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Bab 15 Matriks

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4.5 Pendaraban Dua Matriks (Contoh Soalan)

Soalan 1:
Cari hasil darab bagi setiap pasangan matriks yang berikut.
(a) ( 1   5   2 )( 2 4 3 ) (b) ( 2 8 −3 1 )( 1 0 4 −2 )
(c) ( −3 5 )( 2   6 ) (d) ( 0 4 −1 3 )( 7 −2 ) (e) ( 7   −4 )( −2 0 −1 3 )

Penyelesaian:
(a)  ( 1   5   2 )( 2 4 3 )← Analisis matriks 1×3 dan 3×1  ↘            ↙ = matriks 1×1 =( 1×2 ⊕ 5×4 ⊕ 2×3 ) =( 2+20+6 ) =( 28 )

(b)  ( 2 8 −3 1 )( 1 0 4 −2 )← Analisis matriks 2×2 dan 2×2  ↘            ↙ =matriks 2×2 =( 2×1+8×4   2×0+8×−2 −3×1+1×4   −3×0+1×−2 ) =( 34 −16 1 −2 )

(c) ( −3 5 )( 2   6 )← Analisis matriks 2×1 dan 1×2  ↘            ↙ =matriks 2×2 =( −3×2   −3×6 5×2      5×6 ) =( −6 −18 10 30 )

(d) ( 0 4 −1 3 )( 7 −2 )← Analisis matriks 2×2 dan 2×1  ↘            ↙ =matriks 2×1 =( 0×7+4×−2 −1×7+3×−2 ) =( −8 −13 )

(e) ( 7   −4 )( −2 0 −1 3 )← Analisis matriks 1×2 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)( 3 m )( 1   n )=( 3 12 −2 −8 )
(b)( m 2 −3 1 )( 2 n )=( 12 4+2n )
(c)( m −3 −1 1 )( −1 2 4 n )=( −14 −11 5 3 )

Penyelesaian:
(a) ( 3 m )( 1   4 )=( 3 12 −2 n ) ( 3 12 m 4m )=( 3 12 −2 n )
m= –2,
4m = n
4 (–2) = n
n= –8

(b) ( m 2 −3 1 )( 2 n )=( 12 4+2n ) ( 2m+2n −6+n )=( 12 4+2n )  
–6 + n = 4 + 2n
n= –10
2m + 2n = 12
2m + 2 (–10) = 12
2m – 20 = 12
2m = 32
m = 16

(c) ( m −3 −1 1 )( −1 2 4 n )=( −14 −11 5 3 ) ( −m+(−12) 2m+(−3n) 1+4 −2+n )=( −14 −11 5 3 )
–m – 12 = –14
–m = –2
m = 2
–2 + n = 3
n = 5