Inverse matrices Problem set B Worked solutions

Find the inverse of each matrix, or explain why there is none.

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

   • \(\det = (3)(3) - (2)(4) = 1\); dividing by 1 changes nothing, so \(M^{-1} = \begin{pmatrix}3 & -2 \\ -4 & 3\end{pmatrix}\).

2. (b) \(\begin{pmatrix}4 & 5 \\ 2 & 3\end{pmatrix}\)

   • \(\det = (4)(3) - (5)(2) = 2\)
   • \(M^{-1} = \frac{1}{2}\begin{pmatrix}3 & -5 \\ -2 & 4\end{pmatrix} = \begin{pmatrix}1.5 & -2.5 \\ -1 & 2\end{pmatrix}\)

3. (c) \(\begin{pmatrix}8 & 2 \\ 12 & 3\end{pmatrix}\)

   • \(\det = (8)(3) - (2)(12) = 0\)
   • The determinant is 0, so the matrix is singular and has no inverse.

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

   • \(\det = (1)(-4) - (3)(-2) = 2\)
   • \(M^{-1} = \frac{1}{2}\begin{pmatrix}-4 & -3 \\ 2 & 1\end{pmatrix} = \begin{pmatrix}-2 & -1.5 \\ 1 & 0.5\end{pmatrix}\)

5. (e) \(\begin{pmatrix}0 & -2 \\ 1 & 3\end{pmatrix}\)

   • \(\det = (0)(3) - (-2)(1) = 2\)
   • \(M^{-1} = \frac{1}{2}\begin{pmatrix}3 & 2 \\ -1 & 0\end{pmatrix} = \begin{pmatrix}1.5 & 1 \\ -0.5 & 0\end{pmatrix}\)

\(A = \begin{pmatrix}3 & 1 \\ 5 & 2\end{pmatrix}\) and \(C = \begin{pmatrix}1 & 4 \\ -2 & 6\end{pmatrix}\). Find \(B\) when:

1. (a) \(AB = C\)

   • Multiply on the left by \(A^{-1}\): \(B = A^{-1}C\).
   • \(A^{-1} = \begin{pmatrix}2 & -1 \\ -5 & 3\end{pmatrix}\) (\(\det A = 1\)).
   • \(B = \begin{pmatrix}2 & -1 \\ -5 & 3\end{pmatrix}\begin{pmatrix}1 & 4 \\ -2 & 6\end{pmatrix} = \begin{pmatrix}4 & 2 \\ -11 & -2\end{pmatrix}\)

2. (b) \(BA = C\)

   • Multiply on the right by \(A^{-1}\): \(B = CA^{-1}\).
   • \(B = \begin{pmatrix}1 & 4 \\ -2 & 6\end{pmatrix}\begin{pmatrix}2 & -1 \\ -5 & 3\end{pmatrix} = \begin{pmatrix}-18 & 11 \\ -34 & 20\end{pmatrix}\)

Use your calculator to find the inverse of \(\begin{pmatrix}1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1\end{pmatrix}\).

• Enter the matrix on the calculator and find its inverse.
• From the calculator: \(\det = -2\) and \(M^{-1} = \begin{pmatrix}0.5 & 0.5 & -0.5 \\ 0.5 & -0.5 & 0.5 \\ -0.5 & 0.5 & 0.5\end{pmatrix}\).

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.

• Singular when \(\det = k(k+2) - (4)(2) = k^2 + 2k - 8 = 0\).
• \((k - 2)(k + 4) = 0\), so \(k = 2\) or \(k = -4\).
• Otherwise \(M^{-1} = \frac{1}{k^2 + 2k - 8}\begin{pmatrix}k+2 & -4 \\ -2 & k\end{pmatrix}\).

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.

• The code is \(KW = C\), so \(W = K^{-1}C\).
• \(\det = (3)(2) - (1)(5) = 1\), so \(K^{-1} = \begin{pmatrix}2 & -1 \\ -5 & 3\end{pmatrix}\).
• \(W = \begin{pmatrix}2 & -1 \\ -5 & 3\end{pmatrix}\begin{pmatrix}59 & 11 \\ 105 & 21\end{pmatrix} = \begin{pmatrix}13 & 1 \\ 20 & 8\end{pmatrix}\)
• Reading row by row: 13, 1, 20, 8 is M, A, T, H.

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 \(\begin{pmatrix}58 & 39 \\ 97 & 69\end{pmatrix}\).

1. (a) Encode the word TIDE.

   • Letters: 20, 9, 4, 5. Down the columns: \(\begin{pmatrix}20 & 4 \\ 9 & 5\end{pmatrix}\)
   • \(\begin{pmatrix}3 & 1 \\ 5 & 2\end{pmatrix}\begin{pmatrix}20 & 4 \\ 9 & 5\end{pmatrix} = \begin{pmatrix}69 & 17 \\ 118 & 30\end{pmatrix}\)

2. (b) Find \(K^{-1}\).

   • \(\det K = 3(2) - 1(5) = 1\)
   • \(K^{-1} = \begin{pmatrix}2 & -1 \\ -5 & 3\end{pmatrix}\)

3. (c) Decode the second message.

   • \(\begin{pmatrix}2 & -1 \\ -5 & 3\end{pmatrix}\begin{pmatrix}58 & 39 \\ 97 & 69\end{pmatrix} = \begin{pmatrix}19 & 9 \\ 1 & 12\end{pmatrix}\)
   • Read down the columns: 19, 1, 9, 12, which spell SAIL.

Show that \(\begin{pmatrix}2 & 3c \\ -c & 5\end{pmatrix}\) has an inverse for every value of \(c\).

• \(\det = (2)(5) - (3c)(-c) = 10 + 3c^2\).
• \(c^2 \ge 0\) for every real \(c\), so \(\det \ge 10\).
• The determinant is never zero, so the inverse always exists.

\(P = \begin{pmatrix}1 & 1 \\ 2 & 3\end{pmatrix}\). Find the matrix \(Q\) such that \(P^{-1}QP = \begin{pmatrix}4 & 0 \\ 0 & -2\end{pmatrix}\).

• Multiply on the left by \(P\) and on the right by \(P^{-1}\): \(Q = PDP^{-1}\) where \(D = \begin{pmatrix}4 & 0 \\ 0 & -2\end{pmatrix}\).
• \(P^{-1} = \begin{pmatrix}3 & -1 \\ -2 & 1\end{pmatrix}\) (\(\det P = 1\)).
• \(Q = \begin{pmatrix}4 & -2 \\ 8 & -6\end{pmatrix}\begin{pmatrix}3 & -1 \\ -2 & 1\end{pmatrix} = \begin{pmatrix}16 & -6 \\ 36 & -14\end{pmatrix}\)

By working out \((PQ)(Q^{-1}P^{-1})\), show that \((PQ)^{-1} = Q^{-1}P^{-1}\).

• Matrix multiplication is associative, so the brackets can be moved: \((PQ)(Q^{-1}P^{-1}) = P(QQ^{-1})P^{-1}\).
• \(QQ^{-1} = I\), so this is \(PIP^{-1} = PP^{-1} = I\).
• A matrix that multiplies \(PQ\) to give \(I\) is its inverse, so \((PQ)^{-1} = Q^{-1}P^{-1}\).