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

Solve each pair of equations by a matrix method.

1. (a) \(3x + y = 9,\quad 5x + 2y = 16\)

   • \(\begin{pmatrix}3 & 1 \\ 5 & 2\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}9 \\ 16\end{pmatrix}\)
   • \(\det = (3)(2) - (1)(5) = 1\), so \(\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}2 & -1 \\ -5 & 3\end{pmatrix}\begin{pmatrix}9 \\ 16\end{pmatrix} = \begin{pmatrix}2 \\ 3\end{pmatrix}\)

2. (b) \(2x + 3y = 4,\quad 4x + 5y = 9\)

   • \(\begin{pmatrix}2 & 3 \\ 4 & 5\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}4 \\ 9\end{pmatrix}\)
   • \(\det = (2)(5) - (3)(4) = -2\), so \(\begin{pmatrix}x \\ y\end{pmatrix} = \frac{1}{-2}\begin{pmatrix}5 & -3 \\ -4 & 2\end{pmatrix}\begin{pmatrix}4 \\ 9\end{pmatrix} = \begin{pmatrix}3.5 \\ -1\end{pmatrix}\)

3. (c) \(x - 3y = 5,\quad 2x - 5y = 8\)

   • \(\begin{pmatrix}1 & -3 \\ 2 & -5\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}5 \\ 8\end{pmatrix}\)
   • \(\det = (1)(-5) - (-3)(2) = 1\), so \(\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}-5 & 3 \\ -2 & 1\end{pmatrix}\begin{pmatrix}5 \\ 8\end{pmatrix} = \begin{pmatrix}-1 \\ -2\end{pmatrix}\)

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

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

   • \(\begin{pmatrix}0 & 1 & 2 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix} = \begin{pmatrix}7 \\ 5 \\ 3\end{pmatrix}\)
   • With \(A\) the matrix of coefficients, from the calculator: \(\begin{pmatrix}x \\ y \\ z\end{pmatrix} = A^{-1}\begin{pmatrix}7 \\ 5 \\ 3\end{pmatrix} = \begin{pmatrix}2 \\ 1 \\ 3\end{pmatrix}\).

2. (b) \(2x + y - z = 5,\quad x - 3y + 2z = 4,\quad 3x + 2y + z = 13\)

   • \(\begin{pmatrix}2 & 1 & -1 \\ 1 & -3 & 2 \\ 3 & 2 & 1\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix} = \begin{pmatrix}5 \\ 4 \\ 13\end{pmatrix}\)
   • With \(A\) the matrix of coefficients, from the calculator: \(\begin{pmatrix}x \\ y \\ z\end{pmatrix} = A^{-1}\begin{pmatrix}5 \\ 4 \\ 13\end{pmatrix} = \begin{pmatrix}3 \\ 1 \\ 2\end{pmatrix}\).

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

1. (a) \(kx + 2y = 3,\quad 8x + ky = 1\)

   • No unique solution when the coefficient matrix is singular.
   • \(\det = (k)(k) - (2)(8) = k^2 - 16 = 0\), so \(k^2 = 16\) and \(k = \pm 4\).

2. (b) \((k+2)x + 3y = 1,\quad 7x + (k-2)y = 5\)

   • No unique solution when the coefficient matrix is singular.
   • \(\det = (k + 2)(k - 2) - (3)(7) = k^2 - 25 = 0\), so \(k^2 = 25\) and \(k = \pm 5\).

The equations are \(2x + 5y = 3\) and \(4x + 10y = k\).

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

   • \(\det\begin{pmatrix}2 & 5 \\ 4 & 10\end{pmatrix} = 2(10) - 5(4) = 0\)

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

   • \(4x + 10y = 2(2x + 5y)\), so the second equation is 2 times the first.
   • The equations are the same line when \(k = 2(3) = 6\).

3. (c) State how many solutions there are when \(k = 9\), and what this means for the two lines.

   • The determinant is 0, so there is no unique solution.
   • \(k = 9\) is not 6, so the right-hand sides do not match.
   • The lines are parallel and different: they never meet.

At a school canteen, 3 sandwiches and 2 juices cost £11.10, and 5 sandwiches and 4 juices cost £19.30. Form a matrix equation and find the price of each item.

• \(\begin{pmatrix}3 & 2 \\ 5 & 4\end{pmatrix}\begin{pmatrix}s \\ j\end{pmatrix} = \begin{pmatrix}11.1 \\ 19.3\end{pmatrix}\), with \(s\) and \(j\) the prices in pounds.
• \(\det = (3)(4) - (2)(5) = 2\), so \(\begin{pmatrix}s \\ j\end{pmatrix} = \frac{1}{2}\begin{pmatrix}4 & -2 \\ -5 & 3\end{pmatrix}\begin{pmatrix}11.1 \\ 19.3\end{pmatrix} = \begin{pmatrix}2.9 \\ 1.2\end{pmatrix}\).
• A sandwich costs £2.90 and a juice costs £1.20.

Tomasz invests €60 000 in three funds, X, Y and Z. He puts €8000 more in Z than in X. After a year the funds have grown by 3%, 5% and 2%, and his investment has grown by €2010 in total. Form a matrix equation and find the amount in each fund.

• Total: \(x + y + z = 60000\). Z is 8000 more than X: \(-x + z = 8000\). Growth: \(0.03x + 0.05y + 0.02z = 2010\).
• \(\begin{pmatrix}1 & 1 & 1 \\ -1 & 0 & 1 \\ 0.03 & 0.05 & 0.02\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix} = \begin{pmatrix}60000 \\ 8000 \\ 2010\end{pmatrix}\)
• With \(A\) the matrix of coefficients, from the calculator: \(\begin{pmatrix}x \\ y \\ z\end{pmatrix} = A^{-1}\begin{pmatrix}60000 \\ 8000 \\ 2010\end{pmatrix} = \begin{pmatrix}15000 \\ 22000 \\ 23000\end{pmatrix}\).
• X €15 000, Y €22 000, Z €23 000.

Solve \(kx + y = 3,\quad x + ky = 2\) in terms of \(k\), where \(k \ne \pm 1\).

• \(\begin{pmatrix}k & 1 \\ 1 & k\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}3 \\ 2\end{pmatrix}\), \(\det = k^2 - 1 \ne 0\).
• \(\begin{pmatrix}x \\ y\end{pmatrix} = \frac{1}{k^2 - 1}\begin{pmatrix}k & -1 \\ -1 & k\end{pmatrix}\begin{pmatrix}3 \\ 2\end{pmatrix} = \frac{1}{k^2 - 1}\begin{pmatrix}3k - 2 \\ 2k - 3\end{pmatrix}\)