Higher Engineering Mathematics

(Greg DeLong) #1
THE THEORY OF MATRICES AND DETERMINANTS 269

F

Problem 5. IfA=

(
23
1 − 4

)
andB=

(
− 57
− 34

)

findA×B.

LetA×B=CwhereC=


(
C 11 C 12
C 21 C 22

)

C 11 is the sum of the products of the first row ele-
ments ofAand the first column elements ofBtaken
one at a time,


i.e. C 11 =(2×(−5))+(3×(−3))=− 19


C 12 is the sum of the products of the first row ele-
ments ofAand the second column elements ofB,
taken one at a time,


i.e. C 12 =(2×7)+(3×4)= 26


C 21 is the sum of the products of the second row
elements ofAand the first column elements ofB,
taken one at a time,


i.e. C 21 =(1×(−5))+(− 4 ×(−3))= 7


Finally,C 22 is the sum of the products of the second
row elements ofAand the second column elements
ofB, taken one at a time,


i.e. C 22 =(1×7)+((−4)×4)=− 9


Thus,A×B=


(
− 19 26
7 − 9

)

Problem 6. Simplify
(
340
− 26 − 3
7 − 41

)

×

(
2
5
− 1

)

The sum of the products of the elements of each
row of the first matrix and the elements of the second
matrix, (called acolumn matrix), are taken one at a
time. Thus:
(
340
− 26 − 3
7 − 41


)

×

(
2
5
− 1

)

=

(
(3×2) +(4×5) +(0×(−1))
(− 2 ×2)+(6×5) +(− 3 ×(−1))
(7×2) +(− 4 ×5)+(1×(−1))

)

=

(
26
29
− 7

)

Problem 7. IfA=

(
340
− 26 − 3
7 − 41

)

and

B=

(
2 − 5
5 − 6
− 1 − 7

)

, findA×B.

The sum of the products of the elements of each row
of the first matrix and the elements of each column of
the second matrix are taken one at a time. Thus:
(
340
− 26 − 3
7 − 41

)

×

(
2 − 5
5 − 6
− 1 − 7

)

=

⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝
[(3×2) [(3×(−5))
+(4×5) +(4×(−6))
+(0×(−1))] +(0×(−7))]
[(− 2 ×2) [(− 2 ×(−5))
+(6×5) +(6×(−6))
+(− 3 ×(−1))] +(− 3 ×(−7))]
[(7×2) [(7×(−5))
+(− 4 ×5) +(− 4 ×(−6))
+(1×(−1))] +(1×(−7))]

⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ =

(
26 − 39
29 − 5
− 7 − 18

)

Problem 8. Determine
(
103
212
131

)

×

(
220
132
320

)

The sum of the products of the elements of each row
of the first matrix and the elements of each column of
the second matrix are taken one at a time. Thus:
(
103
212
131

)

×

(
220
132
320

)

=

⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝
[(1×2) [(1×2) [(1×0)
+(0×1) +(0×3) +(0×2)
+(3×3)] +(3×2)] +(3×0)]
[(2×2) [(2×2) [(2×0)
+(1×1) +(1×3) +(1×2)
+(2×3)] +(2×2)] +(2×0)]
[(1×2) [(1×2) [(1×0)
+(3×1) +(3×3) +(3×2)
+(1×3)] +(1×2)] +(1×0)]

⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ =

(
11 8 0
11 11 2
8136

)
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