A-Level Further Maths 9231 / Topic 1.4 / AS & A Level

Rows meet columns: the arithmetic of matrices.

A matrix is a rectangular array of numbers. You add matrices entry by entry, but you multiply them row by column, and that one rule explains everything else: when a product exists, what size it is, and why \(AB\) and \(BA\) are usually different.

Topic 1.4All matrices lessons →

The key idea

A matrix with \(m\) rows and \(n\) columns has order \(m \times n\). Matrices of the same order add and subtract entry by entry, and a scalar multiplies every entry. The product \(AB\) exists only when the number of columns of \(A\) equals the number of rows of \(B\): an \(m \times n\) matrix times an \(n \times p\) matrix is \(m \times p\), and its \((i, j)\) entry is row \(i\) of \(A\) times column \(j\) of \(B\). In general \(AB \neq BA\), but \((AB)C = A(BC)\), \(AI = IA = A\) and \(A + O = A\).

Fig. 1: entry (2, 1) of AB comes from row 2 of A and column 1 of B. Multiply the pairs of entries in order and add: (-1)(2) + (3)(0) + (4)(3) = 10. A has 3 columns and B has 3 rows, so the pairs match up and the product exists

Section 01

Order, sums and scalar multiples.

The order of a matrix is its size, written rows \(\times\) columns: \begin{pmatrix} 2 & -1 & 3 \\ 0 & 4 & 1 \end{pmatrix} has 2 rows and 3 columns, so it is a \(2 \times 3\) matrix. In this course a matrix has at most 3 rows and 3 columns, and it need not be square. Two matrices can be added or subtracted only if they have the same order; you then add or subtract the entries in matching positions. Multiplying by a scalar \(k\) multiplies every entry by \(k\).

Entry \(a_{ij}\) sits in row \(i\) and column \(j\): rows first, then columns, exactly as in the order.
\(A + B = B + A\) and \((A + B) + C = A + (B + C)\): addition behaves like ordinary addition.
The zero matrix \(O\) has every entry \(0\); for each order there is one, for example \(O = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\) for \(2 \times 3\). It satisfies \(A + O = A\) and \(A - A = O\).
Two matrices are equal only if they have the same order and every pair of matching entries is equal, so one matrix equation of order \(2 \times 2\) gives four ordinary equations.

Worked example A combination of two matrices

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

Both are \(2 \times 3\), so the combination is defined and is also \(2 \times 3\).
\(2A = \begin{pmatrix} 4 & -2 & 6 \\ 0 & 8 & 2 \end{pmatrix}\) and \(3B = \begin{pmatrix} 3 & 15 & -6 \\ 9 & -3 & 0 \end{pmatrix}\).
Subtract entry by entry: \(4 - 3 = 1\), \(-2 - 15 = -17\), \(6 - (-6) = 12\), \(0 - 9 = -9\), \(8 - (-3) = 11\), \(2 - 0 = 2\).

Answer: \(2A - 3B = \begin{pmatrix} 1 & -17 & 12 \\ -9 & 11 & 2 \end{pmatrix}\)

Worked example Unknown entries from an equation

Find \(a\), \(b\), \(c\) and \(d\) given that \(\begin{pmatrix} a & 3 \\ 1 & b \end{pmatrix} + 2\begin{pmatrix} 1 & c \\ d & -2 \end{pmatrix} = \begin{pmatrix} 5 & 7 \\ -3 & 0 \end{pmatrix}\).

The left side is \(\begin{pmatrix} a + 2 & 3 + 2c \\ 1 + 2d & b - 4 \end{pmatrix}\).
Equate matching entries: \(a + 2 = 5\), \(3 + 2c = 7\), \(1 + 2d = -3\), \(b - 4 = 0\).
Solve each one separately.

Answer: \(a = 3\), \(b = 4\), \(c = 2\), \(d = -2\)

Your turn

Find the matrix \(X\) such that \(3X + \begin{pmatrix} 1 & 2 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 7 & -4 \\ 6 & 5 \end{pmatrix}\).

Examiner trap

A \(2 \times 3\) matrix and a \(3 \times 2\) matrix have the same number of entries but different orders, so their sum is not defined. And a scalar multiplies every entry: \(3\begin{pmatrix} 1 & 2 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 3 & 6 \\ 0 & -3 \end{pmatrix}\), not just the first row or the first entry.

Section 02

Row by column.

