1.4 - Matrix algebra and powers | Cambridge A-Level Further Mathematics

Some familiar algebraic laws survive for matrices, while others fail because order matters. Learn to keep products in the correct order, use identity and zero matrices safely, and recognise patterns in powers of square matrices.

Order of multiplication

Can two matrices have both products defined and still give $\mathbf{AB}\ne\mathbf{BA}$ ? Yes: even two square matrices of the same size need not commute.

Take $\mathbf A=\begin{pmatrix}1&2\\0&1\end{pmatrix}$ and $\mathbf B=\begin{pmatrix}2&0\\0&3\end{pmatrix}$ . Multiplying on the left by $\mathbf B$ scales rows of $\mathbf A$ ; multiplying on the right by $\mathbf B$ scales columns. There is no reason for those operations to give the same array.

Commuting matrices

Matrices commute when both products are defined and $\mathbf{AB}=\mathbf{BA}$ . Matrix multiplication is not commutative in general. Some particular pairs do commute, including a square matrix and the matching identity.

To show two matrices do not commute, one unequal corresponding entry is enough. Computing both complete products also checks the arithmetic.

Worked example 1: unequal products

Question. For the matrices above, calculate $\mathbf{AB}$ and $\mathbf{BA}$ .

Plan. Keep the row-column rule fixed and change which matrix supplies the rows.

\begin{aligned} \mathbf{AB}&=\begin{pmatrix}1(2)+2(0)&1(0)+2(3)\\0(2)+1(0)&0(0)+1(3)\end{pmatrix} =\begin{pmatrix}2&6\\0&3\end{pmatrix}\\ \mathbf{BA}&=\begin{pmatrix}2(1)+0(0)&2(2)+0(1)\\0(1)+3(0)&0(2)+3(1)\end{pmatrix} =\begin{pmatrix}2&4\\0&3\end{pmatrix}. \end{aligned}

Answer. The matrices do not commute: their top-right entries are $6$ and $4$ .

Check. Apply each sequence to $\begin{pmatrix}0\\1\end{pmatrix}$ . First $\mathbf B$ , then $\mathbf A$ , gives $\begin{pmatrix}6\\3\end{pmatrix}$ ; first $\mathbf A$ , then $\mathbf B$ , gives $\begin{pmatrix}4\\3\end{pmatrix}$ . These are the different second columns found above.

Change just the second diagonal entry of $\mathbf B$ from $3$ to $2$ . The matrix becomes a scalar multiple of the identity; now both rows and columns are scaled equally.

Worked example 2: a pair that commutes

Question. Keeping $\mathbf A=\begin{pmatrix}1&2\\0&1\end{pmatrix}$ , let $\mathbf C=\begin{pmatrix}2&0\\0&2\end{pmatrix}$ . Show that $\mathbf A$ and $\mathbf C$ commute.

Plan. Calculate both products, recognising that $\mathbf C=2\mathbf I$ .

\mathbf{AC}=\begin{pmatrix}2&4\\0&2\end{pmatrix},\qquad \mathbf{CA}=\begin{pmatrix}2&4\\0&2\end{pmatrix}.

For instance, both top-right calculations give $4$ , because the two diagonal scale factors are now equal.

Answer. $\mathbf{AC}=\mathbf{CA}=2\mathbf A$ , so this pair commutes.

Check. For any column $\begin{pmatrix}x\\y\end{pmatrix}$ , both products give $\begin{pmatrix}2x+4y\\2y\end{pmatrix}$ . Equality holds for every input, not just a selected one.

The first example disproves a general commutative law; the second shows why “matrices never commute” would also be false.

Watch out

$(\mathbf A+\mathbf B)^2=\mathbf A^2+2\mathbf{AB}+\mathbf B^2$ is generally wrong. Expansion gives $\mathbf A^2+\mathbf{AB}+\mathbf{BA}+\mathbf B^2$ ; combine the middle terms only when $\mathbf{AB}=\mathbf{BA}$ .

Test order with another pair.

Identity and zero products

The identity behaves like the number one in a product, but a zero matrix product has a surprise: neither factor need be zero.

