changing the shape changes the number of output entries, but each still comes from a complete row-column sum.

Watch out

Wrong: multiplying just the four corresponding pairs of entries. For example, the first example's top-left entry is $1(4)+2(-2)=0$ , not $1(4)=4$ . Matrix multiplication needs the sum across the whole row-column pair.

Keep that pattern even when the product is rectangular.

Order and brackets

You may regroup a matrix product, but you may not usually reverse its factors.

Associative and commutative

An operation is associative if regrouping three objects leaves the result unchanged. It is commutative if exchanging two objects leaves the result unchanged. For conformable matrices,

(\mathbf{AB})\mathbf{C}=\mathbf{A}(\mathbf{BC})

so multiplication is associative. In general $\mathbf{AB}\ne\mathbf{BA}$ , so multiplication is not commutative. These results are not in the formula booklet: learn them.

Why can brackets move? Both groupings make the same products $a_{ir}b_{rs}c_{sj}$ and add them; regrouping finite sums does not change their total. It also lets you calculate efficiently: if a $10$ by $2$ matrix multiplies a $2$ by $10$ matrix and then a $10$ by $1$ column, working out the last two factors first keeps the intermediate matrix small.

Reversing the order is different. It changes the row-column pairs and may even make the product undefined. When both products exist, calculate and compare them rather than declaring that they must differ: some pairs do commute.

Worked example 5: demonstrating non-commutativity

Question. Use $\mathbf{A}=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ and $\mathbf{B}=\begin{pmatrix}2&0\\0&3\end{pmatrix}$ to demonstrate that matrix multiplication is not commutative.

Plan. Calculate both products and compare a pair of corresponding entries.

\begin{aligned} \mathbf{AB}&=\begin{pmatrix}2&3\\0&3\end{pmatrix} && \text{multiply in the given order}\\ \mathbf{BA}&=\begin{pmatrix}2&2\\0&3\end{pmatrix} && \text{reverse the factors}\\ \mathbf{AB}&\ne\mathbf{BA} && \text{top-right entries differ} \end{aligned}

Why this step? A counterexample needs both orders and a conclusion; quoting the general statement alone shows nothing about these matrices.

Answer. The top-right entries are $3$ and $2$ , so the products are unequal and multiplication is not commutative.

Check. The other three entries agree, but one unequal pair is enough.

The second example changes only one diagonal entry into an unknown. Now the comparison finds the exceptional case where the matrices commute.

Worked example 6: when a pair does commute

Question. For $\mathbf{A}=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ and $\mathbf{B}=\begin{pmatrix}2&0\\0&t\end{pmatrix}$ , find $t$ such that $\mathbf{AB}=\mathbf{BA}$ .

Plan. Calculate both products, then equate the entries that can differ.

\begin{aligned} \mathbf{AB}&=\begin{pmatrix}2&t\\0&t\end{pmatrix},\qquad \mathbf{BA}=\begin{pmatrix}2&2\\0&t\end{pmatrix} && \text{row by column}\\ t&=2 && \text{equate top-right entries} \end{aligned}

Why this step? Only the top-right equality creates a condition; the other entries agree for every $t$ .

Answer. $t=2$ .

Check. Then $\mathbf{B}=2\mathbf{I}$ , and both products are $2\mathbf{A}$ .

The distinction is between a general rule and a particular pair: one pair can commute without making multiplication commutative.

Exam tip

For “demonstrate that matrix multiplication is not commutative”, finish with $\mathbf{AB}\ne\mathbf{BA}$ after the calculation. OCR also accepts a correctly calculated unequal pair of corresponding entries, explicitly compared.

Use the comparison to diagnose this claim.

Powers, zero and identity

A positive integer power means repeated matrix multiplication, so square matrices are the ones whose powers you can form.

Powers and special matrices

For a square matrix $\mathbf{A}$ and positive integers $m,n$ ,

\mathbf{A}^{n}=\underbrace{\mathbf{A}\mathbf{A}\cdots\mathbf{A}}_{n\text{ factors}},\qquad \mathbf{A}^{m}\mathbf{A}^{n}=\mathbf{A}^{m+n}

For any matrix $\mathbf{B}$ , with correctly sized zero and identity matrices,

\mathbf{B}+\mathbf{0}=\mathbf{B},\quad \mathbf{B}\mathbf{0}=\mathbf{0},\quad \mathbf{0}\mathbf{B}=\mathbf{0},\quad \mathbf{I}\mathbf{B}=\mathbf{B}=\mathbf{B}\mathbf{I}

Not in the formula booklet: learn it.

