2.5.1 Multiplication of Two Matrices (Examples)


Example 1:
Find the product of the following pairs of matrices.
(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 )

Solution:
(a)  ( 1   5   2 )( 2 4 3 )← Matrices analysis 1×3 and 3×1  ↓   ↓ =1×1 matrix =( 1×2 ⊕ 5×4 ⊕ 2×3 ) =( 2+20+6 ) =( 28 )

(b)
( 2 8 −3 1 )( 1 0 4 −2 )← Matrices analysis 2×2 and 2×2  ↓ ↓ =2×2 matrix =( 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 ) ← Matrices analysis 2 × 1  and 1 × 2   ↓     ↓ = 2 × 2  matrix = ( − 3 × 2    − 3 × 6 5 × 2  5 × 6 ) = ( − 6 − 18 10 30 )

(d)
( 0 4 −1 3 )( 7 −2 )← Matrices analysis 2×2 and 2×1  ↓    ↓ =2×1 matrix =( 0×7+4×−2 −1×7+3×−2 ) =( −8 −13 )

(e)
( 7  −4 )( −2 0 −1 3 )← Matrices analysis 1×2 and 2×2  ↓ ↓ =1×2 matrix =( 7×−2+( −4×−1 ) 7×0+( −4×3 ) ) =( −14+4   0−12 ) =( −10 −12 )



Example 2:
Find the values of m and n in each of the following matrix equations.
( a ) ( 3 m ) ( 1   n ) = ( 3 12 − 2 − 8 ) ( b ) ( m 2 − 3 1 ) ( 2 n ) = ( 12 4 + 2 n ) ( c ) ( m − 3 − 1 1 ) ( − 1 2 4 n ) = ( − 14 − 11 5 3 )
 
Solution:
(a) ( 3 m ) ( 1   n ) = ( 3 12 − 2 − 8 ) ( 3 3 n m m n ) = ( 3 12 − 2 − 8 ) m = − 2 ,   3 n = 12 n = 4

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

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


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