For any $m\times n$ matrix $\mathbf A$ , $\mathbf I_m\mathbf A=\mathbf A=\mathbf A\mathbf I_n$ . On the left, an identity row selects the matching row of $\mathbf A$ ; on the right, an identity column selects its matching column. The identities may have different sizes when $\mathbf A$ is rectangular.

Matrix algebra laws

For compatible dimensions,

\begin{aligned} \mathbf A(\mathbf B+\mathbf C)&=\mathbf{AB}+\mathbf{AC}\\ (\mathbf A+\mathbf B)\mathbf C&=\mathbf{AC}+\mathbf{BC}\\ (\mathbf{AB})\mathbf C&=\mathbf A(\mathbf{BC}). \end{aligned}

Multiplication distributes over addition and is associative: brackets may change, but factor order must stay fixed. Also $\mathbf A\mathbf O=\mathbf O$ and $\mathbf O\mathbf A=\mathbf O$ whenever the products exist. Not in the formula booklet: learn it.

For distributivity, each row-column sum distributes term by term. Associativity says that grouping successive combinations differently gives the same final combination; it does not permit swapping two factors. Addition is commutative for equal-sized matrices, because ordinary entry addition is commutative.

Worked example 3: nonzero factors, zero product

Question. Calculate $\mathbf{AB}$ , where $\mathbf A=\begin{pmatrix}1&0\\0&0\end{pmatrix}$ and $\mathbf B=\begin{pmatrix}0&0\\2&3\end{pmatrix}$ .

Plan. Calculate the row-column products rather than applying the number rule for a zero product.

\mathbf{AB}=\begin{pmatrix}1(0)+0(2)&1(0)+0(3)\\0(0)+0(2)&0(0)+0(3)\end{pmatrix} =\begin{pmatrix}0&0\\0&0\end{pmatrix}.

Answer. $\mathbf{AB}=\mathbf O$ , although $\mathbf A\ne\mathbf O$ and $\mathbf B\ne\mathbf O$ .

Check. Every output of $\mathbf B$ has first coordinate zero. Multiplication by $\mathbf A$ keeps only the first coordinate, so it sends each such output to zero.

Now change only the second row of the right factor. The left matrix still discards that whole row's contribution.

Worked example 4: why cancellation can fail

Question. Let $\mathbf A=\begin{pmatrix}1&0\\0&0\end{pmatrix}$ , $\mathbf B=\begin{pmatrix}0&0\\2&3\end{pmatrix}$ and $\mathbf C=\begin{pmatrix}0&0\\4&5\end{pmatrix}$ . Does $\mathbf{AB}=\mathbf{AC}$ imply $\mathbf B=\mathbf C$ ?

Plan. Compute $\mathbf{AC}$ and compare the right factors.

\mathbf{AC}=\begin{pmatrix}1(0)+0(4)&1(0)+0(5)\\0&0\end{pmatrix}=\mathbf O=\mathbf{AB}.

But $\mathbf B$ and $\mathbf C$ have unequal second rows.

Answer. No: $\mathbf{AB}=\mathbf{AC}$ here, yet $\mathbf B\ne\mathbf C$ .

Check. Subtracting gives $\mathbf A(\mathbf B-\mathbf C)=\mathbf A\begin{pmatrix}0&0\\-2&-2\end{pmatrix}=\mathbf O$ , verifying the equality by a separate product.

Later, an inverse will tell us when cancellation is valid. A nonzero matrix alone is not sufficient. There is no general matrix “division” operation that allows a factor to be crossed out.

Watch out

$\mathbf{AB}=\mathbf O$ does not imply that either factor is zero. Likewise, cancelling a merely nonzero matrix from $\mathbf{AB}=\mathbf{AC}$ can lose valid possibilities.

Find a zero product directly.

Powers of square matrices

A power of a square matrix means repeated matrix multiplication, not raising each entry to that power.

For example, $\begin{pmatrix}1&2\\0&1\end{pmatrix}^2=\begin{pmatrix}1&4\\0&1\end{pmatrix}$ . The top-right entry is a sum of products, $1(2)+2(1)=4$ , not simply $2^2$ ; the agreement with $2^2$ in this entry is accidental.

