Determinants Problem set A Worked solutions

Find the determinant of each matrix.

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

   • \(\det = 5(4) - 2(3) = 14\)

2. (b) \(\begin{pmatrix}-2 & 7 \\ 1 & 3\end{pmatrix}\)

   • \(\det = (-2)(3) - 7(1) = -13\)

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

   • \(\det = 6(2) - (-4)(-3) = 0\)

4. (d) \(\begin{pmatrix}0 & 5 \\ -2 & 9\end{pmatrix}\)

   • \(\det = 0(9) - 5(-2) = 10\)

5. (e) \(\begin{pmatrix}0.5 & 3 \\ 2 & 8\end{pmatrix}\)

   • \(\det = 0.5(8) - 3(2) = -2\)

6. (f) \(\begin{pmatrix}-4 & -1 \\ 6 & -3\end{pmatrix}\)

   • \(\det = (-4)(-3) - (-1)(6) = 18\)

Find the area of each triangle.

1. (a) Vertices \((0,0),\ (6,1),\ (2,5)\)

   • \(\text{Edge vectors from } (0,0): \begin{pmatrix}6 \\ 1\end{pmatrix} \text{ and } \begin{pmatrix}2 \\ 5\end{pmatrix}.\)
   • \(\det = 6(5) - 2(1) = 28\)
   • \(\text{Area} = \tfrac{1}{2}|28| = 14\)

2. (b) Vertices \((2,1),\ (7,4),\ (3,8)\)

   • \(\text{Edge vectors from } (2,1): \begin{pmatrix}5 \\ 3\end{pmatrix} \text{ and } \begin{pmatrix}1 \\ 7\end{pmatrix}.\)
   • \(\det = 5(7) - 1(3) = 32\)
   • \(\text{Area} = \tfrac{1}{2}|32| = 16\)

3. (c) Vertices \((-1,2),\ (4,0),\ (1,6)\)

   • \(\text{Edge vectors from } (-1,2): \begin{pmatrix}5 \\ -2\end{pmatrix} \text{ and } \begin{pmatrix}2 \\ 4\end{pmatrix}.\)
   • \(\det = 5(4) - 2(-2) = 24\)
   • \(\text{Area} = \tfrac{1}{2}|24| = 12\)

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

1. (a) \(\begin{pmatrix}k & 3 \\ 4 & k+1\end{pmatrix}\)

   • \(\det = k(k+1) - 12 = k^2 + k - 12\)
   • \(k^2 + k - 12 = 0 \text{ factorises as } (k - 3)(k + 4) = 0.\)
   • \(k = 3 \text{ or } k = -4\)

2. (b) \(\begin{pmatrix}k-1 & 4 \\ 2 & k+1\end{pmatrix}\)

   • \(\det = (k-1)(k+1) - 8 = k^2 - 9\)
   • \(k^2 - 9 = 0\)
   • \(k = \pm 3\)

3. (c) \(\begin{pmatrix}3 & k \\ k & 12\end{pmatrix}\)

   • \(\det = 3(12) - k^2 = 36 - k^2\)
   • \(36 - k^2 = 0\)
   • \(k = \pm 6\)

Use your calculator to find the determinant of each matrix.

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

   • \(\text{From the calculator: } \det = -45\)

2. (b) \(\begin{pmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{pmatrix}\)

   • \(\text{From the calculator: } \det = 0\)
   • The determinant is zero, so the matrix is singular (the third row is twice the second minus the first).

\(A\) is the point \((3, 2)\). 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

   • \(\det\begin{pmatrix}3 & x \\ 2 & y\end{pmatrix} = 3y - 2x = 0\)
   • \(\text{This holds when } B \text{ is on the line } y = \tfrac{2}{3}x \text{ through } O \text{ and } A.\)
   • \(\text{For example } B = (6, 4): 3(4) - 2(6) = 0.\)

2. (b) negative

   • \(\text{Try } B = (1, 0): \det\begin{pmatrix}3 & 1 \\ 2 & 0\end{pmatrix} = 3(0) - 1(2) = -2.\)
   • \(\text{Any } B \text{ with } 3y - 2x < 0 \text{ works, that is, below the line } OA \text{ (clockwise from A).}\)

3. (c) 12

   • \(3y - 2x = 12\)
   • \(\text{Try } B = (0, 4): 3(4) - 2(0) = 12.\)
   • \(\text{Also } B = (3, 6): 3(6) - 2(3) = 12.\)

Use a determinant to show that the points \((1, 2)\), \((3, 6)\) and \((6, 12)\) lie on one straight line.

• \(\text{Vectors from } (1,2): \begin{pmatrix}2 \\ 4\end{pmatrix} \text{ and } \begin{pmatrix}5 \\ 10\end{pmatrix}.\)
• \(\det\begin{pmatrix}2 & 5 \\ 4 & 10\end{pmatrix} = 2(10) - 5(4) = 0\)
• The triangle has zero area, so the three points lie on one straight line.

A quadrilateral has vertices \((0,0),\ (5,1),\ (6,5)\) and \((1,4)\), in order. Show that it is a parallelogram and find its area.

• \(\overrightarrow{OB} = \begin{pmatrix}5 \\ 1\end{pmatrix},\quad \overrightarrow{OD} = \begin{pmatrix}1 \\ 4\end{pmatrix},\quad \overrightarrow{OC} = \begin{pmatrix}6 \\ 5\end{pmatrix}\)
• \(\overrightarrow{OB} + \overrightarrow{OD} = \overrightarrow{OC}, \text{ so } OBCD \text{ is a parallelogram.}\)
• \(\text{Area} = \left|\det\begin{pmatrix}5 & 1 \\ 1 & 4\end{pmatrix}\right| = 5(4) - 1(1) = 19\)

\(A = \begin{pmatrix}2 & 1 \\ 3 & 4\end{pmatrix}\) and \(B = \begin{pmatrix}1 & -2 \\ 5 & 0\end{pmatrix}\). Find \(\det A\), \(\det B\) and \(\det(AB)\). What do you notice?

• \(\det A = 2(4) - 1(3) = 5,\quad \det B = 1(0) - (-2)(5) = 10\)
• \(AB = \begin{pmatrix}7 & -4 \\ 23 & -6\end{pmatrix}\)
• \(\det(AB) = 7(-6) - (-4)(23) = 50\)
• \(\det(AB) = \det A \times \det B \;(5 \times 10 = 50)\)

For \(M = \begin{pmatrix}a & b \\ c & d\end{pmatrix}\), show that \(\det(kM) = k^2 \det M\). Explain this using area.

• \(kM = \begin{pmatrix}ka & kb \\ kc & kd\end{pmatrix}\)
• \(\det(kM) = (ka)(kd) - (kb)(kc) = k^2 ad - k^2 bc = k^2(ad - bc) = k^2\det M\)
• \(\text{The columns are both } k \text{ times as long, so a length is scaled by } k \text{ in both directions, and the area is scaled by } k \times k = k^2.\)