Latih Kendiri 2.2g (Soalan 1 & 2) – Buku Teks Matematik Tingkatan 5 Bab 2 – Matriks

Soalan 1:
Tuliskan persamaan linear serentak di bawah dalam bentuk matriks.
(a) x – y = 7, x + 3y = 5
(b) 3x + y = 0, 5x + 2y = –14
(c) 7x + 2y = –11, 2x – y = -10
(d) 3x + 2y – 14 = 0, 4y = 5x – 5
(e) 2x + y + 4 = 0, y – 3x = 11
(f) 2x + y = –9, 5x = -12
(g) 2x = 5y, x/5 + 2y = 3
(h) x/y = 4, 0.8(x + 5) = 3y

Penyelesaian:
(a)
$$\begin{aligned} & x-y=7 \\ & x+3 y=5 \\ & {\left[\begin{array}{cc} 1 & -1 \\ 1 & 3 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 7 \\ 5 \end{array}\right]} \end{aligned}$$
(b)
$$\begin{aligned} & 3 x+y=0 \\ & 5 x+2 y=-14 \\ & {\left[\begin{array}{ll} 3 & 1 \\ 5 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} 0 \\ -14 \end{array}\right]} \end{aligned}$$

(c)
$$\begin{aligned} & 7 x+2 y=-11 \\ & 2 x-y=-10 \\ & {\left[\begin{array}{cc} 7 & 2 \\ 2 & -1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} -11 \\ -10 \end{array}\right]} \end{aligned}$$

(d)
$$\begin{gathered} 3 x+2 y-14=0 \\ 3 x+2 y=14 \\ 4 y=5 x-5 \\ 5 x-4 y=5 \\ {\left[\begin{array}{cc} 3 & 2 \\ 5 & -4 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} 14 \\ 5 \end{array}\right]} \end{gathered}$$

(e)
$$\begin{gathered} 2 x+y+4=0 \\ 2 x+y=-4 \\ y-3 x=11 \\ -3 x+y=11 \\ {\left[\begin{array}{cc} 2 & 1 \\ -3 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} -4 \\ 11 \end{array}\right]} \end{gathered}$$

(f)
$$\begin{aligned} & 2 x+y=-9 \\ & 5 x=-12 \\ & {\left[\begin{array}{ll} 2 & 1 \\ 5 & 0 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} -9 \\ -12 \end{array}\right]} \end{aligned}$$

(g)
$$\begin{aligned} & 2 x=5 y \\ & 2 x-5 y=0 \\ & \frac{x}{5}+2 y=3 \\ & {\left[\begin{array}{cc} 2 & -5 \\ \frac{1}{5} & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 0 \\ 3 \end{array}\right]} \end{aligned}$$

(h)
$$\begin{gathered} \frac{x}{y}=4 \\ x=4 y \\ x-4 y=0 \\ 0.8(x+5)=3 y \\ 0.8 x+4=3 y \\ 0.8 x-3 y=-4 \\ {\left[\begin{array}{cc} 1 & -4 \\ 0.8 & -3 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} 0 \\ -4 \end{array}\right]} \end{gathered}$$

Soalan 2:
Dalam suatu pertandingan catur, jumlah peserta ialah 100 orang. Bilangan peserta lelaki, x, ialah 14 orang kurang daripada 2 kali bilangan peserta perempuan, y. Tuliskan persamaan linear serentak yang mewakili maklumat di atas dalam bentuk matriks.

Penyelesaian:
$$\begin{aligned} & x+y=100 \\ & x=2 y-14 \\ & x-2 y=-14 \\ & {\left[\begin{array}{cc} 1 & 1 \\ 1 & -2 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} 100 \\ -14 \end{array}\right] } \end{aligned}$$