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Cambridge International AS & A Level · Topic 1.4
A-Level Further Maths 9231 / Topic 1.4 / AS & A Level

One number decides whether a matrix can be undone.

Every square matrix has a determinant, a single number built from its entries. If it is zero the matrix is singular and has no inverse; if not, the determinant is exactly what you divide by to find the inverse, for 2 by 2 and 3 by 3 matrices alike.

Topic 1.4All matrices lessons →
The key idea

For \(\mathbf{M} = \begin{pmatrix}a&b\\c&d\end{pmatrix}\), \(\det\mathbf{M} = ad - bc\) and \(\mathbf{M}^{-1} = \dfrac{1}{ad - bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}\). A 3 by 3 determinant is found by expanding along the first row with signs \(+\ -\ +\); the inverse is \(\dfrac{1}{\det\mathbf{M}}\) times the adjugate (the transposed matrix of cofactors). \(\mathbf{M}\) is singular when \(\det\mathbf{M} = 0\), and then it has no inverse. For non-singular matrices, \((\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}\) and \(\det(\mathbf{AB}) = \det\mathbf{A}\det\mathbf{B}\).

The minor of a a b c d e f g h i minor of a = ei − fh delete the row and column of a Cofactor signs + − + − + − + − + cofactor = sign × minor row 1: +a, −b, +c
Fig. 1: left, the minor of an entry is the 2 by 2 determinant left when you delete that entry's row and column. Right, the sign pattern that turns minors into cofactors; expanding along the first row uses + - +
Section 01

The 2 by 2 inverse.

The inverse of a square matrix \(\mathbf{M}\) is the matrix \(\mathbf{M}^{-1}\) with \(\mathbf{M}\mathbf{M}^{-1} = \mathbf{M}^{-1}\mathbf{M} = \mathbf{I}\). Not every matrix has one. The test is the determinant, written \(\det\mathbf{M}\): if \(\det\mathbf{M} = 0\) the matrix is singular (no inverse); if \(\det\mathbf{M} \ne 0\) it is non-singular and the inverse exists.

For a 2 by 2 matrix:

\[\mathbf{M} = \begin{pmatrix}a&b\\c&d\end{pmatrix}, \qquad \det\mathbf{M} = ad - bc\]

\[\mathbf{M}^{-1} = \frac{1}{ad - bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}\]

Worked example Swap, negate, divide
Find the inverse of \(\mathbf{A} = \begin{pmatrix}5&3\\4&2\end{pmatrix}\) and check your answer.
  1. \(\det\mathbf{A} = 5 \times 2 - 3 \times 4 = 10 - 12 = -2\), which is not zero, so \(\mathbf{A}\) is non-singular.
  2. Swap 5 and 2, negate 3 and 4: \(\begin{pmatrix}2&-3\\-4&5\end{pmatrix}\).
  3. Divide by \(-2\): \(\mathbf{A}^{-1} = -\dfrac{1}{2}\begin{pmatrix}2&-3\\-4&5\end{pmatrix} = \begin{pmatrix}-1&\frac{3}{2}\\2&-\frac{5}{2}\end{pmatrix}\).
  4. Check: \(\mathbf{A}\mathbf{A}^{-1} = \begin{pmatrix}-5 + 6 & \frac{15}{2} - \frac{15}{2}\\-4 + 4 & 6 - 5\end{pmatrix} = \begin{pmatrix}1&0\\0&1\end{pmatrix}\).
Answer: \(\mathbf{A}^{-1} = \begin{pmatrix}-1&\frac{3}{2}\\2&-\frac{5}{2}\end{pmatrix}\)
Worked example When is it singular?
Find the values of \(k\) for which \(\begin{pmatrix}k&6\\2&k - 1\end{pmatrix}\) is singular.
  1. Singular means the determinant is zero: \(k(k - 1) - 6 \times 2 = 0\).
  2. \(k^{2} - k - 12 = 0\), so \((k - 4)(k + 3) = 0\).
  3. Check \(k = 4\): \(\begin{pmatrix}4&6\\2&3\end{pmatrix}\) has determinant \(12 - 12 = 0\). Its rows are multiples of each other, the sign of a singular matrix.
Answer: \(k = 4\) or \(k = -3\)
Your turn
Find the inverse of \(\begin{pmatrix}3&1\\5&2\end{pmatrix}\).
Examiner trap

Swap the leading diagonal, not the other one: \(\begin{pmatrix}5&3\\4&2\end{pmatrix}\) becomes \(\begin{pmatrix}2&-3\\-4&5\end{pmatrix}\), not \(\begin{pmatrix}5&-4\\-3&2\end{pmatrix}\). And watch the signs in \(ad - bc\) when \(b\) or \(c\) is negative: \(\begin{pmatrix}3&-2\\4&1\end{pmatrix}\) has determinant \(3 - (-8) = 11\). Multiply your answer by the original matrix; if you do not get \(\mathbf{I}\), something is wrong.

Section 02

The 3 by 3 determinant.