These effects follow from the row-column rule. A zero column gives only zero products. An identity column selects exactly one entry of each row. For an $m$ by $n$ matrix, the two identity sizes differ: $\mathbf{I}_m\mathbf{B}=\mathbf{B}\mathbf{I}_n=\mathbf{B}$ . Associativity makes repeated multiplication unambiguous.

Worked example 7: a matrix whose square is zero

Question. For $\mathbf{N}=\begin{pmatrix}0&2\\0&0\end{pmatrix}$ , find $\mathbf{N}^2$ and $\mathbf{N}^3$ .

Plan. Multiply $\mathbf{N}$ by itself, then multiply the result by $\mathbf{N}$ .

\begin{aligned} \mathbf{N}^2&=\begin{pmatrix}0(0)+2(0)&0(2)+2(0)\\0&0\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix} && \text{row by column}\\ \mathbf{N}^3&=\mathbf{N}^2\mathbf{N}=\mathbf{0}\mathbf{N}=\mathbf{0} && \text{zero times any conformable matrix} \end{aligned}

Why this step? Squaring a matrix means a product of matrices, so the non-zero entry need not survive.

Answer. Both powers are the $2$ by $2$ zero matrix.

Check. $\mathbf{N}\ne\mathbf{0}$ , yet every row-column sum in its square is zero. This answers the opening question.

The second example changes the matrix size to three dimensions, keeping the same idea of entries moving along a chain.

Worked example 8: a three-step chain

Question. For $\mathbf{P}=\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}$ , find $\mathbf{P}^2$ and $\mathbf{P}^3$ .

Plan. Use the row-column rule twice, keeping the order unchanged.

\begin{aligned} \mathbf{P}^2&=\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix} && \text{multiply }\mathbf{P}\text{ by itself}\\ \mathbf{P}^3&=\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix} =\begin{pmatrix}0&0&0\\0&0&0\\0&0&0\end{pmatrix} && \text{multiply once more} \end{aligned}

Why this step? Multiplying the first row of $\mathbf{P}^2$ by $\mathbf{P}$ selects its third row, which is zero.

Answer. $\mathbf{P}^2=\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}$ and $\mathbf{P}^3=\mathbf{0}$ .

Check. Acting on $\begin{pmatrix}x\\y\\z\end{pmatrix}$ , $\mathbf{P}$ gives $\begin{pmatrix}y\\z\\0\end{pmatrix}$ , then $\begin{pmatrix}z\\0\\0\end{pmatrix}$ , then zero.

The chain calculation explains why powers can vanish even though the starting matrix contains a non-zero entry.

Watch out

Wrong: $\begin{pmatrix}0&2\\0&0\end{pmatrix}^{2}=\begin{pmatrix}0&4\\0&0\end{pmatrix}$ . This squares individual entries. A matrix square uses row-column multiplication and is zero here. Also, $\mathbf{AB}=\mathbf{0}$ does not force either factor to be zero.

A final variation lengthens the links but keeps the same shape.

Quick Recap

Add or subtract corresponding entries; a scalar multiplies every entry.
A matrix product uses row-column sums, and its dimensions are the outer dimensions.
Matrix multiplication is associative, so brackets can move; it is not commutative in general, so factor order matters.
A matrix power is repeated multiplication. The identity leaves a conformable matrix unchanged; multiplying by a zero matrix gives a zero matrix.

Yes: the non-zero matrix $\begin{pmatrix}0&2\\0&0\end{pmatrix}$ multiplied by itself gives the zero matrix.

Last Accessed

Pure Core

Statistics

Mechanics

Discrete Mathematics

Additional Pure Mathematics

4.03b, 4.03c - Adding and multiplying matrices

Addition and scalar multiples

Row-by-column multiplication

Order and brackets

Powers, zero and identity

Quick Recap

Learn faster with Nova