Matematik SPM 2018, Kertas 2 (Soalan 7 & 8)

Soalan 7 (6 markah):
Rajah 6 menunjukkan dua garis lurus, JK dan LM, dilukis pada suatu satah Cartes.
Garis lurus KM adalah selari dengan paksi-x.

Rajah 6

Cari
(a) persamaan garis lurus KM,
(b) persamaan garis lurus LM,
(c) nilai bagi k.

Penyelesaian:
(a)
Persamaan garis lurus KM ialah y = 3

(b)
$\begin{array}{l}\text{Diberi persamaan }JK:\\ 2y=4x+3\\ y=2x+\frac{3}{2}\\ \text{Oleh itu, }{m}_{JK}=2\\ m_{LM}=m_{JK}=2\\ \\ y=mx+c\\ \text{Pada }M\left(5,3\right)\\ 3=2\left(5\right)+c\\ 3=10+c\\ c=-7\\ \\ \therefore \text{ Persamaan garis lurus }LM\text{ ialah}\\ y=2x-7.\end{array}$


(c)
$\begin{array}{l}\text{Gantikan }\left(k,0\right)\text{ ke dalam }y=2x-7\\ 0=2\left(k\right)-7\\ 7=2k\\ k=\frac{7}{2}\end{array}$

Soalan 8 (6 markah):
$\begin{array}{l}\text{Diberi }A=\left(\begin{array}{l}4-2\\ 3-1\end{array}\right),B=m\left(\begin{array}{l}-1n\\ -34\end{array}\right)\\ \text{dan }I=\left(\begin{array}{l}\text{1 0}\\ \text{0}1\end{array}\right).\end{array}$
(a) Cari nilai m dan nilai n jika AB = I.
(b) Tulis persamaan linear serentak berikut dalam bentuk persamaan matriks:
4x – 2y = 3
3x – y = 2
Seterusnya, menggunakan kaedah matriks, hitung nilai x dan nilai y.

Penyelesaian:
(a)
$\begin{array}{l}\text{Jika }AB=I,\text{ maka }B=A^{-1}\\ A^{-1}=\frac{1}{4\left(-1\right)-\left(-2\right)\left(3\right)}\left(\begin{array}{l}-12\\ -3\text{4}\end{array}\right)\\ A^{-1}=\frac{1}{2}\left(\begin{array}{l}-12\\ -3\text{4}\end{array}\right)\\ \\ \text{Secara perbandingan:}\\ B=A^{-1}\\ m\left(\begin{array}{l}-1n\\ -34\end{array}\right)=\frac{1}{2}\left(\begin{array}{l}-12\\ -3\text{4}\end{array}\right)\\ m=\frac{1}{2};n=2\end{array}$

(b)
$\begin{array}{l}4x-2y=3\\ 3x-y=2\\ \left(\begin{array}{l}4-2\\ 3-1\end{array}\right)\left(\begin{array}{l}x\\ y\end{array}\right)=\left(\begin{array}{l}3\\ 2\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\frac{1}{2}\left(\begin{array}{l}-12\\ -3\text{4}\end{array}\right)\left(\begin{array}{l}3\\ 2\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\frac{1}{2}\left(\begin{array}{l}\left(-1\right)\left(3\right)\text{+}\left(2\right)\left(2\right)\\ \left(-3\right)\left(3\right)\text{+}\left(4\right)\left(2\right)\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\frac{1}{2}\left(\begin{array}{l}1\\ -1\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\left(\begin{array}{l}\frac{1}{2}\\ -\frac{1}{2}\end{array}\right)\\ x=\frac{1}{2}\text{ dan }y=-\frac{1}{2}\end{array}$