EXAMINATION QUESTION PAPER - Written examination
GRA 60353
Mathematics
Department of Economics
Start date: 27.11.2019 Time 09.00
Finish date: 27.11.2019 Time 12.00
Weight: 80% of GRA 6035
Total no. of pages: 2 incl. front page
Answer sheets: Squares
Examination support BI-approved exam calculator. Simple calculator.
materials permitted: Bilingual dictionary.
Exam Final exam in GRA 6035 Mathematics
Date November 27th, 2019 at 0900 - 1200
This exam consists of 12+1 problems (one additional problem is for extra credits, and can be skipped).
Each problem has a maximal score of 6p, and 72p (12 solved problems) is marked as 100% score.
You must give reasons for your answers. Precision and clarity will be emphasized when
evaluating your answers.
Question 1.
We consider the matrix A, and the column vectors v1 , v2 , v3 , v4 of A, given by
2 1 5 9 2 1 5 9
−1 1 2 −3 −1 1 2 −3
A= 3 0 1 10 , v1 = 3 , v2 = 0 , v3 = 1 , v4 = 10
0 3 0 −6 0 3 0 −6
(a) (6p) Compute the rank of A.
(b) (6p) Find dim Null(A) and a base for Null(A).
(c) (6p) If possible, express v4 as a linear expression of the vectors {v1 , v2 , v3 }.
Question 2.
We consider the matrix A and the vector v given by
4 0 6 −3
A = −1 3 0 , v = −1
1 1 2 2
(a) (6p) Show that v is an eigenvector of A.
(b) (6p) Find the eigenvalues of A.
(c) (6p) Determine whether A is diagonalizable.
Question 3.
(a) (6p) Solve the differential equation y 0 − 4y = 10e−t .
(b) (6p) Solve differential equation 2t + 2ty 2 + (2y + 2yt2 )y 0 = 0.
(c) (6p) Solve the linear system of differential equations:
0
y1 4 0 6 y1
y20 = −1 3 0 · y2
y30 1 1 2 y3
Question 4.
We consider the Lagrange problem given by
min f (x, y, z) = x2 + y 2 + z 2 − xy + xz − yz subject to x + y + z = 11
(a) (6p) Determine whether f is convex or concave.
(b) (6p) Solve the Lagrange problem, and find the minimum value.
(c) (6p) Use the envelope theorem to estimate the minimum value of the Lagrange problem
min f (x, y, z) = x2 + y 2 + z 2 − xy + xz − yz subject to x + y + z = 10
Question 5.
Extra credit (6p) Find the particular solution of the system of difference equations that satisfies
the given initial condition:
4 0 6 1
yt+1 = −1 3 0 · yt , y0 = 1
1 1 2 1
1