SPM Mathematics Trial 2021 (Kelantan), Paper 2 (Question 12)

Question 12:
(a) Given $M(\begin{array}{l}9\\ -8\end{array}\begin{array}{l}3\\ -6\end{array})=(\begin{array}{l}\text{1}\\ 0\end{array}\begin{array}{l}\text{0}\\ 1\end{array})$ . Find matrix M. [2 Marks]

(b) An H truck is capable of carrying x kg of sand and a T truck is capable of carrying y kg of sand. 8 H trucks and 6 T Trucks are capable of carrying a total of 5 200 kg of sand at a time while 10 H lorries and 11 T lorries are capable of carrying 90% of the total 8 tons of sand.

(i) Write 2 linear equations that represent the above information. [2 marks]
(ii) Next using the matrix method, calculate the values of x and y. [5 marks]


Solution:
(a)

$\begin{array}{l}M\text{ is inverse matrix to }(\begin{array}{l}9\\ -8\end{array}\begin{array}{l}3\\ -6\end{array})\\ \\ \text{Inverse matrix}=\frac{1}{ad-bc}\left(\begin{array}{l}d-b\\ -ca\end{array}\right)\\ \\ M=\frac{1}{9\left(-6\right)-\left(-8\right)\left(3\right)}\left(\begin{array}{l}-6-3\\ 89\end{array}\right)\\ \\ M=-\frac{1}{30}\left(\begin{array}{l}-6-3\\ 89\end{array}\right)\\ \\ =\left(\begin{array}{l}\frac{1}{5}\frac{1}{10}\\ -\frac{4}{15}-\frac{3}{10}\end{array}\right)\end{array}$


(b)(i)

$\begin{array}{l}\text{Equation 1,}\\ 8x+6y=5200\\ \\ \text{Equation 2,}\\ 10x+11y=\frac{90}{100}\times 8000\leftarrow \boxed{\text{1 ton = 1000 kg}}\\ 10x+11y=7200\end{array}$


(b)(ii)

$\begin{array}{l}\left(\begin{array}{l}8\text{ 6}\\ 1011\end{array}\right)\left(\begin{array}{l}x\\ y\end{array}\right)=\left(\begin{array}{l}5200\\ 7200\end{array}\right)\\ \\ \left(\begin{array}{l}x\\ y\end{array}\right)=\frac{1}{8\left(11\right)-6\left(10\right)}\left(\begin{array}{l}11-6\\ -10\text{ 8}\end{array}\right)\left(\begin{array}{l}5200\\ 7200\end{array}\right)\\ \\ \left(\begin{array}{l}x\\ y\end{array}\right)=\frac{1}{28}\left(\begin{array}{l}11\left(5200\right)+\left(-6\right)\left(7200\right)\\ -10\left(5200\right)+8\left(7200\right)\end{array}\right)\\ \\ \left(\begin{array}{l}x\\ y\end{array}\right)=\frac{1}{28}\left(\begin{array}{l}14000\\ 5600\end{array}\right)\\ \\ \left(\begin{array}{l}x\\ y\end{array}\right)=\left(\begin{array}{l}500\\ 200\end{array}\right)\\ \\ x=500\text{ and }y=200\end{array}$