Matrix powers

For a square matrix $\mathbf A$ , $\mathbf A^n$ is the product of $n$ copies of $\mathbf A$ , for positive integer $n$ . Define $\mathbf A^0=\mathbf I$ . Thus $\mathbf A^r\mathbf A^s=\mathbf A^{r+s}$ for nonnegative integers $r,s$ . A non-square matrix has no ordinary square power because its self-product is incompatible.

Repeated multiplication may reveal a short pattern. Check the mechanism that continues it; the first two values alone do not prove a general formula.

Worked example 5: an accumulating entry

Question. For $\mathbf U=\begin{pmatrix}1&2\\0&1\end{pmatrix}$ , calculate $\mathbf U^2$ and $\mathbf U^3$ , then determine $\mathbf U^{12}$ .

Plan. Multiply successively and identify how each additional factor changes the upper-right entry.

\mathbf U^2=\begin{pmatrix}1&4\\0&1\end{pmatrix},\qquad \mathbf U^3=\begin{pmatrix}1&6\\0&1\end{pmatrix}.

For any number $a$ ,

\begin{pmatrix}1&a\\0&1\end{pmatrix}\mathbf U =\begin{pmatrix}1&a+2\\0&1\end{pmatrix}.

Each new factor therefore adds $2$ , leaving the other entries unchanged. Starting from $\mathbf U^0=\mathbf I$ , twelve factors give an upper-right entry $24$ .

Answer. $\mathbf U^{12}=\begin{pmatrix}1&24\\0&1\end{pmatrix}$ .

Check. Three blocks of $\mathbf U^4=\begin{pmatrix}1&8\\0&1\end{pmatrix}$ combine to give upper-right entry $8+8+8=24$ , independently grouping the twelve factors.

Change just the sign of the upper-right entry. The same mechanism now subtracts $2$ at each multiplication.

Worked example 6: a negative increment

Question. For $\mathbf V=\begin{pmatrix}1&-2\\0&1\end{pmatrix}$ , find $\mathbf V^2$ , $\mathbf V^3$ and $\mathbf V^{12}$ .

Plan. Track the upper-right entry while preserving matrix multiplication.

\mathbf V^2=\begin{pmatrix}1&-4\\0&1\end{pmatrix},\qquad \mathbf V^3=\begin{pmatrix}1&-6\\0&1\end{pmatrix}.

Multiplying $\begin{pmatrix}1&a\\0&1\end{pmatrix}$ by $\mathbf V$ changes $a$ to $a-2$ , so twelve copies give $-24$ .

Answer. $\mathbf V^{12}=\begin{pmatrix}1&-24\\0&1\end{pmatrix}$ .

Check. Direct calculation gives $\mathbf{UV}=\mathbf{VU}=\mathbf I$ . Multiplying the proposed twelfth powers also gives $\mathbf I$ , as twelve cancelling pairs must.

Other powers cycle instead of growing. For $\mathbf R=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ , multiplication gives $\mathbf R^2=-\mathbf I$ and hence $\mathbf R^4=\mathbf I$ . A block of four factors can then be removed: $\mathbf R^{10}=(\mathbf R^4)^2\mathbf R^2=-\mathbf I$ . Always establish the short identity before reducing a large power.

Problem-solving

Look for a stable entry pattern or a small power equal to $\mathbf I$ , $-\mathbf I$ or $\mathbf O$ . Keep factors in order; powers of the same matrix commute because they are copies of that one matrix.

Use the accumulating-entry method with a fresh increment.

Quick Recap

Matrix multiplication is associative and distributive; it is generally not commutative.
Matching identities leave a matrix unchanged, but zero products can have nonzero factors.
Do not cancel a matrix just because it is nonzero.
Powers use repeated matrix multiplication; justify any pattern you use.

The opening pair has $\mathbf{AB}\ne\mathbf{BA}$ , even though both are defined. Order remains essential in every later calculation.

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1.4 - Matrix algebra and powers

Order of multiplication

Identity and zero products

Powers of square matrices

Quick Recap

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