QBUS1040 Final Exam Semester 1, 2022
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QBUS1040 Final Exam
Semester 1, 2022
Questions 1–4 involve short computations. You only need to provide the answer (number or vector); no working is needed.
1. (5 points) Let
A = ] , x = 「(l)3 , y = [ 1] .
2. (5 points) Let
A = ] , B = 「(l) 2 .
3. (5 points) Let
R = 「(l)2 1 1 , b = 「(l) .
4. (5 points) Suppose that a particular computer takes 3 seconds to perform the Gram-Schmidt algorithm on 20 500-vectors (i.e., 20 vectors each of length 500). Estimate how long it would take for the same computer to solve a least squares problem
minimize ∥Ax − b∥2
where A is a 1000 × 300 matrix.
Questions 5– 10 involve longer responses. You need to provide justification/working for all of your answers.
5. (10 points) Let x,y be two n-vectors, and α,β be two scalars such that α + β = 1. Define µ = αx + βy . Show that
α(x − µ)(x − µ)T + β(y − µ)(y − µ)T = αxxT + βyyT − µµT .
A = ] .
Determine whether the rows of A are linearly independent or not.
7. (10 points) Let A be an m × n matrix with linearly independent rows. Let B be an n × n orthogonal matrix. Does the pseudoinverse of AB exist? If so, give an expression for the pseudoinverse of AB; if not, provide a counterexample of A and B where the pseudoinverse does not exist.
Assume that w1 , . . . ,wn > 0. Consider the following variant of least squares:
n
minimize 工 wi (bi − ai(T)x)2 .
i=1
Show that this can be rewritten as an ordinary least squares problem, and give a formula for the solution in terms of A, b and W .
9. (10 points) Consider a small least squares problem
minimize ∥Ax − b∥2
where
lb1 」
A = [1 a2 a3 a4 ] , b =
minimize ∥x − b∥2 ,
where
= [1 a2 a3 4] ,
and 4 is the demeaned version of a4 , i.e., 4 = a4 −(1T a4 )1/n. Provide a formula for the optimal solution = (1 , 2 , 3 , 4 ) of this modified problem in terms of the optimal x* for the original problem.
Hint: there is an easy way to deduce what is without computing any pseudoinverses.
10. (10 points) Compute the solution and optimal value of the following optimization problem:
minimize (x1 − 1)2 + (x2 − 2)2 + (x3 − 3)2
subject to 4x1 − x2 + x3 = 1.
Show your working.
2022-11-11