345
Chapter 6 Test R3 + R1 and R3 + R2
5. Write the augmented matrix 3
-1 3
2
-3
-10 26
3R1+R2and4R1+R3 3
-1 0
→
-3
2
-1 0
3
87 -10
8
3
1
12.
-3
−32 = Thus,z=4,y= −3, and 417 5 − 3−8+9 x= =2 −1 7. det= −3(2) − 6(−1) = 0, the matrix is not invertible
10.
11.
− 0.2(3.4)] 3[2.4(7.3) − + −−1.3(0.5)] −41.283,−−the is 1.3(0.2)] invertible det = 0.5[0.5(7.3) 8(−1)] −=(−5)[3(4) 8(1)]matrix + 0[3(−1) − 1(1)] = 1.2[2.4(3.4) det = 2[4(1) 44, the matrix is invertible − 4(1) = −1 det = 3(1) Thus,theinverseisgivenby 0 -1 -2 1 0 0 1 1 2 0 1 0 0 1 1 0 0 1
Swap R1 and R2 → 1 0 0
1 -1 1
2 0 -2 1 1 0
1 0 0
0 0 1
1 0 0
0 1 0
Swap R2 and R3 → 1 0 0
1 1 -1
2 0 1 0 -2 1
−R2 + R1 and R2 + R3 → 1 0 0
0 1 0
1 1 -1
0 0 1
1 -1
1 0 1 0 0
0 2 1
0 -1
1 0 0
-1 1 1
[
-1
1 -1
1 0
[ 11-r
0 -1 20
]
The eigenvector]sare[−5t,3t]and[−2t,t]respectively 15-r 4 13. [
6. det = 2(4) − 1(8) = 0, the matrix is not invertible
9.
0 0
0 1
1 1
r2 − 1 = 0 The eigenvalues are r = −1andr=1
17
8.
1 0
0 -1 0
Thus, the inverse is given by
2
−1R3→
-3
2
0 0
−R3→
−14R2 + 5R3 →
1 0
→
4
]
r2 −4r+3=0 The eigenv [aluesarer=]1[and]r=[3 14 4 x Forr = ]=1,[ -42] -12 y x -2t = [ [7t 4 x 12 Forr [ -14 ] [ y ] ]=3,[ -42] x -t [ 3t -5-r 14. 2
-4 -2
0 0
] ]
0 0
-8 -1-R
r3 + r2−9r−9=0 The eigenvalues are r = −3,r=−1andr=3 The eigenvectors are [2t,t, −t], [t, t, −t] and [0, −2t, t] resp ectively -1-r 0 15. 4 0
-8 11-r
r3 −3r2−49r+147=0 The eigenvalues are r = −7,r=3andr=7 6
0
-8
4 x 0 = y t 0 z
0
18
Forr=−7,
x
0
z 0 y=0
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