In Exercises findmatinversefirst - findmatinverselast, find the inverse of the matrix or state that the matrix is not invertible.
\(A = \left [ \begin{array}{rr} 1 & 2 \\ 3 & 4 \end{array} \right ]\)

\(A^{-1} = \left [ \begin{array}{rr} -2 & 1 \\ \frac {3}{2} & -\frac {1}{2} \end{array} \right ]\)
\(B = \left [ \begin{array}{rr} 12 & -7 \\ -5 & 3 \end{array} \right ]\)

\(B^{-1} = \left [ \begin{array}{rr} \answer {3} & \answer {7} \\ \answer {5} & \answer {12} \end{array} \right ]\)

\(C = \left [ \begin{array}{rr} 6 & 15 \\ 14 & 35 \end{array} \right ]\)

\(C\) is not invertible
\(D = \left [ \begin{array}{rr} 2 & -1 \\ 16 & -9 \end{array} \right ]\)

\(D^{-1} = \left [ \begin{array}{rr} \answer {\frac {9}{2}} & \answer {-\frac {1}{2}} \\ \answer {8} & \answer {-1} \end{array} \right ]\)

\(E = \left [ \begin{array}{rrr} 3 & 0 & 4 \\ 2 & -1 & 3 \\ -3 & 2 & -5 \end{array} \right ]\)

\(E^{-1} = \left [ \begin{array}{rrr} -1 & 8 & 4 \\ 1 & -3 & -1 \\ 1 & -6 & -3 \end{array} \right ]\)
\(F = \left [ \begin{array}{rrr} 4 & \hphantom {-}6 & -3 \\ 3 & 4 & -3 \\ 1 & 2 & 6 \end{array} \right ]\)

\(F^{-1} = \left [ \begin{array}{rrr} -\frac {5}{2} & \frac {7}{2} & \frac {1}{2} \\ \frac {7}{4} & -\frac {9}{4} & -\frac {1}{4} \\ -\frac {1}{6} & \frac {1}{6} & \frac {1}{6} \end{array} \right ]\)
\(G = \left [ \begin{array}{rrr} 1 & \hphantom {1}2 & 3 \\ 2 & 3 & 11 \\ 3 & 4 & 19 \end{array} \right ]\)

\(G\) is not invertible
\(H = \left [ \begin{array}{rrrr} 1 & 0 & -3 & \hphantom {-}0 \\ 2 & -2 & 8 & 7 \\ -5 & 0 & 16 & 0 \\ 1 & 0 & 4 & 1 \end{array} \right ]\)

\(H^{-1} = \left [ \begin{array}{rrrr} 16 & 0 & 3 & 0 \\ -90 & -\frac {1}{2} & -\frac {35}{2} & \frac {7}{2} \\ 5 & 0 & 1 & 0 \\ -36 & 0 & -7 & \hphantom {-}1 \end{array} \right ]\)
In Exercises 2by2inversefirst - 2by2inverselast, use one matrix inverse to solve the following systems of linear equations.

The coefficient matrix is \(B^{-1}\) from Exercise matrixB above so the inverse we need is \((B^{-1})^{-1} = B\).
\(\left \{ \begin{array}{rcr} 3x + 7y & = & 26 \\ 5x + 12y & = & 39 \end{array} \right .\)

\(\left [ \begin{array}{rr} 12 & -7 \\ -5 & 3 \end{array} \right ] \left [ \begin{array}{r} 26 \\ 39 \end{array} \right ] = \left [ \begin{array}{r} 39 \\ -13 \end{array} \right ] \;\) So \(x = 39\) and \(y = -13\).
\(\left \{ \begin{array}{rcr} 3x + 7y & = & 0 \\ 5x + 12y & = & -1 \end{array} \right .\)

\(x = \answer {7}\) and \(y = \answer {-3}\)
\(\left \{ \begin{array}{rcr} 3x + 7y & = & -7 \\ 5x + 12y & = & 5 \end{array} \right .\)

\(\left [ \begin{array}{rr} 12 & -7 \\ -5 & 3 \end{array} \right ] \left [ \begin{array}{r} -7 \\ 5 \end{array} \right ] = \left [ \begin{array}{r} -119 \\ 50 \end{array} \right ] \;\) So \(x = -119\) and \(y = 50\).
In Exercises 3by3inversefirst - 3by3inverselast, use the inverse of \(E\) from Exercise matrixE above to solve the following systems of linear equations.

