4.5 Pendaraban Dua Matriks (Contoh Soalan)
Soalan 1:
Cari hasil darab bagi setiap pasangan matriks yang berikut.
(a) (1 5 2)(243)(b) (28−31)(104−2)
(c) (−35)(2 6)(d) (04−13)(7−2)(e) (7 −4)(−20−13)
Penyelesaian:
(a) (1 5 2)(243)←Analisis matriks1×3 dan 3×1 ↘ ↙= matriks 1×1=(1×2 ⊕ 5×4 ⊕ 2×3)=(2+20+6)=(28)(b) (28−31)(104−2)←Analisis matriks2×2 dan 2×2 ↘ ↙=matriks 2×2=(2×1+8×4 2×0+8×−2−3×1+1×4 −3×0+1×−2)=(34−161−2)
(c)(−35)(2 6)←Analisis matriks2×1 dan 1×2 ↘ ↙=matriks 2×2=(−3×2 −3×65×2 5×6)=(−6−181030)
(d)(04−13)(7−2)←Analisis matriks2×2 dan 2×1 ↘ ↙=matriks 2×1=(0×7+4×−2−1×7+3×−2)=(−8−13)
(e)(7 −4)(−20−13)←Analisis matriks1×2 dan 2×2 ↘ ↙=matriks 1×2=(7×−2+(−4×−1) 7×0+(−4×3))=(−14+4 0−12)=(−10 −12)
Contoh 2:
Cari nilai mdan nilai n dalam setiap persamaan matriks yang berikut:
(a)(3m)(1 n)=(312−2−8)
(b)(m2−31)(2n)=(124+2n)
(c)(m−3−11)(−124n)=(−14−1153)
Penyelesaian:
(a)(3m)(1 4)=(312−2n)(312m4m)=(312−2n)
m= –2,
4m = n
4 (–2) = n
n= –8
(b)(m2−31)(2n)=(124+2n)(2m+2n−6+n)=(124+2n)
–6 + n = 4 + 2n
n= –10
2m + 2n = 12
2m + 2 (–10) = 12
2m – 20 = 12
2m = 32
m = 16
(c)(m−3−11)(−124n)=(−14−1153)(−m+(−12)2m+(−3n)1+4−2+n)=(−14−1153)
–m – 12 = –14
–m = –2
m = 2
–2 + n = 3
n = 5