Description

Patrick Gardocki
8.2.33
2 −3 1  −12 −1 2 2 −3 1 
A = 1 1 −1; B =  1 ; X =  4  + c1 3 AX = B → 1 1 −1
4 −1 −1 10 2 5 4 −1 −1
2 −3 1 1 −12 c1 = 0, AX = 1 1 −17 =  1 
4 −1 −1 7 10
If X is solution for AX = 0, then X is a solution for any constant, c1.
2 −3 1  1 −3/2 1/2  1 0 −2/5
R1/2 R2*2/5 R3-5R2
A = 1 1 −1 −−−−→ 0 5/2 −3/2 −−−−−−−→ 0 1 −3/5 −−−−−→
4 −1 −1 R2-R1 4 −1 −1 R1+R2*3/2 0 −5 −3
RREF(A)
Rank(A) = 2
Nullity(A) = # of Columns – Rank ∴ Nullity(A) = 3 − 2 = 1
8.3.5
1 1 1 1 1 1  1 1 1  1 1
-(-R1 + R2) R3-3R2 R3/9
A = 1 0 4 −−−−−−−−→ 0 1 −3 −−−−−→ 0 1 −3 −−−→ 0 1 1 4 1 0 3 0 0 0 9 0 0
1 0 0
0 1 0 = RREF(A)
0 0 1
Rank = number of non-zero rows in RREF = 3
1 0 0x1 0
0 1 0x2 = 0
0 0 1 x3 0
The null space is a zero vector, therefore the nullity is 0.
8.4.24

1 1 1 y z − 1 · 2 +x x 4 +z z + 1 · x
2 +x x 3 +y y 4 +z z = 1 · 3 + y 4 + z 2 + x
x1 −12 x
 2 =  1  If
x3 10
1
0

0 0
1
0 −2/5
−3/5 =
0
1 
−3
1 −−−−−−−R2+3R3→
R1−R2−R3
y 3 + y

3 +y y 4 +z z = 4y + yz − 3z − yz 2 +x x 4 +z z = 4x + xz − z − xz

2 +x x 3 +y y = 3x + xy − 2y − xy
1 1 1

x y z = −x + 2y − z
2 + x 3 + y 4 + z
8.5.26
det(AB) = detA ∗ detB = detB ∗ detA = det(BA)
8.6.31
4 −34 −3 1 0 16 − 3x 0
If AA = I, A is its own inverse. = → 16 − 3x = 1 → x = 5 x 4 x 4 0 1 0 16 − 3x
Since AA = I, A is its own inverse.
8.6.36
If either A or B is singular, then det(A) = 0 ∴ det(AB) = det(A) ∗ det(B) = 0 Thus AB is singular if A or B is singular
8.12.36
S = PDPT
√1 √1 √1  1 T  8
3263
A = √−130 √260 3 0√−13 0 √26 =  4 11 
√1√−1 √10 0 5 √1 √−1 √1
3 2 6 3 2 6
8.9.5
8 − λ 5
8.10.16
0 1 1 0 − λ 1 1  √

1 1 1 →  1 1 − λ 1  = 0; λ = −1, 1 ± 2
1 1 0 1 1 0 − λ
1 1 1 1 1 1  1 
Gauss-Jordan Elimination λ = −1; 1 2 1 → 0 1 0 ∴ K1 =  0 
1 1 1 0 0 0 −1
√
 
√1 2 1   1 
λ = 1 + 2;  → 
0 0 0
√
 
√1 − 2 1   1 
λ = 1 − 2;  → 0 1
0 0 0
√
Norm of K1 is 2, K2 and K3 is 2,
8.10.21
a and b
0
AK1 = 2
2 2
0
2 2 1  −2  1 
2−1 =  2  = −2−1 = λK1; λ1 = −2
0 0 0 0
0
AK2 = 2
2 2
0
2 2 1  −2  1 
2 0  =  0  = −2 0  = λK2; λ2 = −2
0 −1 2 −1
0
AK2 = 2
2 2
0
2 21 4 1
21 = 4 = 44 = λK2; λ3 = 4
0 1 4 1
c
K ̸ are not orthogonal. K orthogonal K orthogonal
Gram-Schmidt Process: V1 = K1
T  1   1   12 
V2 = K2 − KV12TVV11V1 =  0  − 21 −1 =  12 
−1 0 −1
Orthogonal set of eigenvectors: (V1,V2,K3)
Norms:
√
||V||K3|| = 3
 
P 2 6 3 2 − 2 1