The coefficient matrix is \(E = \left [ \begin{array}{rrr} 3 & 0 & 4 \\ 2 & -1 & 3 \\ -3 & 2 & -5 \end{array} \right ]\) from Exercise matrixE, so \(E^{-1} = \left [ \begin{array}{rrr} -1 & 8 & 4 \\ 1 & -3 & -1 \\ 1 & -6 & -3 \end{array} \right ]\)
\(\left \{ \begin{array}{rcr} 3x + 4z & = & 1 \\ 2x - y + 3z & = & 0 \\ \!-3x + 2y - 5z & = & 0 \end{array} \right .\)

\(\left [ \begin{array}{rrr} -1 & 8 & 4 \\ 1 & -3 & -1 \\ 1 & -6 & -3 \end{array} \right ] \left [ \begin{array}{r} 1 \\ 0 \\ 0 \end{array} \right ] = \left [ \begin{array}{r} -1 \\ 1 \\ 1 \end{array} \right ] \;\) So \(x = -1\), \(y = 1\) and \(z = 1\).
\(\left \{ \begin{array}{rcr} 3x + 4z & = & 0 \\ 2x - y + 3z & = & 1 \\ \!-3x + 2y - 5z & = & 0 \end{array} \right .\)

\(\left [ \begin{array}{rrr} -1 & 8 & 4 \\ 1 & -3 & -1 \\ 1 & -6 & -3 \end{array} \right ] \left [ \begin{array}{r} 0 \\ 1 \\ 0 \end{array} \right ] = \left [ \begin{array}{r} 8 \\ -3 \\ -6 \end{array} \right ] \;\) So \(x = 8\), \(y = -3\) and \(z = -6\).
\(\left \{ \begin{array}{rcr} 3x + 4z & = & 0 \\ 2x - y + 3z & = & 0 \\ \!-3x + 2y - 5z & = & 1 \end{array} \right .\)

\(\left [ \begin{array}{rrr} -1 & 8 & 4 \\ 1 & -3 & -1 \\ 1 & -6 & -3 \end{array} \right ] \left [ \begin{array}{r} 0 \\ 0 \\ 1 \end{array} \right ] = \left [ \begin{array}{r} 4 \\ -1 \\ -3 \end{array} \right ] \;\) So \(x = 4\), \(y = -1\) and \(z = -3\).
This exercise is a continuation of Example rotationmatrixex in Section MatArithmetic and gives another application of matrix inverses. Recall that given the position matrix \(P\) for a point in the plane, the matrix \(RP\) corresponds to a point rotated \(45^{\circ }\) counterclockwise from \(P\) where
\[R = \left [ \begin{array}{rr} \frac {\sqrt {2}}{2} & -\frac {\sqrt {2}}{2} \\[3pt] \frac {\sqrt {2}}{2} & \frac {\sqrt {2}}{2} \\ \end{array} \right ]\]
  1. Find \(R^{-1}\).
  2. If \(RP\) rotates a point counterclockwise \(45^{\circ }\), what should \(R^{-1}P\) do? Check your answer by finding \(R^{-1}P\) for various points on the coordinate axes and the lines \(y=\pm x\).
  3. Find \(R^{-1}P\) where \(P\) corresponds to a generic point \(P(x,y)\). Verify that this takes points on the curve \(y=\frac {2}{x}\) to points on the curve \(x^2-y^2=4\).
A Sasquatch’s diet consists of three primary foods: Ippizuti Fish, Misty Mushrooms, and Sun Berries. Each serving of Ippizuti Fish is 500 calories, contains 40 grams of protein, and has no Vitamin X. Each serving of Misty Mushrooms is 50 calories, contains 1 gram of protein, and 5 milligrams of Vitamin X. Finally, each serving of Sun Berries is 80 calories, contains no protein, but has 15 milligrams of Vitamin X.
  1. If an adult male Sasquatch requires 3200 calories, 130 grams of protein, and 275 milligrams of Vitamin X daily, use a matrix inverse to find how many servings each of Ippizuti Fish, Misty Mushrooms, and Sun Berries he needs to eat each day.

    The adult male Sasquatch needs: 3 servings of Ippizuti Fish, 10 servings of Misty Mushrooms, and 15 servings of Sun Berries daily.
  2. An adult female Sasquatch requires 3100 calories, 120 grams of protein, and 300 milligrams of Vitamin X daily. Use the matrix inverse you found in part (a) to find how many servings each of Ippizuti Fish, Misty Mushrooms, and Sun Berries she needs to eat each day.

