Inverse matrices Problem set A Worked solutions

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

1. (a) \(\begin{pmatrix}2 & 1 \\ 5 & 3\end{pmatrix}\)

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

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

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

3. (c) \(\begin{pmatrix}6 & 9 \\ 2 & 3\end{pmatrix}\)

   • \(\det = 6(3) - 9(2) = 0\)
   • The determinant is zero, so there is no inverse.

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

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

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

   • \(\det = 0(5) - 2(-1) = 2\)
   • \(M^{-1} = \frac{1}{2}\begin{pmatrix}5 & -2 \\ 1 & 0\end{pmatrix} = \begin{pmatrix}2.5 & -1 \\ 0.5 & 0\end{pmatrix}\)

\(A = \begin{pmatrix}2 & 1 \\ 5 & 3\end{pmatrix}\) and \(C = \begin{pmatrix}4 & -1 \\ 9 & 0\end{pmatrix}\). Find \(B\) when:

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

   • \(A^{-1} = \begin{pmatrix}3 & -1 \\ -5 & 2\end{pmatrix}\)
   • \(AB = C \Rightarrow B = A^{-1}C \text{ (multiply on the left)}\)
   • \(B = \begin{pmatrix}3 & -1 \\ -5 & 2\end{pmatrix}\begin{pmatrix}4 & -1 \\ 9 & 0\end{pmatrix} = \begin{pmatrix}3 & -3 \\ -2 & 5\end{pmatrix}\)

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

   • \(BA = C \Rightarrow B = CA^{-1} \text{ (multiply on the right)}\)
   • \(B = \begin{pmatrix}4 & -1 \\ 9 & 0\end{pmatrix}\begin{pmatrix}3 & -1 \\ -5 & 2\end{pmatrix} = \begin{pmatrix}17 & -6 \\ 27 & -9\end{pmatrix}\)

Use your calculator to find the inverse of \(\begin{pmatrix}1 & 0 & 2 \\ 0 & 1 & 1 \\ 1 & 1 & 0\end{pmatrix}\).

• \(\text{From the calculator: } \det = -3,\; \text{so the inverse exists.}\)
• \(\text{From the calculator: } M^{-1} = \begin{pmatrix}\frac{1}{3} & -\frac{2}{3} & \frac{2}{3} \\ -\frac{1}{3} & \frac{2}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} & -\frac{1}{3}\end{pmatrix}\)
• \(\text{In fractions of } \tfrac{1}{3}: \; M^{-1} = \tfrac{1}{3}\begin{pmatrix}1 & -2 & 2 \\ -1 & 2 & 1 \\ 1 & 1 & -1\end{pmatrix}\)

Find the values of \(k\) for which \(\begin{pmatrix}k & 2 \\ 3 & k-1\end{pmatrix}\) is singular. For all other values of \(k\), write down its inverse.

• \(\det = k(k-1) - 2(3) = k^2 - k - 6\)
• \(\text{Singular when } k^2 - k - 6 = 0, \text{ that is } (k - 3)(k + 2) = 0.\)
• \(k = 3 \text{ or } k = -2\)
• \(\text{Otherwise } M^{-1} = \frac{1}{k^2 - k - 6}\begin{pmatrix}k-1 & -2 \\ -3 & k\end{pmatrix}\)

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 \(\begin{pmatrix}2 & 1 \\ 5 & 3\end{pmatrix}\). The coded matrix is \(\begin{pmatrix}28 & 26 \\ 76 & 73\end{pmatrix}\). Find the word.

• \(\text{Coded} = K \times \text{Word, so Word} = K^{-1} \times \text{Coded.}\)
• \(K^{-1} = \frac{1}{6 - 5}\begin{pmatrix}3 & -1 \\ -5 & 2\end{pmatrix} = \begin{pmatrix}3 & -1 \\ -5 & 2\end{pmatrix}\)
• \(\text{Word} = \begin{pmatrix}3 & -1 \\ -5 & 2\end{pmatrix}\begin{pmatrix}28 & 26 \\ 76 & 73\end{pmatrix} = \begin{pmatrix}8 & 5 \\ 12 & 16\end{pmatrix}\)
• \(8, 5, 12, 16 = \text{H, E, L, P}\)
• The word is HELP.

Two friends send each other messages 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 \(K = \begin{pmatrix}2 & 1 \\ 5 & 3\end{pmatrix}\). They then send the four numbers of the product. Alma encodes the word SAND. Her friend's reply is coded with the same key as \(\begin{pmatrix}31 & 31 \\ 85 & 80\end{pmatrix}\).

1. (a) Encode the word SAND.

   • Letters: 19, 1, 14, 4. Down the columns: \(\begin{pmatrix}19 & 14 \\ 1 & 4\end{pmatrix}\)
   • \(\begin{pmatrix}2 & 1 \\ 5 & 3\end{pmatrix}\begin{pmatrix}19 & 14 \\ 1 & 4\end{pmatrix} = \begin{pmatrix}39 & 32 \\ 98 & 82\end{pmatrix}\)

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

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

3. (c) Decode the second message.

   • \(\begin{pmatrix}3 & -1 \\ -5 & 2\end{pmatrix}\begin{pmatrix}31 & 31 \\ 85 & 80\end{pmatrix} = \begin{pmatrix}8 & 13 \\ 15 & 5\end{pmatrix}\)
   • Read down the columns: 8, 15, 13, 5, which spell HOME.

Show that \(\begin{pmatrix}1 & 2c \\ -c & 3\end{pmatrix}\) has an inverse for every value of \(c\).

• \(\det = 1(3) - (2c)(-c) = 3 + 2c^2\)
• \(c^2 \ge 0, \text{ so } 3 + 2c^2 \ge 3.\)
• The determinant is never zero, so the inverse exists for every value of c.

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

• \(P^{-1}QP = D \text{, so multiply on the left by } P \text{ and on the right by } P^{-1}: Q = PDP^{-1}.\)
• \(\det P = 3 - 2 = 1,\quad P^{-1} = \begin{pmatrix}3 & -2 \\ -1 & 1\end{pmatrix}\)
• \(PD = \begin{pmatrix}3 & -2 \\ 3 & -3\end{pmatrix}\)
• \(Q = \begin{pmatrix}3 & -2 \\ 3 & -3\end{pmatrix}\begin{pmatrix}3 & -2 \\ -1 & 1\end{pmatrix} = \begin{pmatrix}11 & -8 \\ 12 & -9\end{pmatrix}\)

By working out \((AB)(B^{-1}A^{-1})\), show that \((AB)^{-1} = B^{-1}A^{-1}\).

• \((AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1}\)
• \(= AIA^{-1} = AA^{-1} = I\)
• \(\text{So } B^{-1}A^{-1} \text{ is the inverse of } AB: (AB)^{-1} = B^{-1}A^{-1}.\)