Soalan 3:
Selesaikan persamaan linear serentak di bawah dengan menggunakan kaedah matriks.
(a) x – 2y = 5, 2x – 3y = 10
(b) 2x – 5y = 1, 3x – y = –5
(c) 2x – y = 8, x + y = 1
(d) 3x + 2y = 4, 9x + 4y = 14
(e) 4x + 3y = 11, 2y = 9 – 6x
(f) 5x – 5y – 6 = 0, 2x – 2.1 = 3y
(g) p + 3q = 4, 3 + p/2 = q
(h) m + n = 5, m/2 – n/4 = 1
Penyelesaian:
(a)
$$ \begin{gathered} x-2 y=5 \\ 2 x-3 y=10 \end{gathered} $$
$$ \begin{aligned} {\left[\begin{array}{ll} 1 & -2 \\ 2 & -3 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right] } & =\left[\begin{array}{c} 5 \\ 10 \end{array}\right] \\ {\left[\begin{array}{l} x \\ y \end{array}\right] } & =\frac{1}{1(-3)-(-2)(2)}\left[\begin{array}{ll} -3 & 2 \\ -2 & 1 \end{array}\right]\left[\begin{array}{c} 5 \\ 10 \end{array}\right] \\ & =1\left[\begin{array}{c} -3(5)+2(10) \\ -2(5)+1(10) \end{array}\right] \\ & =1\left[\begin{array}{c} 5 \\ 10 \end{array}\right] \\ & =\left[\begin{array}{c} 5 \\ 0 \end{array}\right] \\ & x=5, y=0 \end{aligned} $$
(b)
$$ \begin{aligned} 2 x-5 y & =1 \\ 3 x-y & =-5 \end{aligned} $$
$$ \begin{aligned} {\left[\begin{array}{ll} 2 & -5 \\ 3 & -1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right] } & =\left[\begin{array}{c} 1 \\ -5 \end{array}\right] \\ {\left[\begin{array}{l} x \\ y \end{array}\right] } & =\frac{1}{2(-1)-(-5)(3)}\left[\begin{array}{ll} -1 & 5 \\ -3 & 2 \end{array}\right]\left[\begin{array}{c} 1 \\ -5 \end{array}\right] \\ & =\frac{1}{13}\left[\begin{array}{l} -1(1)+5(-5) \\ -3(1)+2(-5) \end{array}\right] \\ & =\frac{1}{13}\left[\begin{array}{l} -26 \\ -13 \end{array}\right] \\ & =\left[\begin{array}{l} -2 \\ -1 \end{array}\right] \\ x=-2, y= & -1 \end{aligned} $$
(c)
$$ \begin{array}{r} 2 x-y=8 \\ x+y=1 \end{array} $$
$$ \begin{aligned} {\left[\begin{array}{cc} 2 & -1 \\ 1 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right] } & =\left[\begin{array}{l} 8 \\ 1 \end{array}\right] \\ {\left[\begin{array}{l} x \\ y \end{array}\right] } & =\frac{1}{2(1)-(-1)(1)}\left[\begin{array}{cc} 1 & 1 \\ -1 & 2 \end{array}\right]\left[\begin{array}{l} 8 \\ 1 \end{array}\right] \\ & =\frac{1}{3}\left[\begin{array}{c} 1(8)+1(1) \\ -1(8)+2(1) \end{array}\right] \\ & =\frac{1}{3}\left[\begin{array}{c} 9 \\ -6 \end{array}\right] \\ & =\left[\begin{array}{c} 3 \\ -2 \end{array}\right] \\ x=3, y & =-2 \end{aligned} $$
(d)
$$ \begin{gathered} 3 x+2 y=4 \\ 9 x+4 y=14 \end{gathered} $$
$$ \begin{aligned} {\left[\begin{array}{ll} 3 & 2 \\ 9 & 4 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right] } & =\left[\begin{array}{c} 4 \\ 14 \end{array}\right] \\ {\left[\begin{array}{l} x \\ y \end{array}\right] } & =\frac{1}{3(4)-(2)(9)}\left[\begin{array}{cc} 4 & -2 \\ -9 & 3 \end{array}\right]\left[\begin{array}{c} 4 \\ 14 \end{array}\right] \\ & =-\frac{1}{6}\left[\begin{array}{c} 4(4)+(-2)(14) \\ -9(4)+(3)(14) \end{array}\right] \\ & =-\frac{1}{6}\left[\begin{array}{c} -12 \\ 6 \end{array}\right] \\ & =\left[\begin{array}{c} 2 \\ -1 \end{array}\right] \\ x=2, y & =-1 \end{aligned} $$
(e)
