The inverse
- \(M^{-1}\) undoes \(M\): \(MM^{-1} = M^{-1}M = I\). It is the matrix version of a reciprocal.
- Only a square matrix can have an inverse.
AHL 5.4
Inverse matrices
AHL 5.4
Inverse matrices
By the end of this section
- Find the inverse of a 2 × 2 matrix by hand, and of larger matrices on the calculator
- Recognise a singular matrix
- Solve AB = C and BA = C
Key points
2 × 2 by hand
- \(\begin{pmatrix}a & b \\ c & d\end{pmatrix}^{-1} = \dfrac{1}{ad - bc}\begin{pmatrix}d & -b \\ -c & a\end{pmatrix}\)
- Swap the leading diagonal, change the signs of the other two, divide by the determinant.
- Check your answer on the calculator.
Singular matrices
- If the determinant is 0 there is no inverse: you cannot divide by 0. The matrix is called singular.
Solving matrix equations
- If \(AB = C\) then \(B = A^{-1}C\): multiply on the left of both sides.
- If \(BA = C\) then \(B = CA^{-1}\): multiply on the right of both sides.
- The side matters, because order matters.
Explainer
Deep dive into the core concept.On the calculator
Inverse
a^-1
- A singular matrix gives an error.
- Larger matrices: always the calculator.
Screenshot and video to come
Worked examples
Problems like the ones in class. Pick one, try it, then open the working.An inverse by hand
Watch ExplainerFind the inverse of \(\begin{pmatrix}3 & 5 \\ 1 & 2\end{pmatrix}\).
Working
Have a go first. Then open the working one line at a time.
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Determinant
\(3(2) - 5(1) = 1\)
Find this first.
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Swap and change signs
\(\begin{pmatrix}2 & -5 \\ -1 & 3\end{pmatrix}\)
Swap 3 and 2. Change the signs of 5 and 1.
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Divide by the determinant
\(\begin{pmatrix}2 & -5 \\ -1 & 3\end{pmatrix}\)
Dividing by 1 changes nothing.
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Check
\(\begin{pmatrix}3 & 5 \\ 1 & 2\end{pmatrix}\begin{pmatrix}2 & -5 \\ -1 & 3\end{pmatrix} = \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}\)
The product is \(I\).
A determinant that is not 1
Watch ExplainerFind the inverse of \(\begin{pmatrix}4 & 2 \\ 3 & 2\end{pmatrix}\).
Working
Have a go first. Then open the working one line at a time.
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Determinant
\(4(2) - 2(3) = 2\)
Not zero, so there is an inverse.
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Swap and change signs
\(\begin{pmatrix}2 & -2 \\ -3 & 4\end{pmatrix}\)
Swap 4 and 2 on the leading diagonal. Change the signs of 2 and 3.
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Divide by 2
\(\frac{1}{2}\begin{pmatrix}2 & -2 \\ -3 & 4\end{pmatrix} = \begin{pmatrix}1 & -1 \\ -1.5 & 2\end{pmatrix}\)
Either form is fine.
No inverse
Watch ExplainerShow that \(\begin{pmatrix}6 & 4 \\ 9 & 6\end{pmatrix}\) has no inverse.
Working
Have a go first. Then open the working one line at a time.
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Determinant
\(6(6) - 4(9) = 0\)
36 − 36.
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Conclusion
The matrix is singular
There is no inverse, because the formula divides by the determinant. The two columns lie along the same line.
Solve AB = C
Watch Explainer\(A = \begin{pmatrix}3 & 5 \\ 1 & 2\end{pmatrix}\) and \(C = \begin{pmatrix}11 & 1 \\ 4 & 0\end{pmatrix}\). Given that \(AB = C\), find \(B\).
Working
Have a go first. Then open the working one line at a time.
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Which side?
\(B = A^{-1}C\)
\(A\) is on the left of \(B\), so multiply both sides on the left by \(A^{-1}\).
