Determinants Problem set B Worked solutions

Find the determinant of each matrix.

1. (a) \(\begin{pmatrix}4 & 3 \\ 2 & 7\end{pmatrix}\)

   • \(\det = (4)(7) - (3)(2) = 22\)

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

   • \(\det = (-3)(4) - (5)(2) = -22\)

3. (c) \(\begin{pmatrix}8 & -6 \\ -4 & 3\end{pmatrix}\)

   • \(\det = (8)(3) - (-6)(-4) = 0\)

4. (d) \(\begin{pmatrix}0 & -4 \\ 3 & 6\end{pmatrix}\)

   • \(\det = (0)(6) - (-4)(3) = 12\)

5. (e) \(\begin{pmatrix}2.5 & 2 \\ 3 & 4\end{pmatrix}\)

   • \(\det = (2.5)(4) - (2)(3) = 4\)

6. (f) \(\begin{pmatrix}-5 & -2 \\ 4 & -3\end{pmatrix}\)

   • \(\det = (-5)(-3) - (-2)(4) = 23\)

Find the area of each triangle.

1. (a) Vertices \((0,0),\ (7,2),\ (3,6)\)

   • Take vectors from the first vertex: \(\begin{pmatrix}7 \\ 2\end{pmatrix}\) and \(\begin{pmatrix}3 \\ 6\end{pmatrix}\).
   • \(\text{area} = \tfrac{1}{2}\left|(7)(6) - (2)(3)\right| = \tfrac{1}{2}\left|36\right| = 18\)

2. (b) Vertices \((1,2),\ (6,3),\ (2,7)\)

   • Take vectors from the first vertex: \(\begin{pmatrix}5 \\ 1\end{pmatrix}\) and \(\begin{pmatrix}1 \\ 5\end{pmatrix}\).
   • \(\text{area} = \tfrac{1}{2}\left|(5)(5) - (1)(1)\right| = \tfrac{1}{2}\left|24\right| = 12\)

3. (c) Vertices \((-2,1),\ (3,-1),\ (1,5)\)

   • Take vectors from the first vertex: \(\begin{pmatrix}5 \\ -2\end{pmatrix}\) and \(\begin{pmatrix}3 \\ 4\end{pmatrix}\).
   • \(\text{area} = \tfrac{1}{2}\left|(5)(4) - (-2)(3)\right| = \tfrac{1}{2}\left|26\right| = 13\)

Find the values of \(k\) for which each matrix has determinant zero.

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

   • \(\det = (k)(k + 3) - (5)(2) = k^2 + 3k - 10 = 0\)
   • \((k + 5)(k - 2) = 0\), so \(k = 2\) or \(k = -5\).

2. (b) \(\begin{pmatrix}k + 3 & 2 \\ 8 & k - 3\end{pmatrix}\)

   • \(\det = (k + 3)(k - 3) - (2)(8) = k^2 - 25 = 0\)
   • \(k^2 = 25\), so \(k = 5\) or \(k = -5\).

3. (c) \(\begin{pmatrix}3 & k \\ k & 27\end{pmatrix}\)

   • \(\det = (3)(27) - (k)(k) = 81 - k^2 = 0\)
   • \(k^2 = 81\), so \(k = 9\) or \(k = -9\).

Use your calculator to find the determinant of each matrix.

1. (a) \(\begin{pmatrix}3 & 1 & 0 \\ 2 & -1 & 4 \\ 1 & 5 & -2\end{pmatrix}\)

   • From the calculator: \(\det = -46\).

2. (b) \(\begin{pmatrix}1 & 3 & 5 \\ 2 & 4 & 6 \\ 3 & 5 & 7\end{pmatrix}\)

   • From the calculator: \(\det = 0\), so the matrix is singular.

\(A\) is the point \((5, 2)\). Give the coordinates of a point \(B\) so that the determinant of the matrix with columns \(\overrightarrow{OA}\) and \(\overrightarrow{OB}\) is:

1. (a) zero

   • The matrix is \(\begin{pmatrix}5 & b_1 \\ 2 & b_2\end{pmatrix}\) with \(\det = 5b_2 - 2b_1\).
   • For zero, \(B\) must be a multiple of \(A\): \(B = (10, 4)\) gives \((5)(4) - (10)(2) = 0\).

2. (b) negative

   • Try \(B = (2, 0)\): \(\det = (5)(0) - (2)(2) = -4\), which is negative.

3. (c) 12

   • \(5b_2 - 2b_1 = 12\). Try \(b_1 = 4\): \(5b_2 = 20\) so \(b_2 = 4\), giving \(B = (4, 4)\).
   • Another is \(b_2 = 0\), \(b_1 = -6\), giving \((-6, 0)\).

Use a determinant to show that the points \((1, 4)\), \((4, 6)\) and \((7, 8)\) lie on one straight line.

• Vectors from \((1, 4)\): \(\begin{pmatrix}3 \\ 2\end{pmatrix}\) and \(\begin{pmatrix}6 \\ 4\end{pmatrix}\).
• \(\det\begin{pmatrix}3 & 6 \\ 2 & 4\end{pmatrix} = (3)(4) - (6)(2) = 0\)
• The triangle has no area, so the three points lie on one straight line.

A quadrilateral has vertices \((0,0),\ (6,2),\ (8,6)\) and \((2,4)\), in order. Show that it is a parallelogram and find its area.

• The sides from the origin are \(\begin{pmatrix}6 \\ 2\end{pmatrix}\) and \(\begin{pmatrix}2 \\ 4\end{pmatrix}\). Their sum is \(\begin{pmatrix}8 \\ 6\end{pmatrix}\), the fourth vertex, so the shape is a parallelogram.
• Area = \(|\det| = |(6)(4) - (2)(2)| = 20\).

\(A = \begin{pmatrix}3 & 1 \\ 2 & 5\end{pmatrix}\) and \(B = \begin{pmatrix}2 & -3 \\ 4 & 1\end{pmatrix}\). Find \(\det A\), \(\det B\) and \(\det(AB)\). What do you notice?

• \(\det A = (3)(5) - (1)(2) = 13\) and \(\det B = (2)(1) - (-3)(4) = 14\).
• \(AB = \begin{pmatrix}(3)(2) + (1)(4) & (3)(-3) + (1)(1) \\ (2)(2) + (5)(4) & (2)(-3) + (5)(1)\end{pmatrix} = \begin{pmatrix}10 & -8 \\ 24 & -1\end{pmatrix}\)
• \(\det(AB) = (10)(-1) - (-8)(24) = 182\), which is \(13 \times 14\), so \(\det(AB) = \det A \times \det B\).

For \(N = \begin{pmatrix}p & q \\ r & s\end{pmatrix}\), show that \(\det(3N) = 9 \det N\). Explain this using area.

• \(3N = \begin{pmatrix}3p & 3q \\ 3r & 3s\end{pmatrix}\).
• \(\det(3N) = (3p)(3s) - (3q)(3r) = 9ps - 9qr = 9(ps - qr) = 9\det N\).
• Both columns are 3 times as long, so lengths scale by 3 in both directions and areas by \(3^2 = 9\).