$$ \begin{aligned} 4 x+3 y & =11 \\ 2 y & =9-6 x \\ 6 x+2 y & =9 \end{aligned} $$
$$ \begin{aligned} {\left[\begin{array}{ll} 4 & 3 \\ 6 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right] } & =\left[\begin{array}{c} 11 \\ 9 \end{array}\right] \\ {\left[\begin{array}{l} x \\ y \end{array}\right] } & =\frac{1}{4(2)-(3)(6)}\left[\begin{array}{cc} 2 & -3 \\ -6 & 4 \end{array}\right]\left[\begin{array}{c} 11 \\ 9 \end{array}\right] \\ & =-\frac{1}{10}\left[\begin{array}{c} 2(11)+(-3)(9) \\ -6(11)+4(9) \end{array}\right] \\ & =-\frac{1}{10}\left[\begin{array}{c} -5 \\ -30 \end{array}\right] \\ & =\left[\begin{array}{c} 0.5 \\ 3 \end{array}\right] \\ x=0.5, y & =3 \end{aligned} $$
(f)
$$ \begin{gathered} 5 x-5 y-6=0 \\ 5 x-5 y=6 \\ 2 x-2.1=3 y \\ 2 x-3 y=2.1 \end{gathered} $$
$$ \begin{aligned} {\left[\begin{array}{ll} 5 & -5 \\ 2 & -3 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right] } & =\left[\begin{array}{c} 6 \\ 2.1 \end{array}\right] \\ {\left[\begin{array}{l} x \\ y \end{array}\right] } & =\frac{1}{5(-3)-(-5)(2)}\left[\begin{array}{ll} -3 & 5 \\ -2 & 5 \end{array}\right]\left[\begin{array}{c} 6 \\ 2.1 \end{array}\right] \\ & =-\frac{1}{5}\left[\begin{array}{l} -3(6)+5(2.1) \\ -2(6)+5(2.1) \end{array}\right] \\ & =-\frac{1}{5}\left[\begin{array}{l} -7.5 \\ -1.5 \end{array}\right] \\ & =\left[\begin{array}{l} 1.5 \\ 0.3 \end{array}\right] \\ x=1.5, y & =0.3 \end{aligned} $$
(g)
$$ \begin{aligned} p+3 q & =4 \\ 3+\frac{p}{2} & =q \\ \frac{1}{2} p-q & =-3 \end{aligned} $$
$$ \begin{aligned} {\left[\begin{array}{cc} 1 & 3 \\ \frac{1}{2} & -1 \end{array}\right]\left[\begin{array}{l} p \\ q \end{array}\right] } & =\left[\begin{array}{c} 4 \\ -3 \end{array}\right] \\ {\left[\begin{array}{l} p \\ q \end{array}\right] } & =\frac{1}{1(-1)-3\left(\frac{1}{2}\right)}\left[\begin{array}{cc} -1 & -3 \\ -\frac{1}{2} & 1 \end{array}\right]\left[\begin{array}{c} 4 \\ -3 \end{array}\right] \\ & =-\frac{2}{5}\left[\begin{array}{c} -1(4)+(-3)(-3) \\ -\frac{1}{2}(4)+1(-3) \end{array}\right] \\ & =-\frac{2}{5}\left[\begin{array}{c} 5 \\ -5 \end{array}\right] \\ & =\left[\begin{array}{c} -2 \\ 2 \end{array}\right] \\ p=-2, q & =2 \end{aligned} $$
(h)
$$ \begin{aligned} m+n & =5 \\ \frac{m}{2}-\frac{n}{4} & =1 \end{aligned} $$
$$ \begin{aligned} {\left[\begin{array}{cc} 1 & 1 \\ \frac{1}{2} & -\frac{1}{4} \end{array}\right]\left[\begin{array}{c} m \\ n \end{array}\right] } & =\left[\begin{array}{l} 5 \\ 1 \end{array}\right] \\ {\left[\begin{array}{c} m \\ n \end{array}\right] } & =\frac{1}{1\left(-\frac{1}{4}\right)-(1)\left(\frac{1}{2}\right)}\left[\begin{array}{cc} -\frac{1}{4} & -1 \\ -\frac{1}{2} & 1 \end{array}\right]\left[\begin{array}{l} 5 \\ 1 \end{array}\right] \\ & =-\frac{4}{3}\left[\begin{array}{r} -\frac{1}{4}(5)+(-1)(1) \\ -\frac{1}{2}(5)+1(1) \end{array}\right] \\ & =-\frac{4}{3}\left[\begin{array}{r} -\frac{9}{4} \\ -\frac{3}{2} \end{array}\right] \\ & =\left[\begin{array}{l} 3 \\ 2 \end{array}\right] \end{aligned} $$
$$ m=3, n=2 $$
Selesaikan persamaan linear serentak di bawah dengan menggunakan kaedah matriks.