To multiply \(A\) by \(B\), take row \(i\) of \(A\) and column \(j\) of \(B\), multiply the pairs of entries in order and add: that number is the \((i, j)\) entry of \(AB\) (Fig. 1). The pairs only match up when the number of columns of \(A\) equals the number of rows of \(B\). The inside numbers of the orders must agree, and the outside numbers give the order of the product.

\[(m \times n)(n \times p) = m \times p\]

Order matters. \(AB\) and \(BA\) can both exist and still differ; they may even have different orders. With \(A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\) and \(B = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}\), \(AB = \begin{pmatrix} 7 & 2 \\ 3 & 1 \end{pmatrix}\) but \(BA = \begin{pmatrix} 1 & 2 \\ 3 & 7 \end{pmatrix}\). So say 'pre-multiply by \(A\)' (\(AB\)) or 'post-multiply by \(A\)' (\(BA\)) carefully.
Brackets do not matter. \((AB)C = A(BC)\) whenever the products exist, so a product of three matrices is written \(ABC\) and you may work out either pair first, keeping the left-to-right order.
Identity. The identity (unit) matrix \(I\) is square with 1s down the leading diagonal and 0s elsewhere: \(I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) or \(\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\). For a square \(A\) of the same order, \(AI = IA = A\), just as \(1 \times a = a\) for numbers. Also \(AO = OA = O\).

Worked example Both products, different orders

Given \(A = \begin{pmatrix} 1 & 2 & 0 \\ -1 & 3 & 4 \end{pmatrix}\) and \(B = \begin{pmatrix} 2 & 1 \\ 0 & -1 \\ 3 & 2 \end{pmatrix}\), find \(AB\) and \(BA\).

\(A\) is \(2 \times 3\) and \(B\) is \(3 \times 2\): \(AB\) is \(2 \times 2\), and \(BA\) is \(3 \times 3\).
Row 1 of \(A\) with the columns of \(B\): \(1(2) + 2(0) + 0(3) = 2\) and \(1(1) + 2(-1) + 0(2) = -1\).
Row 2 of \(A\): \(-1(2) + 3(0) + 4(3) = 10\) and \(-1(1) + 3(-1) + 4(2) = 4\). So \(AB = \begin{pmatrix} 2 & -1 \\ 10 & 4 \end{pmatrix}\).
For \(BA\), row 1 of \(B\) is \((2, 1)\): \(2(1) + 1(-1) = 1\), \(2(2) + 1(3) = 7\), \(2(0) + 1(4) = 4\). Rows 2 and 3 likewise.
\[BA = \begin{pmatrix} 1 & 7 & 4 \\ 1 & -3 & -4 \\ 1 & 12 & 8 \end{pmatrix}\]

Answer: \(AB = \begin{pmatrix} 2 & -1 \\ 10 & 4 \end{pmatrix}\) (\(2 \times 2\)) and \(BA = \begin{pmatrix} 1 & 7 & 4 \\ 1 & -3 & -4 \\ 1 & 12 & 8 \end{pmatrix}\) (\(3 \times 3\))

Worked example A product of three

Given \(A = \begin{pmatrix} 1 & 2 & -1 \end{pmatrix}\), \(B = \begin{pmatrix} 1 & 0 \\ 2 & 1 \\ 0 & 3 \end{pmatrix}\) and \(C = \begin{pmatrix} 2 \\ -1 \end{pmatrix}\), find \(ABC\), and check that \((AB)C = A(BC)\).

Orders: \((1 \times 3)(3 \times 2)(2 \times 1)\), so \(ABC\) is \(1 \times 1\).
\(AB = \begin{pmatrix} 1 + 4 + 0 & 0 + 2 - 3 \end{pmatrix} = \begin{pmatrix} 5 & -1 \end{pmatrix}\), so \((AB)C = \begin{pmatrix} 10 + 1 \end{pmatrix} = \begin{pmatrix} 11 \end{pmatrix}\), since \(5 \times 2 = 10\) and \((-1) \times (-1) = 1\).
\(BC = \begin{pmatrix} 2 \\ 4 - 1 \\ 0 - 3 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ -3 \end{pmatrix}\), so \(A(BC) = \begin{pmatrix} 2 + 6 + 3 \end{pmatrix} = \begin{pmatrix} 11 \end{pmatrix}\).
Both routes agree, as associativity promises; \(CBA\), in contrast, is not even defined, since \(C\) is \(2 \times 1\) and \(B\) is \(3 \times 2\).

Answer: \(ABC = \begin{pmatrix} 11 \end{pmatrix}\)

Your turn

With \(A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\) and \(B = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}\), work out \(AB\) and \(BA\), and so show that \(AB \neq BA\).

Examiner trap

Never multiply entry by entry: \(\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\) is \(\begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}\), not \(\begin{pmatrix} 0 & 2 \\ 3 & 0 \end{pmatrix}\). Check the orders before you start, because \(AB\) can exist while \(BA\) does not. And the \((i, j)\) entry always uses a row of the left matrix and a column of the right one.

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\(A\) is a \(2 \times 3\) matrix and \(B\) is a \(3 \times 4\) matrix. What is the order of \(AB\)?

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Find \(\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\).

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\(A\) is a \(3 \times 3\) matrix. Which matrix \(M\) satisfies \(AM = MA = A\) for every such \(A\)?

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\(A\) is \(2 \times 3\), \(B\) is \(3 \times 3\) and \(C\) is \(3 \times 1\). What is the order of \(ABC\)?

Section 03

Powers and unknown entries.