For a 3 by 3 matrix, each entry has a minor: the 2 by 2 determinant left when you delete that entry's row and column (Fig. 1, left). Its cofactor is the minor with a sign from the chessboard pattern (Fig. 1, right): \(C_{ij} = (-1)^{i + j}M_{ij}\) for row \(i\), column \(j\). The determinant is the sum of each first-row entry times its cofactor.

Expanding along the first row:

\[\begin{aligned}\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix} &= a\begin{vmatrix}e&f\\h&i\end{vmatrix}\\ &\quad - b\begin{vmatrix}d&f\\g&i\end{vmatrix}\\ &\quad + c\begin{vmatrix}d&e\\g&h\end{vmatrix}\end{aligned}\]

so \(\det\mathbf{M} = a(ei - fh) - b(di - fg) + c(dh - eg)\). The middle term carries a minus sign because the entry \(b\) sits in row 1, column 2, and \((-1)^{1 + 2} = -1\). Vertical bars \(|\ |\) mean the determinant of the array inside them. Your calculator can find a determinant, but on Paper 1 you must show the expansion to earn the method marks.

Worked example Expand along the first row
Find \(\det\mathbf{A}\), where \(\mathbf{A} = \begin{pmatrix}2&1&3\\0&-1&4\\1&2&1\end{pmatrix}\).
  1. First term: \(2\begin{vmatrix}-1&4\\2&1\end{vmatrix} = 2\big((-1)(1) - 4 \times 2\big) = 2(-9) = -18\).
  2. Second term (minus sign): \(-1\begin{vmatrix}0&4\\1&1\end{vmatrix} = -1(0 - 4) = 4\).
  3. Third term: \(3\begin{vmatrix}0&-1\\1&2\end{vmatrix} = 3\big(0 - (-1)\big) = 3\).
  4. Add: \(-18 + 4 + 3\).
Answer: \(\det\mathbf{A} = -11\)
Worked example A value of k that makes it singular
Find the values of \(k\) for which \(\mathbf{M} = \begin{pmatrix}1&k&2\\-1&1&1\\1&2&k\end{pmatrix}\) is singular.
  1. First term: \(1\begin{vmatrix}1&1\\2&k\end{vmatrix} = k - 2\).
  2. Second term: \(-k\begin{vmatrix}-1&1\\1&k\end{vmatrix} = -k(-k - 1) = k^{2} + k\).
  3. Third term: \(2\begin{vmatrix}-1&1\\1&2\end{vmatrix} = 2(-2 - 1) = -6\).
  4. \(\det\mathbf{M} = k - 2 + k^{2} + k - 6 = k^{2} + 2k - 8 = (k - 2)(k + 4)\).
  5. Singular when \(\det\mathbf{M} = 0\).
Answer: \(k = 2\) or \(k = -4\); for every other value \(\mathbf{M}\) is non-singular
Your turn
Find the determinant of \(\begin{pmatrix}3&0&1\\2&1&-1\\0&4&2\end{pmatrix}\).
Examiner trap

The commonest slip is the sign of the middle term. It is \(-b(di - fg)\): the minus belongs to the position, whatever the sign of \(b\) itself. With \(b = -1\) the term becomes \(-(-1)(di - fg) = +(di - fg)\). Bracket every negative entry as you substitute.

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Check your understanding

Four quick checks. Each correct answer earns XP and records this skill on your profile.

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

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Which is the inverse of \(\begin{pmatrix}4&3\\2&2\end{pmatrix}\)?

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In a 3 by 3 matrix, the cofactor \(C_{23}\) of the entry in row 2, column 3 is

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For non-singular square matrices \(\mathbf{A}\) and \(\mathbf{B}\) of the same size, \((\mathbf{AB})^{-1}\) equals

Section 03

The 3 by 3 inverse, and products.

For a non-singular 3 by 3 matrix, build the inverse in four steps. Then a short proof gives the inverse of a product, which reverses the order, just as you take off your shoes before your socks when you undo getting dressed.

  1. Determinant. Find \(\det\mathbf{M}\); if it is zero, stop: there is no inverse.
  2. Minors. Replace every entry by its minor.
  3. Cofactors. Apply the sign pattern \(\begin{pmatrix}+&-&+\\-&+&-\\+&-&+\end{pmatrix}\).
  4. Adjugate. Transpose (rows become columns) to get \(\operatorname{adj}\mathbf{M}\), then \(\mathbf{M}^{-1} = \dfrac{1}{\det\mathbf{M}}\operatorname{adj}\mathbf{M}\).

The inverse of a product. If \(\mathbf{A}\) and \(\mathbf{B}\) are non-singular and the same size, then

\[\begin{aligned}(\mathbf{AB})(\mathbf{B}^{-1}\mathbf{A}^{-1}) &= \mathbf{A}(\mathbf{B}\mathbf{B}^{-1})\mathbf{A}^{-1}\\ &= \mathbf{A}\mathbf{I}\mathbf{A}^{-1} = \mathbf{A}\mathbf{A}^{-1} = \mathbf{I}\end{aligned}\]

using the associative law. So \(\mathbf{B}^{-1}\mathbf{A}^{-1}\) is the inverse of \(\mathbf{AB}\): \((\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}\). A useful check on any product is \(\det(\mathbf{AB}) = \det\mathbf{A}\det\mathbf{B}\), which also shows that \(\mathbf{AB}\) is non-singular whenever \(\mathbf{A}\) and \(\mathbf{B}\) are.