(a) x – 2y = 5, 2x – 3y = 10
(b) 2x – 5y = 1, 3x – y = –5
(c) 2x – y = 8, x + y = 1
(d) 3x + 2y = 4, 9x + 4y = 14
(e) 4x + 3y = 11, 2y = 9 – 6x
(f) 5x – 5y – 6 = 0, 2x – 2.1 = 3y
(g) p + 3q = 4, 3 + p/2 = q
(h) m + n = 5, m/2 – n/4 = 1
Penyelesaian:
(a)
$$ \begin{gathered} x-2 y=5 \\ 2 x-3 y=10 \end{gathered} $$
$$ \begin{aligned} {\left[\begin{array}{ll} 1 & -2 \\ 2 & -3 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right] } & =\left[\begin{array}{c} 5 \\ 10 \end{array}\right] \\ {\left[\begin{array}{l} x \\ y \end{array}\right] } & =\frac{1}{1(-3)-(-2)(2)}\left[\begin{array}{ll} -3 & 2 \\ -2 & 1 \end{array}\right]\left[\begin{array}{c} 5 \\ 10 \end{array}\right] \\ & =1\left[\begin{array}{c} -3(5)+2(10) \\ -2(5)+1(10) \end{array}\right] \\ & =1\left[\begin{array}{c} 5 \\ 10 \end{array}\right] \\ & =\left[\begin{array}{c} 5 \\ 0 \end{array}\right] \\ & x=5, y=0 \end{aligned} $$
(b)
$$ \begin{aligned} 2 x-5 y & =1 \\ 3 x-y & =-5 \end{aligned} $$
$$ \begin{aligned} {\left[\begin{array}{ll} 2 & -5 \\ 3 & -1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right] } & =\left[\begin{array}{c} 1 \\ -5 \end{array}\right] \\ {\left[\begin{array}{l} x \\ y \end{array}\right] } & =\frac{1}{2(-1)-(-5)(3)}\left[\begin{array}{ll} -1 & 5 \\ -3 & 2 \end{array}\right]\left[\begin{array}{c} 1 \\ -5 \end{array}\right] \\ & =\frac{1}{13}\left[\begin{array}{l} -1(1)+5(-5) \\ -3(1)+2(-5) \end{array}\right] \\ & =\frac{1}{13}\left[\begin{array}{l} -26 \\ -13 \end{array}\right] \\ & =\left[\begin{array}{l} -2 \\ -1 \end{array}\right] \\ x=-2, y= & -1 \end{aligned} $$
(c)
$$ \begin{array}{r} 2 x-y=8 \\ x+y=1 \end{array} $$
$$ \begin{aligned} {\left[\begin{array}{cc} 2 & -1 \\ 1 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right] } & =\left[\begin{array}{l} 8 \\ 1 \end{array}\right] \\ {\left[\begin{array}{l} x \\ y \end{array}\right] } & =\frac{1}{2(1)-(-1)(1)}\left[\begin{array}{cc} 1 & 1 \\ -1 & 2 \end{array}\right]\left[\begin{array}{l} 8 \\ 1 \end{array}\right] \\ & =\frac{1}{3}\left[\begin{array}{c} 1(8)+1(1) \\ -1(8)+2(1) \end{array}\right] \\ & =\frac{1}{3}\left[\begin{array}{c} 9 \\ -6 \end{array}\right] \\ & =\left[\begin{array}{c} 3 \\ -2 \end{array}\right] \\ x=3, y & =-2 \end{aligned} $$
(d)
$$ \begin{gathered} 3 x+2 y=4 \\ 9 x+4 y=14 \end{gathered} $$
$$ \begin{aligned} {\left[\begin{array}{ll} 3 & 2 \\ 9 & 4 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right] } & =\left[\begin{array}{c} 4 \\ 14 \end{array}\right] \\ {\left[\begin{array}{l} x \\ y \end{array}\right] } & =\frac{1}{3(4)-(2)(9)}\left[\begin{array}{cc} 4 & -2 \\ -9 & 3 \end{array}\right]\left[\begin{array}{c} 4 \\ 14 \end{array}\right] \\ & =-\frac{1}{6}\left[\begin{array}{c} 4(4)+(-2)(14) \\ -9(4)+(3)(14) \end{array}\right] \\ & =-\frac{1}{6}\left[\begin{array}{c} -12 \\ 6 \end{array}\right] \\ & =\left[\begin{array}{c} 2 \\ -1 \end{array}\right] \\ x=2, y & =-1 \end{aligned} $$
(e)
