Soalan Percubaan Matematik SPM 2021 (Kelantan), Kertas 2 – Soalan 12

Soalan 12:
(a) Diberi $M(\begin{array}{l}9\\ -8\end{array}\begin{array}{l}3\\ -6\end{array})=(\begin{array}{l}\text{1}\\ 0\end{array}\begin{array}{l}\text{0}\\ 1\end{array})$ . Cari matriks M. [2 Markah]

(b) Sebuah lori H mampu membawa x kg pasir dan sebuah lori T mampu membawa y kg pasir. 8 buah lori H dan 6 buah lori T mampu membawa sejumlah 5 200 kg pasir dalam satu-satu masa manakala 10 buah lori H dan 11 buah lori T mampu membawa 90% daripada sejumlah 8 tan pasir.

(i) Tulis 2 persamaan linear yang mewakili maklumat di atas. [2 markah]
(ii) Seterusnya dengan menggunakan kaedah matriks, hitung nilai x dan y. [5 markah]

Penyelesaian:
(a)

$\begin{array}{l}M\text{ adalah matriks songsang kepada }(\begin{array}{l}9\\ -8\end{array}\begin{array}{l}3\\ -6\end{array})\\ \\ \text{matriks songsang}=\frac{1}{ad-bc}\left(\begin{array}{l}d-b\\ -ca\end{array}\right)\\ \\ M=\frac{1}{9\left(-6\right)-\left(-8\right)\left(3\right)}\left(\begin{array}{l}-6-3\\ 89\end{array}\right)\\ \\ M=-\frac{1}{30}\left(\begin{array}{l}-6-3\\ 89\end{array}\right)\\ \\ =\left(\begin{array}{l}\frac{1}{5}\frac{1}{10}\\ -\frac{4}{15}-\frac{3}{10}\end{array}\right)\end{array}$


(b)(i)

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


(b)(ii)

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