Determinants Problem set A Answers

1. (a) 14
2. (b) -13
3. (c) 0
4. (d) 10
5. (e) -2
6. (f) 18

1. (a) 14
2. (b) 16
3. (c) 12

1. (a) \(k = 3\) or \(k = -4\)
2. (b) \(k = \pm 3\)
3. (c) \(k = \pm 6\)

1. (a) −45
2. (b) 0

1. (a) Any point on the line through O and A, for example \((6, 4)\).
2. (b) Any point clockwise from A, for example \((1, 0)\), which gives −2.
3. (c) Any point with \(3y - 2x = 12\), for example \((0, 4)\) or \((3, 6)\).

From \((1,2)\) the two vectors are \(\begin{pmatrix}2 \\ 4\end{pmatrix}\) and \(\begin{pmatrix}5 \\ 10\end{pmatrix}\). \(2(10) - 5(4) = 0\): no area, so the points are in line.

\(\begin{pmatrix}5 \\ 1\end{pmatrix} + \begin{pmatrix}1 \\ 4\end{pmatrix} = \begin{pmatrix}6 \\ 5\end{pmatrix}\), so it is a parallelogram. Area \(= 5(4) - 1(1) = 19\).

\(\det A = 5,\; \det B = 10,\; AB = \begin{pmatrix}7 & -4 \\ 23 & -6\end{pmatrix},\; \det(AB) = 50\). \(\det(AB) = \det A \times \det B\).

\(\det(kM) = (ka)(kd) - (kb)(kc) = k^2(ad - bc)\). Both vectors are \(k\) times as long, so the area is \(k^2\) times as big.