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Material Type: Exam; Class: Applied Linear Algebra; Subject: Mathematics; University: Arizona State University - Tempe; Term: Summer II 2008;
Typology: Exams
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2.a.
x 1 x 2 x 3
=
b.
.
2 z = 6 yields z = 3. Back-substitution: −y + z = − 2 yields y = −(− 3 − 2) = 5, and −x + y = 3 yields x = −(−5 + 3) = 2.
c. E 1 =
.^ d.^ E^2 =
. E^123 =^ E^2 E^1 =
.
e. L = E− 1 1 E 2 − 1 =
=
. U^ =
(from^ b.).
10 · cos γ yields γ ≈ 31. 948 ◦. b. Calculate the normal vector N = (2, 3 , 1) as a nonzero solution of M N = 0 where M is the matrix with rows CA and CB (alternatively, calculate the cross product of CA and CB). The desired distance is the length of the projection of A onto N , i.e. d = A · N/‖N ‖ ≈ 2 .1381. c. Write AB = B − A = (− 6 , 4 , 0) and AC = −CA. The closest point PB is the projection of AB onto AC, translated by A. PB = A + ABAC··ACAC AC = 125 (2, 0 , 1) = (4. 8 , 0 , 2 .4).
(†) (^) U is the upper triangular matrix of the LU-decomposition of A.
(vi.) A = (1 0), B = (0 2)T^ , C = (0 3)T^. (vii.)
( 1 1 0 1
( 1 − 1 0 1
) is not lower triangular. (viii.) The only nonzero entry of L^2 is in its bottom left corner, but the third column of L is zero. Hence all columns of L^3 = L · L^2 are zero. ♣