Systems of equations with the inverse matrix Problem set A Worked solutions

Solve each pair of equations by a matrix method.

1. (a) \(2x + y = 7,\quad 5x + 3y = 18\)

   • \(\begin{pmatrix}2 & 1 \\ 5 & 3\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}7 \\ 18\end{pmatrix}\)
   • \(\det = 1,\quad A^{-1} = \begin{pmatrix}3 & -1 \\ -5 & 2\end{pmatrix}\)
   • \(\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}3 & -1 \\ -5 & 2\end{pmatrix}\begin{pmatrix}7 \\ 18\end{pmatrix} = \begin{pmatrix}3 \\ 1\end{pmatrix}\)
   • \(x = 3,\; y = 1\)

2. (b) \(4x + 3y = 2,\quad 2x + 2y = 3\)

   • \(\begin{pmatrix}4 & 3 \\ 2 & 2\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}2 \\ 3\end{pmatrix}\)
   • \(\det = 2,\quad A^{-1} = \begin{pmatrix}1 & -1.5 \\ -1 & 2\end{pmatrix}\)
   • \(\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}1 & -1.5 \\ -1 & 2\end{pmatrix}\begin{pmatrix}2 \\ 3\end{pmatrix} = \begin{pmatrix}-2.5 \\ 4\end{pmatrix}\)
   • \(x = -2.5,\; y = 4\)

3. (c) \(x - 2y = 4,\quad 3x - 4y = 10\)

   • \(\begin{pmatrix}1 & -2 \\ 3 & -4\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}4 \\ 10\end{pmatrix}\)
   • \(\det = 2,\quad A^{-1} = \begin{pmatrix}-2 & 1 \\ -1.5 & 0.5\end{pmatrix}\)
   • \(\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}-2 & 1 \\ -1.5 & 0.5\end{pmatrix}\begin{pmatrix}4 \\ 10\end{pmatrix} = \begin{pmatrix}2 \\ -1\end{pmatrix}\)
   • \(x = 2,\; y = -1\)

Solve each system with your calculator, using the inverse matrix.

1. (a) \(x + 2z = 5,\quad y + z = 3,\quad x + y = 2\)

   • \(\begin{pmatrix}1 & 0 & 2 \\ 0 & 1 & 1 \\ 1 & 1 & 0\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix} = \begin{pmatrix}5 \\ 3 \\ 2\end{pmatrix}\)
   • \(\text{From the calculator: } \begin{pmatrix}x \\ y \\ z\end{pmatrix} = A^{-1}\begin{pmatrix}5 \\ 3 \\ 2\end{pmatrix} = \begin{pmatrix}1 \\ 1 \\ 2\end{pmatrix}\)
   • \(x = 1,\; y = 1,\; z = 2\)

2. (b) \(2x - y + 3z = 9,\quad x + 4y - z = 6,\quad 3x + y + 2z = 11\)

   • \(\begin{pmatrix}2 & -1 & 3 \\ 1 & 4 & -1 \\ 3 & 1 & 2\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix} = \begin{pmatrix}9 \\ 6 \\ 11\end{pmatrix}\)
   • \(\text{From the calculator: } \begin{pmatrix}x \\ y \\ z\end{pmatrix} = A^{-1}\begin{pmatrix}9 \\ 6 \\ 11\end{pmatrix} = \begin{pmatrix}1 \\ 2 \\ 3\end{pmatrix}\)
   • \(x = 1,\; y = 2,\; z = 3\)

Find the values of \(k\) for which each system does not have a unique solution.

1. (a) \(kx + 3y = 1,\quad 12x + ky = 5\)

   • \(\text{No unique solution when } \det = 0.\)
   • \(\det = k(k) - 3(12) = k^2 - 36 = 0\)
   • \(k = \pm 6\)

2. (b) \((k-1)x + 2y = 3,\quad 4x + (k+1)y = 1\)

   • \(\det = (k-1)(k+1) - 2(4) = k^2 - 9 = 0\)
   • \(k = \pm 3\)

The equations are \(3x - 2y = 4\) and \(6x - 4y = k\).

