Download Linear Algebra Homework Solutions: Finding Inverse Matrices and Regression Lines and more Assignments Linear Algebra in PDF only on Docsity!
Homework Solutions Section 5.
(^) restart:
(^) with(linalg):
Warning, the protected names norm and trace have been redefined and
unprotected
1 Problem 1
1.1 Problem 1a)
(^) A1 := array(1..3,1..2,[[1,1],[2,-3],[0,0]]);
A1 :=
(^) b1 := array(1..3,1..1,[[3],[1],[2]]);
b1 :=
(^) inv := multiply(
(^) inverse(multiply(transpose(A1),A1)), transpose(A1));
inv :=
(^) x1 := multiply(inv,b1);
x1 :=
[
]
1.2 Problem 1b)
(^) A2 := array(1..3,1..2,[[-1,1],[2,1],[1,-2]]);
A2 :=
(^) b2 := array(1..3,1..1,[[10],[5],[20]]);
b2 :=
(^) inv := multiply(
(^) inverse(multiply(transpose(A2),A2)), transpose(A2));
inv :=
(^) x2 := multiply(inv,b2);
x2 :=
1.3 Problem 1c)
(^) A3 := array(1..4,1..3,[[1,1,1],[-1,1,1],[0,-1,1],[1,0,1]]);
A3 :=
(^) b3 := array(1..4,1..1,[[4],[0],[1],[2]]);
b3 :=
(^) inv := multiply(
(^) inverse(multiply(transpose(A3),A3)), transpose(A3));
inv :=
(^) x3 := multiply(inv,b3);
x3 :=
2 Problem 2
2.1 Problem 2a)
(^) p1 := multiply(A1,x1);
p1 :=
(^) r1 := matadd(p1,-b1);
r1 :=
(^) transpose(A1);
(^) multiply(transpose(A3),r3);
3 Problem 3
3.1 Problem 3a)
(^) A := array(1..3,1..2,[[1,2],[2,4],[-1,-2]]);
A :=
(^) rank(A);
deficient rank, so A
T A is singular!
(^) b := array(1..3,1..1,[[3],[2],[1]]);
b :=
(^) ATA := multiply(transpose(A),A);
ATA :=
[
]
(^) ATb := multiply(transpose(A),b);
ATb :=
[
]
(^) aug := augment(ATA,ATb);
aug :=
[
]
(^) gaussjord(aug); [ 1 2 1
0 0 0
]
(^) solve( x1+2*x2=1,{x1,x2});
{x2 = x2 , x1 = − 2 x2 + 1}
(^) x1 = -2*alpha + 1, x2 = alpha;
x1 = − 2 α + 1, x2 = α
3.2 Problem 3b)
(^) A := array(1..3,1..3,[[1,1,3],[-1,3,1],[1,2,4]]);
A :=
(^) rank(A);
deficient rank, so A T A is singular!
(^) b := array(1..3,1..1,[[-2],[0],[8]]);
b :=
(^) ATA := multiply(transpose(A),A);
ATA :=
(^) ATb := multiply(transpose(A),b);
ATb :=
(^) aug := augment(ATA,ATb);
aug :=
(^) gaussjord(aug);
solve( {x1+2*x3=2,x2+x3=1},{x1,x2,x3});
{x2 = 1 − x3 , x1 = 2 − 2 x3 , x3 = x3 }
(^) x1 = 2-2*alpha, x2 = 1-alpha, x3=alpha;
x1 = 2 − 2 α, x2 = 1 − α, x3 = α
4 Problem 7
(^) A := array(1..5,1..2,[[x1,1],[x2,1],[x3,1],[x4,1],[xN,1]]);
A :=
x1 1
x2 1
x3 1
x4 1
xN 1
6 Problem 13
(^) A := array(1..4,1..3,[[-12,-22,1],[02,2.42,1],
[1.12,-42,1],[2.42,-1.62,1]]);
A :=
(^) b := array(1..4,1..1,[[1ˆ2+2ˆ2],[0ˆ2+2.4ˆ2],
(^) [1.1ˆ2+4ˆ2],[2.4ˆ2+1.6ˆ2]]);
b :=
(^) inv := multiply(
(^) inverse(multiply(transpose(A),A)), transpose(A));
inv :=
(^) c := multiply(inv,b);
c :=
(^) r := sqrt(c[3,1]+c[1,1]ˆ2+c[2,1]ˆ2);
r := 2. 725157057
The equation of the circle of best fit is therefore
(^) (x-c[1,1])ˆ2+(y-c[2,1])ˆ2 = rˆ2;
(x − .5754905350)
2
2 = 7. 426480985
