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IB Mathematics: Applications and Interpretation HL · AHL 5: Matrices · AHL 5.4
Name Date
Inverse matrices Problem set B
Work in your exercise book. Find 2 × 2 inverses by hand, then check with the calculator.
Each question matches the question with the same number in set A. Do the ones you got wrong there.
Difficulty: first skills applying them stretch
Skills
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Find the inverse of each matrix, or explain why there is none.
- (a) \(\begin{pmatrix}3 & 2 \\ 4 & 3\end{pmatrix}\)
- (b) \(\begin{pmatrix}4 & 5 \\ 2 & 3\end{pmatrix}\)
- (c) \(\begin{pmatrix}8 & 2 \\ 12 & 3\end{pmatrix}\)
- (d) \(\begin{pmatrix}1 & 3 \\ -2 & -4\end{pmatrix}\)
- (e) \(\begin{pmatrix}0 & -2 \\ 1 & 3\end{pmatrix}\)
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\(A = \begin{pmatrix}3 & 1 \\ 5 & 2\end{pmatrix}\) and \(C = \begin{pmatrix}1 & 4 \\ -2 & 6\end{pmatrix}\). Find \(B\) when:
- (a) \(AB = C\)
- (b) \(BA = C\)
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Use your calculator to find the inverse of \(\begin{pmatrix}1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1\end{pmatrix}\).
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Find the values of \(k\) for which \(\begin{pmatrix}k & 4 \\ 2 & k+2\end{pmatrix}\) is singular. For all other values of \(k\), write down its inverse.
In context
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A four-letter word is written as numbers (A = 1, B = 2, …) in a 2 × 2 matrix, row by row. It is put into code by multiplying on the left by \(\begin{pmatrix}3 & 1 \\ 5 & 2\end{pmatrix}\). The coded matrix is \(\begin{pmatrix}59 & 11 \\ 105 & 21\end{pmatrix}\). Find the word.
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A treasure hunt gives its clues in code. A word is turned into numbers (A = 1, B = 2, up to Z = 26), written in pairs down the columns of a matrix and multiplied on the left by the key \(K = \begin{pmatrix}3 & 1 \\ 5 & 2\end{pmatrix}\). The numbers of the product are painted on a rock. The first clue is the word TIDE. The second clue is coded with the same key as \(\begin{pmatrix}58 & 39 \\ 97 & 69\end{pmatrix}\).
- (a) Encode the word TIDE.
- (b) Find \(K^{-1}\).
- (c) Decode the second message.
Stretch
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Show that \(\begin{pmatrix}2 & 3c \\ -c & 5\end{pmatrix}\) has an inverse for every value of \(c\).
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\(P = \begin{pmatrix}1 & 1 \\ 2 & 3\end{pmatrix}\). Find the matrix \(Q\) such that \(P^{-1}QP = \begin{pmatrix}4 & 0 \\ 0 & -2\end{pmatrix}\).
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By working out \((PQ)(Q^{-1}P^{-1})\), show that \((PQ)^{-1} = Q^{-1}P^{-1}\).
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