pembahasan matrik
3
1. a) | 5 6 −1 3 | =15 +6=21 b) | 2 −3 4 1 1 1 3 0 −5 | 2 −3 1 1 3 0 =¿ ¿ (−10 ) +(−9 ) +( 0) −( 12 ) −( 0 ) −( 15) ¿ −19−15 =−34 2. a) ( 3 2 10 7 ) = 1 21−20 ( 7 −2 −10 3 ) = ( 7 −2 −10 3 ) b) ( 4 8 3 6 ) = 1 24 −24 ( 6 −8 −3 4 ) =tidakmemiliki invers matriks c) ( 2 4 4 7 ) = 1 14 −16 ( 7 −4 −4 2 ) = −1 2 ( 7 −4 −4 2 ) = ( −7 2 2 2 − 1 ) 3. P = ( 2 3 −4 0 −4 2 1 −1 5 ) C 11 =(−1) 1+ 1 | −4 2 −1 5 | =1 x ( −20 −(−2 ) )=−20 +2=−18 C 12 =(−1) 1+ 2 | 0 2 1 5 | =−1 x (0−( 2) )=0 + 2=−2 C 13 =(−1) 1+ 3 | 0 −4 1 −1 | =1 x (0−( −4) )=0+4=4 C 21 =(−1) 2+ 1 | 3 −4 −1 5 | =−1 x ( 15−( 4 ) )=−15 + 4=−11 C 22 =(−1) 2+ 2 | 2 −4 1 5 | =1 x (10 −(−4 ) )=10 +4=14 C 23 =(−1) 2+ 3 | 2 3 1 −1 | =−1 x (−2−( 3) )=2+ 3=5 C 31 =(−1) 3+ 1 | 3 −4 −4 2 | =1 x ( 6−( 16 ) )=6−16 =−10 C 32 =(−1) 3+ 2 | 2 −4 0 2 | =−1 x (4−( 0) )=−4+ 0=−4
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Transcript of pembahasan matrik
1. a) | 5 6−1 3|=15+6=21
b) |2 −3 41 1 13 0 −5|
2 −31 13 0
=¿
¿ (−10 )+(−9 )+(0 )− (12 )−(0 )−(15 ) ¿−19−15=−34
2. a)( 3 210 7)= 1
21−20 ( 7 −2−10 3 )=( 7 −2
−10 3 )b)(4 8
3 6)= 124−24 ( 6 −8
−3 4 )=tidak memiliki invers matriks
c) (2 44 7)= 1
14−16 ( 7 −4−4 2 )=−1
2 ( 7 −4−4 2 )=(−7
22
2 −1)
3. P = (2 3 −40 −4 21 −1 5 )
C11=(−1)1+1|−4 2−1 5|=1x (−20− (−2 ) )=−20+2=−18
C12=(−1)1+2|0 21 5|=−1 x ( 0−(2 ) )=0+2=−2
C13=(−1)1+3|0 −41 −1|=1 x (0−(−4 ) )=0+4=4
C21=(−1)2+1| 3 −4−1 5 |=−1 x (15−(4 ) )=−15+4=−11
C22=(−1)2+2|2 −41 5 |=1 x (10−(−4 ) )=10+4=14
C23=(−1)2+3|2 31 −1|=−1 x (−2−(3 ) )=2+3=5
C31=(−1)3+1| 3 −4−4 2 |=1 x (6−(16 ))=6−16=−10
C32=(−1)3+2|2 −40 2 |=−1 x (4−(0 ))=−4+0=−4
C33=(−1)3+3|2 30 −4|=1x (−8− (6 ) )=−8−6=−14
Jadi Adj(P)=(C11 C12 C13
C21 C22 C23
C31 C32 C33)=(−18 −2 4
−11 14 5−10 −4 −14)
4. a) Adj (A) = [+|4 52 −1| −|1 5
0 −1| +|1 40 2|
−|0 32 −1| +|2 3
0 −1| −|2 00 2|
+|0 34 5| −|2 3
1 5| +|0 34 5|]=[−14 1 2
6 −2 4−12 −7 −12]
|2 0 31 4 50 2 −1|
2 01 40 2
=(−8 )+(0 )+ (6 )−(0 )−(20 )−(0 )=−22
A−1= 1det A
adj ( A )=−122 [−18 −2 4
−11 14 5−10 −4 −14 ]=[
911
111
−211
12
−711
−522
511
211
711
]5. b) Adj (B) = [+|
3 71 6| −|2 7
4 6| +|2 34 1|
−|0 11 6| +|1 1
4 6| −|1 04 1|
+|0 13 7| −|1 1
2 7| +|1 02 3|]=[ 11 16 −10
1 2 −1−3 −5 −3 ]
|1 0 12 3 74 1 6|
1 02 34 1
=(12 )+(7 )+(0 )−(18 )−(0 )−(2 )=−1
B−1= 1det B
adj (B )=−11 [ 11 16 −10
1 2 −1−3 −5 −3 ]=[−11 −16 10
−1 −2 13 5 3 ]
6. A . X=BA−1 . X=A−1BX=A−1B
¿ 1det A
adj (A ) .B=15 ( 4 −1
−3 2 ) x ( 5 711 3)=(
45
−15
−35
25
)x ( 5 711 3)=(
95
5
75
−3)7. X . A=BX . A . A−1=B . A−1
X=B A−1
B .1
det Aadj (A )=( 5 7
11 3)x 15 ( 4 −1
−3 2 )=( 5 711 3)x (
45
−15
−35
25
)=( 15
95
7 −1)