    The adult female Sasquatch needs: 3 servings of Ippizuti Fish and 20 servings of Sun Berries daily. (No Misty Mushrooms are needed!)
  3. An adolescent Sasquatch requires 5000 calories, 400 grams of protein daily, but no Vitamin X daily. Use the matrix inverse you found in part (a) to find how many servings each of Ippizuti Fish, Misty Mushrooms, and Sun Berries she needs to eat each day.

    The adolescent Sasquatch requires 10 servings of Ippizuti Fish daily. (No Misty Mushrooms or Sun Berries are needed!)
Matrices can be used in cryptography. Suppose we wish to encode the message ‘BIGFOOT LIVES’. We start by assigning a number to each letter of the alphabet, say \(A=1\), \(B=2\) and so on. We reserve \(0\) to act as a space. Hence, our message ‘BIGFOOT LIVES’ corresponds to the string of numbers ‘2, 9, 7, 6, 15, 15, 20, 0, 12, 9, 22, 5, 19.’ To encode this message, we use an invertible matrix. Any invertible matrix will do, but for this exercise, we choose
\[ A = \left [ \begin{array}{rrr} 2 & -3 & 5 \\ 3 & 1 &-2 \\ -7 & 1 & -1 \end{array} \right ] \]

Since \(A\) is \(3 \times 3\) matrix, we encode our message string into a matrix \(M\) with \(3\) rows. To do this, we take the first three numbers, 2 9 7, and make them our first column, the next three numbers, 6 15 15, and make them our second column, and so on. We put \(0\)’s to round out the matrix.

\[ M = \left [ \begin{array}{rrrrr} 2 & 6 & 20 & 9 & 19 \\ 9 & 15 & 0 & 22 & 0 \\ 7 & 15 & 12 & 5 & 0 \end{array} \right ] \]

To encode the message, we find the product \(AM\)

\[AM = \left [ \begin{array}{rrr} 2 & -3 & 5 \\ 3 & 1 &-2 \\ -7 & 1 & -1 \end{array} \right ]\left [ \begin{array}{rrrrr} 2 & 6 & 20 & 9 & 19 \\ 9 & 15 & 0 & 22 & 0 \\ 7 & 15 & 12 & 5 & 0 \end{array} \right ] = \left [ \begin{array}{rrrrr} 12 & 42 & 100 & -23 & 38 \\ 1 & 3 & 36 & 39 & 57 \\ -12 & -42 & -152 & -46 & -133 \end{array} \right ]\]

So our coded message is ‘12, 1, \(-12\), 42, 3, \(-42\), 100, 36, \(-152\), \(-23\), 39, \(-46\), 38, 57, \(-133\).’ To decode this message, we start with this string of numbers, construct a message matrix as we did earlier (we should get the matrix \(AM\) again) and then multiply by \(A^{-1}\).

  1. Find \(A^{-1}\).

    \(A^{-1} = \left [ \begin{array}{rrr} 1 & 2 & 1 \\ 17 & 33 & 19 \\ 10 & 19 & 11 \end{array} \right ] \)
  2. Use \(A^{-1}\) to decode the message and check this method actually works.

    \( \left [ \begin{array}{rrr} 1 & 2 & 1 \\ 17 & 33 & 19 \\ 10 & 19 & 11 \end{array} \right ] \left [ \begin{array}{rrrrr} 12 & 42 & 100 & -23 & 38 \\ 1 & 3 & 36 & 39 & 57 \\ -12 & -42 & -152 & -46 & -133 \end{array} \right ] = \left [ \begin{array}{rrrrr} 2 & 6 & 20 & 9 & 19 \\ 9 & 15 & 0 & 22 & 0 \\ 7 & 15 & 12 & 5 & 0 \end{array} \right ] \quad \checkmark \)
  3. Decode the message ‘14, 37, \(-76\), 128, 21, \(-151\), 31, 65, \(-140\)

    ‘LOGS RULE’
  4. Choose another invertible matrix and encode and decode your own messages.
Using the matrices \(A\) from Exercise findmatinversefirst, \(B\) from Exercise matrixB and \(D\) from Exercise matrixD, show \(AB = D\) and \(D^{-1} = B^{-1}A^{-1}\). That is, show that \((AB)^{-1} = B^{-1}A^{-1}\).
Let \(M\) and \(N\) be invertible \(n \times n\) matrices. Show that \((MN)^{-1} = N^{-1}M^{-1}\) and compare your work to Exercise fcircginverse in Section InverseFunctions.