(a) Find the magnitude of the vector \(\binom{-1}{7}\).
(b) The determinant of the matrix \(\begin{pmatrix}6 &2m \\ 5 & m\end{pmatrix}\) is 24. Find the value of \(m\).
(c) Given: \[ L = \begin{pmatrix}2 &5 \\ 3 & 9\end{pmatrix}, \quad M = \binom{-4}{2}, \quad N = \begin{pmatrix}1 & 7\end{pmatrix} \] Work out the following:
(i) \(NM\)
(ii) \(LM\)
(iii) \(L^{2}\)
(iv) \(L^{-1}\)
▶️ Answer/Explanation
(a) Ans: \(5\sqrt{2}\) or 7.071 (to 3 d.p.)
Magnitude = \(\sqrt{(-1)^2 + 7^2} = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2}\).
(b) Ans: \(m = -6\)
Determinant = \(6m – 2m \times 5 = 24\) → \(6m – 10m = 24\) → \(-4m = 24\) → \(m = -6\).
(c)(i) Ans: \(10\)
\(NM = \begin{pmatrix}1 & 7\end{pmatrix}\binom{-4}{2} = (1 \times -4) + (7 \times 2) = -4 + 14 = 10\).
(c)(ii) Ans: \(\binom{2}{6}\)
\(LM = \begin{pmatrix}2 &5 \\ 3 &9\end{pmatrix}\binom{-4}{2} = \binom{(2 \times -4) + (5 \times 2)}{(3 \times -4) + (9 \times 2)} = \binom{-8 + 10}{-12 + 18} = \binom{2}{6}\).
(c)(iii) Ans: \(\begin{pmatrix}19 & 55\\ 33 & 96\end{pmatrix}\)
\(L^2 = L \times L = \begin{pmatrix}2 &5 \\ 3 &9\end{pmatrix}\begin{pmatrix}2 &5 \\ 3 &9\end{pmatrix} = \begin{pmatrix}4+15 &10+45 \\ 6+27 &15+81\end{pmatrix} = \begin{pmatrix}19 &55 \\ 33 &96\end{pmatrix}\).
(c)(iv) Ans: \(\frac{1}{3}\begin{pmatrix}9 & -5 \\ -3 & 2\end{pmatrix}\)
First, find \(\det(L) = (2 \times 9) – (5 \times 3) = 18 – 15 = 3\).
Then, \(L^{-1} = \frac{1}{3}\begin{pmatrix}9 & -5 \\ -3 & 2\end{pmatrix}\).
(a) Given vectors:
\(\vec{OA}=\begin{pmatrix}4\\ 3\end{pmatrix}\), \(\vec{AB}=\begin{pmatrix}8\\ -7\end{pmatrix}\), \(\vec{AC}=\begin{pmatrix}-3\\ 6\end{pmatrix}\)
Find:
(i) \(\left| \vec{OB} \right|\),
(ii) \(\left| \vec{BC} \right|\).
(b) 
PQRS is a parallelogram with diagonals PR and SQ intersecting at X.
\(\vec{PQ}= a\) and \(\vec{PS}=b\).
Find \(\vec{QX}\) in terms of \(a\) and \(b\).
(c) Given matrix \(M=\begin{pmatrix}2 & 5 \\ 1 & 8\end{pmatrix}\), calculate:
(i) \(M^{2}\),
(ii) \(M^{-1}\).
▶️ Answer/Explanation
(a)(i) Ans: 12.6 or 12.64 to 12.65
First find \(\vec{OB} = \vec{OA} + \vec{AB} = \begin{pmatrix}4\\ 3\end{pmatrix} + \begin{pmatrix}8\\ -7\end{pmatrix} = \begin{pmatrix}12\\ -4\end{pmatrix}\).
Then calculate magnitude: \(\sqrt{12^2 + (-4)^2} = \sqrt{144 + 16} = \sqrt{160} ≈ 12.65\).
(a)(ii) Ans: \(\begin{pmatrix}-11\\ 13\end{pmatrix}\)
Find \(\vec{BC} = \vec{AC} – \vec{AB} = \begin{pmatrix}-3\\ 6\end{pmatrix} – \begin{pmatrix}8\\ -7\end{pmatrix} = \begin{pmatrix}-11\\ 13\end{pmatrix}\).
(b) Ans: \(\frac{1}{2}(b – a)\)
In a parallelogram, diagonals bisect each other. Thus, \(\vec{QX} = \frac{1}{2}\vec{QS} = \frac{1}{2}(\vec{PS} – \vec{PQ}) = \frac{1}{2}(b – a)\).
(c)(i) Ans: \(\begin{pmatrix}9 & -50\\ 10 & 69\end{pmatrix}\)
Matrix multiplication gives \(M^2 = \begin{pmatrix}2×2+5×1 & 2×5+5×8\\ 1×2+8×1 & 1×5+8×8\end{pmatrix} = \begin{pmatrix}9 & -50\\ 10 & 69\end{pmatrix}\).
(c)(ii) Ans: \(\frac{1}{11}\begin{pmatrix}8 & -5\\ -1 & 2\end{pmatrix}\)
For \(M = \begin{pmatrix}a & b\\ c & d\end{pmatrix}\), the inverse is \(\frac{1}{ad-bc}\begin{pmatrix}d & -b\\ -c & a\end{pmatrix}\). Here, \(ad-bc = 11\).