For a square matrix, \(A^2 = AA\), \(A^3 = A^2A = AA^2\), and so on; powers exist only for square matrices, because \(AA\) needs columns of \(A\) to equal rows of \(A\). A matrix equation such as \(AB = BA\) or \(A^2 = kA + I\) is really several ordinary equations, one for each entry: multiply out both sides and equate matching entries.

\(A^2\) is not the matrix of squared entries: for \(A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\), \(A^2 = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\).
Ordinary algebra still works if you keep the order: \(A(B + C) = AB + AC\) and \(A^2 A = A A^2\). But \((A + B)^2 = A^2 + AB + BA + B^2\), which is \(A^2 + 2AB + B^2\) only when \(AB = BA\).
A product can be zero with neither factor zero: \(\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix} = O\). So \(AB = O\) does not force \(A = O\) or \(B = O\).
Once you know \(A^2 = pA + qI\), every higher power reduces to the same form: multiply by \(A\) and replace \(A^2\) again.

Worked example A square written with A and I

Given \(A = \begin{pmatrix} 2 & 1 \\ 3 & 1 \end{pmatrix}\), show that \(A^2 = 3A + I\). Hence express \(A^3\) in the form \(pA + qI\), and write \(A^3\) out in full.

\[A^2 = \begin{pmatrix} 2 & 1 \\ 3 & 1 \end{pmatrix}\begin{pmatrix} 2 & 1 \\ 3 & 1 \end{pmatrix}\] \[= \begin{pmatrix} 4 + 3 & 2 + 1 \\ 6 + 3 & 3 + 1 \end{pmatrix} = \begin{pmatrix} 7 & 3 \\ 9 & 4 \end{pmatrix}\]
\(3A + I = \begin{pmatrix} 6 & 3 \\ 9 & 3 \end{pmatrix} + \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 7 & 3 \\ 9 & 4 \end{pmatrix}\), the same matrix, so \(A^2 = 3A + I\).
\(A^3 = A \cdot A^2 = A(3A + I) = 3A^2 + A = 3(3A + I) + A = 10A + 3I\).
\[A^3 = \begin{pmatrix} 20 & 10 \\ 30 & 10 \end{pmatrix} + \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 23 & 10 \\ 30 & 13 \end{pmatrix}\]

Answer: \(A^3 = 10A + 3I = \begin{pmatrix} 23 & 10 \\ 30 & 13 \end{pmatrix}\) (multiplying \(A\) by \(\begin{pmatrix} 7 & 3 \\ 9 & 4 \end{pmatrix}\) directly gives the same)

Worked example Matrices that commute

Given \(A = \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}\) and \(B = \begin{pmatrix} a & 1 \\ b & 2 \end{pmatrix}\), find \(a\) and \(b\) such that \(AB = BA\).

\(AB = \begin{pmatrix} a + 2b & 1 + 4 \\ 3b & 6 \end{pmatrix} = \begin{pmatrix} a + 2b & 5 \\ 3b & 6 \end{pmatrix}\).
\(BA = \begin{pmatrix} a & 2a + 3 \\ b & 2b + 6 \end{pmatrix}\).
Equate entries: \(a + 2b = a\) gives \(b = 0\); \(5 = 2a + 3\) gives \(a = 1\); \(3b = b\) and \(6 = 2b + 6\) both agree with \(b = 0\).
Check: with \(B = \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix}\), both products are \(\begin{pmatrix} 1 & 5 \\ 0 & 6 \end{pmatrix}\).

Answer: \(a = 1\), \(b = 0\)

Your turn

Let \(A = \begin{pmatrix} 2 & k \\ 0 & -2 \end{pmatrix}\), where \(k\) is a constant. Show that \(A^2 = 4I\) for every value of \(k\), and hence find \(A^4\).

Examiner trap

When you replace \(A^2\) by \(3A + I\), multiply every term: \(A(3A + I) = 3A^2 + A\), not \(3A^2 + I\), because \(AI = A\). Write the constant term as a matrix, \(3I\), never as the number \(3\): you cannot add a number to a matrix.

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\(A = \begin{pmatrix} 1 & k \\ 0 & 2 \end{pmatrix}\) and \(B = \begin{pmatrix} 3 & 1 \\ 0 & 1 \end{pmatrix}\), where \(k\) is a constant. Find the value of \(k\) for which \(AB = BA\).

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\(A = \begin{pmatrix} 2 & 1 \\ 3 & 1 \end{pmatrix}\). Given that \(A^2 = pA + qI\), find \(p\) and \(q\).

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\(A\) is \(3 \times 2\), \(B\) is \(2 \times 3\) and \(C\) is \(3 \times 3\). Which of these products is not defined?

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Find \(ABC\), where \(A = \begin{pmatrix} 1 & -2 \end{pmatrix}\), \(B = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix}\) and \(C = \begin{pmatrix} 2 \\ 1 \end{pmatrix}\).

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Six original Cambridge-style questions

Combinations of non-square matrices, both products of a pair, unknown entries from a square, powers through \(A^2 = pA + qI\), why \((A + B)^2 \neq A^2 + 2AB + B^2\), and a \(3 \times 3\) cube.