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Multiply
\(\begin{pmatrix}2 & -5 \\ -1 & 3\end{pmatrix}\begin{pmatrix}11 & 1 \\ 4 & 0\end{pmatrix} = \begin{pmatrix}2 & 2 \\ 1 & -1\end{pmatrix}\)
\(A^{-1}\) from the first example.
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Check
\(AB = C\)
Multiply back on the calculator.
Solve BA = C
Watch Explainer\(A = \begin{pmatrix}3 & 5 \\ 1 & 2\end{pmatrix}\) and \(C = \begin{pmatrix}7 & 12 \\ 0 & 1\end{pmatrix}\). Given that \(BA = C\), find \(B\).
Working
Have a go first. Then open the working one line at a time.
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Which side?
\(B = CA^{-1}\)
\(A\) is on the right of \(B\), so multiply both sides on the right by \(A^{-1}\).
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Multiply
\(\begin{pmatrix}7 & 12 \\ 0 & 1\end{pmatrix}\begin{pmatrix}2 & -5 \\ -1 & 3\end{pmatrix} = \begin{pmatrix}2 & 1 \\ -1 & 3\end{pmatrix}\)
\(C\) first, then \(A^{-1}\).
Decode a message
Watch ExplainerA word is turned into numbers using A = 1, B = 2, up to Z = 26. The numbers are written two at a time as the columns of a matrix, and that matrix is multiplied on the left by the key \(K = \begin{pmatrix}2 & 3 \\ 3 & 5\end{pmatrix}\). The coded message is \(\begin{pmatrix}27 & 37 \\ 41 & 58\end{pmatrix}\). Find the word.
Working
Have a go first. Then open the working one line at a time.
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Undo the key
\(K^{-1} \times\) coded message
The key was multiplied on the left, so its inverse goes on the left.
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The inverse of the key
\(K^{-1} = \begin{pmatrix}5 & -3 \\ -3 & 2\end{pmatrix}\)
\(\det K = 2(5) - 3(3) = 1\). Swap 2 and 5. Change the signs of the 3s.
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Multiply
\(\begin{pmatrix}5 & -3 \\ -3 & 2\end{pmatrix}\begin{pmatrix}27 & 37 \\ 41 & 58\end{pmatrix} = \begin{pmatrix}12 & 11 \\ 1 & 5\end{pmatrix}\)
Top left: \(5(27) + (-3)(41) = 12\).
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Read down each column
12, 1, 11, 5
L is the 12th letter, A the 1st, K the 11th, E the 5th.
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Answer
LAKE
Check: \(K\) times the decoded matrix gives the coded message.
Encode a message
Watch ExplainerA word is turned into numbers using A = 1, B = 2, up to Z = 26. The numbers are written two at a time down the columns of a matrix, and the matrix is multiplied on the left by the key \(K = \begin{pmatrix}3 & 2 \\ 4 & 3\end{pmatrix}\). Encode the word MATH, then find the matrix the receiver multiplies by to decode it.
Working
Have a go first. Then open the working one line at a time.
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Letters to numbers
M = 13, A = 1, T = 20, H = 8
Write them in pairs down the columns: \(\begin{pmatrix}13 & 20 \\ 1 & 8\end{pmatrix}\).
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Multiply by the key
\(\begin{pmatrix}3 & 2 \\ 4 & 3\end{pmatrix}\begin{pmatrix}13 & 20 \\ 1 & 8\end{pmatrix} = \begin{pmatrix}41 & 76 \\ 55 & 104\end{pmatrix}\)
The key goes on the left. Top left: \(3(13) + 2(1) = 41\).
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The coded message
Send 41, 55, 76, 104
Read down each column.
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The decoding matrix
\(\det K = 3(3) - 2(4) = 1,\quad K^{-1} = \begin{pmatrix}3 & -2 \\ -4 & 3\end{pmatrix}\)
The receiver multiplies the coded matrix on the left by \(K^{-1}\).