$$ \begin{aligned} 4 x+3 y & =11 \\ 2 y & =9-6 x \\ 6 x+2 y & =9 \end{aligned} $$
$$ \begin{aligned} {\left[\begin{array}{ll} 4 & 3 \\ 6 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right] } & =\left[\begin{array}{c} 11 \\ 9 \end{array}\right] \\ {\left[\begin{array}{l} x \\ y \end{array}\right] } & =\frac{1}{4(2)-(3)(6)}\left[\begin{array}{cc} 2 & -3 \\ -6 & 4 \end{array}\right]\left[\begin{array}{c} 11 \\ 9 \end{array}\right] \\ & =-\frac{1}{10}\left[\begin{array}{c} 2(11)+(-3)(9) \\ -6(11)+4(9) \end{array}\right] \\ & =-\frac{1}{10}\left[\begin{array}{c} -5 \\ -30 \end{array}\right] \\ & =\left[\begin{array}{c} 0.5 \\ 3 \end{array}\right] \\ x=0.5, y & =3 \end{aligned} $$
(f)
$$ \begin{gathered} 5 x-5 y-6=0 \\ 5 x-5 y=6 \\ 2 x-2.1=3 y \\ 2 x-3 y=2.1 \end{gathered} $$
$$ \begin{aligned} {\left[\begin{array}{ll} 5 & -5 \\ 2 & -3 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right] } & =\left[\begin{array}{c} 6 \\ 2.1 \end{array}\right] \\ {\left[\begin{array}{l} x \\ y \end{array}\right] } & =\frac{1}{5(-3)-(-5)(2)}\left[\begin{array}{ll} -3 & 5 \\ -2 & 5 \end{array}\right]\left[\begin{array}{c} 6 \\ 2.1 \end{array}\right] \\ & =-\frac{1}{5}\left[\begin{array}{l} -3(6)+5(2.1) \\ -2(6)+5(2.1) \end{array}\right] \\ & =-\frac{1}{5}\left[\begin{array}{l} -7.5 \\ -1.5 \end{array}\right] \\ & =\left[\begin{array}{l} 1.5 \\ 0.3 \end{array}\right] \\ x=1.5, y & =0.3 \end{aligned} $$
(g)
$$ \begin{aligned} p+3 q & =4 \\ 3+\frac{p}{2} & =q \\ \frac{1}{2} p-q & =-3 \end{aligned} $$
$$ \begin{aligned} {\left[\begin{array}{cc} 1 & 3 \\ \frac{1}{2} & -1 \end{array}\right]\left[\begin{array}{l} p \\ q \end{array}\right] } & =\left[\begin{array}{c} 4 \\ -3 \end{array}\right] \\ {\left[\begin{array}{l} p \\ q \end{array}\right] } & =\frac{1}{1(-1)-3\left(\frac{1}{2}\right)}\left[\begin{array}{cc} -1 & -3 \\ -\frac{1}{2} & 1 \end{array}\right]\left[\begin{array}{c} 4 \\ -3 \end{array}\right] \\ & =-\frac{2}{5}\left[\begin{array}{c} -1(4)+(-3)(-3) \\ -\frac{1}{2}(4)+1(-3) \end{array}\right] \\ & =-\frac{2}{5}\left[\begin{array}{c} 5 \\ -5 \end{array}\right] \\ & =\left[\begin{array}{c} -2 \\ 2 \end{array}\right] \\ p=-2, q & =2 \end{aligned} $$
(h)
$$ \begin{aligned} m+n & =5 \\ \frac{m}{2}-\frac{n}{4} & =1 \end{aligned} $$
$$ \begin{aligned} {\left[\begin{array}{cc} 1 & 1 \\ \frac{1}{2} & -\frac{1}{4} \end{array}\right]\left[\begin{array}{c} m \\ n \end{array}\right] } & =\left[\begin{array}{l} 5 \\ 1 \end{array}\right] \\ {\left[\begin{array}{c} m \\ n \end{array}\right] } & =\frac{1}{1\left(-\frac{1}{4}\right)-(1)\left(\frac{1}{2}\right)}\left[\begin{array}{cc} -\frac{1}{4} & -1 \\ -\frac{1}{2} & 1 \end{array}\right]\left[\begin{array}{l} 5 \\ 1 \end{array}\right] \\ & =-\frac{4}{3}\left[\begin{array}{r} -\frac{1}{4}(5)+(-1)(1) \\ -\frac{1}{2}(5)+1(1) \end{array}\right] \\ & =-\frac{4}{3}\left[\begin{array}{r} -\frac{9}{4} \\ -\frac{3}{2} \end{array}\right] \\ & =\left[\begin{array}{l} 3 \\ 2 \end{array}\right] \end{aligned} $$
$$ m=3, n=2 $$