Worked example A full 3 by 3 inverse
Find \(\mathbf{A}^{-1}\), where \(\mathbf{A} = \begin{pmatrix}1&0&2\\2&1&0\\1&1&1\end{pmatrix}\).
  1. Determinant, along row 1: \(1(1 - 0) - 0 + 2(2 - 1) = 1 + 2 = 3\). Non-zero, so the inverse exists.
  2. Minors (for example, deleting row 2 and column 1 leaves \(\begin{vmatrix}0&2\\1&1\end{vmatrix} = -2\)): \(\begin{pmatrix}1&2&1\\-2&-1&1\\-2&-4&1\end{pmatrix}\).
  3. Cofactors, with the signs \(+\ -\ +\,/\,-\ +\ -\,/\,+\ -\ +\): \(\begin{pmatrix}1&-2&1\\2&-1&-1\\-2&4&1\end{pmatrix}\).
  4. Transpose for the adjugate: \(\operatorname{adj}\mathbf{A} = \begin{pmatrix}1&2&-2\\-2&-1&4\\1&-1&1\end{pmatrix}\).
  5. Divide by the determinant: \(\mathbf{A}^{-1} = \dfrac{1}{3}\begin{pmatrix}1&2&-2\\-2&-1&4\\1&-1&1\end{pmatrix}\).
  6. Check: row 1 of \(\mathbf{A}\), \((1,\ 0,\ 2)\), times the columns of \(\operatorname{adj}\mathbf{A}\) gives \(1 + 2 = 3\), \(2 - 2 = 0\), \(-2 + 2 = 0\). The other rows give \((0,\ 3,\ 0)\) and \((0,\ 0,\ 3)\), so \(\mathbf{A}\operatorname{adj}\mathbf{A} = 3\mathbf{I}\) and \(\mathbf{A}\mathbf{A}^{-1} = \mathbf{I}\).
Answer: \(\mathbf{A}^{-1} = \dfrac{1}{3}\begin{pmatrix}1&2&-2\\-2&-1&4\\1&-1&1\end{pmatrix}\)
Worked example Inverse of a product
Given \(\mathbf{A}^{-1} = \begin{pmatrix}2&1\\1&1\end{pmatrix}\) and \(\mathbf{B}^{-1} = \begin{pmatrix}1&-1\\0&2\end{pmatrix}\), find \((\mathbf{AB})^{-1}\).
  1. Reverse the order: \((\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}\). There is no need to find \(\mathbf{A}\) or \(\mathbf{B}\).
  2. \(\begin{pmatrix}1&-1\\0&2\end{pmatrix}\begin{pmatrix}2&1\\1&1\end{pmatrix} = \begin{pmatrix}2 - 1 & 1 - 1\\0 + 2 & 0 + 2\end{pmatrix}\).
  3. Check with determinants: \(\det\mathbf{B}^{-1}\det\mathbf{A}^{-1} = 2 \times 1 = 2\), and \(\begin{vmatrix}1&0\\2&2\end{vmatrix} = 2\).
Answer: \((\mathbf{AB})^{-1} = \begin{pmatrix}1&0\\2&2\end{pmatrix}\) (the wrong order, \(\mathbf{A}^{-1}\mathbf{B}^{-1}\), gives \(\begin{pmatrix}2&0\\1&1\end{pmatrix}\))
Your turn
\(\mathbf{A}\) and \(\mathbf{B}\) are 3 by 3 matrices with \(\det\mathbf{A} = 3\) and \(\det\mathbf{B} = -2\). Find \(\det(\mathbf{AB})\) and \(\det\big((\mathbf{AB})^{-1}\big)\).
Examiner trap

Two different transposes go wrong here. Forgetting to transpose the cofactor matrix gives a matrix that looks plausible but fails the check \(\mathbf{A}\mathbf{A}^{-1} = \mathbf{I}\). And \((\mathbf{AB})^{-1}\) is \(\mathbf{B}^{-1}\mathbf{A}^{-1}\), never \(\mathbf{A}^{-1}\mathbf{B}^{-1}\): matrix multiplication is not commutative, so the order matters.

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

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Find the value of \(k\) for which \(\begin{pmatrix}2&1&k\\1&k&1\\0&1&1\end{pmatrix}\) is singular.

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\(\mathbf{P} = \begin{pmatrix}1&2&1\\0&1&3\\2&1&0\end{pmatrix}\), with \(\det\mathbf{P} = 7\). Find the entry in row 1, column 2 of \(\mathbf{P}^{-1}\).

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

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Six original Cambridge-style questions
2 by 2 inverses, values of k that make a matrix singular, a matrix that is never singular, a full 3 by 3 inverse with a check, the inverse of a product, and an inverse found from a matrix equation.