1. (a) Find the determinant of the matrix of coefficients.

   • \(\det\begin{pmatrix}3 & -2 \\ 6 & -4\end{pmatrix} = 3(-4) - -2(6) = 0\)

2. (b) Find the value of \(k\) for which there are infinitely many solutions.

   • \(6x - 4y = 2(3x - 2y)\), so the second equation is 2 times the first.
   • The equations are the same line when \(k = 2(4) = 8\).

3. (c) State how many solutions there are when \(k = 7\), and what this means for the two lines.

   • The determinant is 0, so there is no unique solution.
   • \(k = 7\) is not 8, so the right-hand sides do not match.
   • The lines are parallel and different: they never meet.

At a café, 3 coffees and 2 teas cost €15.80, and 4 coffees and 5 teas cost €26.20. Form a matrix equation and find the price of each drink.

• \(\text{Let } c = \text{coffee, } t = \text{tea: } \begin{pmatrix}3 & 2 \\ 4 & 5\end{pmatrix}\begin{pmatrix}c \\ t\end{pmatrix} = \begin{pmatrix}15.8 \\ 26.2\end{pmatrix}\)
• \(\det = 3(5) - 2(4) = 7,\quad A^{-1} = \tfrac{1}{7}\begin{pmatrix}5 & -2 \\ -4 & 3\end{pmatrix}\)
• \(\begin{pmatrix}c \\ t\end{pmatrix} = \tfrac{1}{7}\begin{pmatrix}5 & -2 \\ -4 & 3\end{pmatrix}\begin{pmatrix}15.8 \\ 26.2\end{pmatrix} = \begin{pmatrix}3.8 \\ 2.2\end{pmatrix}\)
• A coffee costs €3.80 and a tea costs €2.20.

Amira invests $40 000 in three funds, A, B and C. She puts $5000 more in A than in C. After a year the funds have grown by 4%, 2.5% and 6%, and her investment has grown by $1575 in total. Form a matrix equation and find the amount in each fund.

• \(\text{Let } a, b, c \text{ be the amounts in A, B, C.}\)
• \(a + b + c = 40000;\quad a - c = 5000;\quad 0.04a + 0.025b + 0.06c = 1575\)
• \(\begin{pmatrix}1 & 1 & 1 \\ 1 & 0 & -1 \\ 0.04 & 0.025 & 0.06\end{pmatrix}\begin{pmatrix}a \\ b \\ c\end{pmatrix} = \begin{pmatrix}40000 \\ 5000 \\ 1575\end{pmatrix}\)
• \(\text{From the calculator: } \begin{pmatrix}a \\ b \\ c\end{pmatrix} = A^{-1}\begin{pmatrix}40000 \\ 5000 \\ 1575\end{pmatrix} = \begin{pmatrix}15000 \\ 15000 \\ 10000\end{pmatrix}\)
• A: $15 000, B: $15 000, C: $10 000.

Solve \(kx + y = 2,\quad x + ky = 1\) in terms of \(k\), where \(k \ne \pm 1\).

• \(\begin{pmatrix}k & 1 \\ 1 & k\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}2 \\ 1\end{pmatrix},\quad \det = k^2 - 1\)
• \(A^{-1} = \frac{1}{k^2 - 1}\begin{pmatrix}k & -1 \\ -1 & k\end{pmatrix}\)
• \(\begin{pmatrix}x \\ y\end{pmatrix} = \frac{1}{k^2 - 1}\begin{pmatrix}k & -1 \\ -1 & k\end{pmatrix}\begin{pmatrix}2 \\ 1\end{pmatrix} = \frac{1}{k^2 - 1}\begin{pmatrix}2k - 1 \\ k - 2\end{pmatrix}\)
• \(x = \frac{2k - 1}{k^2 - 1},\quad y = \frac{k - 2}{k^2 - 1}\)