UNIVERSITY OF BAHRAIN
COLLEGE OF SCIENCE
DEPARTMENT OF MATHEMATICS
SECOND SEMESTER 2023 -2024
TEST 2 (B) - Key
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MATHS. 211 DATE: 8 - 5 - 2024
TIME: 11:00 PM - 12:00 PM
EXAM INSTRUCTIONS:
1) Number of pages is 5 including the cover page.
2) Show all your work, except for question one.
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NAME : I.D. NUMBER: SECTION: SERIAL NUMBER:
Q.1 Q.2 Q.3 Q.4 OUT OF
16 10 6 8 40
GOOD LUCK
1
QUESTION 1 (2 marks for each question)
Fill in the blanks to make a correct statement:
1) Suppose that (p)B = (1 , 3 , -3) where B = {1 + 𝑥, 1 − 𝑥 2 , 𝑥 − 𝑥 2 } is a basis of P2 .
then
p = 4 - 2x
2) Suppose that S = {u1 , u2, u3, u4, u5} is linearly independent in Rn . Is n 5 or n < 5?
n5
3) Let S = {sin2(x), 3cos2(x)}. Is S linearly dependent or linearly independent?
linearly independent
1 −1 2 0
4) If 𝐴′ = [0 1 4 3] is a row-echelon form matrix of A, then a basis of the row space of A is
0 0 0 0
{ (1, -1, 2, 0) , (0, 1, 4, 3) }
5) If A is a matrix of size 5 6, then the largest possible value of rank(A) is equal to
5
1 1 1
6) If 𝐴 = [1 2 −1], then the rank of A is equal to
2 3 0
2
7) If A is a matrix of size 6 5 and nullity(AT) = 3, then
nullity(A) = 2
8) If { 1 +2 X + X2 , 3 + X , a + X} is a basis of P2 , then
a 3
2
QUESTION 2 (4 + 4 + 2 marks) (Show your work)
a) Show that B = {u , v , w} is a basis of R3 , where
u = (2, 1, 0) , v = (1, 0, 1) , w = (0, -1, 1)
Solution: Since B has 3 elements and dim R3 = 3, it is sufficient to prove that B is linearly
independent.
au + bv + cw = 0 a(2, 1, 0) + b(1, 0, 1)+ c(0, -1, 1) = (0, 0, 0) 2
2a + b = 0 1
a -c = 0
b+c = 0
a= b =c = 0 1
b) Find the coordinates of h = (1, 2, 3) relative to B.
Solution:
au + bv + cw = h a(2, 1, 0) + b(1, 0, 1)+ c(0, -1, 1) = (1, 2, 3) 1
2a + b = 1
a -c = 2
b+c = 3
a = -4, b = 9 , c = -6 2
(h)B = (-4 , 9 , -6) 1
c) Without any calculations, determine whether {u, v, w, h} is linearly independent.
Solution:
The set {p1, p2, p3, p} is linearly dependent because it consists of 4 vectors in the vector space R3 with
dimension 3. 2
3
QUESTION 3 (6 marks)
Find the dimension of V = Span{f1 , f2 , f3}, where f1 = x , f2 = = 𝑒 𝑥 , f3 = 𝑒 −𝑥
Solution:
The set {f1 , f2 , f3} is a spanning set of V. To prove that it a basis, it suffices to show that it is linearly
independent, equivalently to prove that 𝑊(𝑥) is not the zero function. 1
We have
𝑥 𝑒𝑥 𝑒 −𝑥
𝑊(𝑥) = |1 𝑒𝑥 −𝑒 −𝑥 | 1
0 𝑒𝑥 𝑒 −𝑥
𝑥 1 1
= 𝑒 𝑒𝑥 −𝑥
|1 1 −1| 1
0 1 1
So
𝑥 0 0
𝑊(𝑥) = |1 1 −1| = 2x 1
0 1 1
That is,
𝑊(𝑥) = 2𝑥 ≠ zero function. 1
Hence, dimV = 3 1
4
QUESTION 4 (4 + 4 marks)
1 2 −1 2 1
Let 𝐴′ = [ 0 0 1 1 1 ] be a row-echelon form of A.
0 0 0 1 1
a) Find a basis of the nullspace of A.
Solution:
Let us transform the matrix A' to a reduced row-echelon form
𝟏 2 −1 0 −1 𝟏 2 0 0 −1
𝐴′ ~ [0 𝟎 𝟏 0 0 ] ~ [ 0 𝟎 𝟏 0 0 ] = 𝐴′′ 1
0 0 0 𝟏 1 0 0 0 𝟏 1
𝟏 2 0 0 −1
[0 𝟎 𝟏 0 0 ] 𝑋 = 𝑂 x1 + 2 x2 – x5 = 0 ; x3 = 0 ; x4 + x5 = 0 1
0 0 0 𝟏 1
Let x2 = r , x5 = s be free variables, then the leading variables are x1 = -2r + s, x3 = 0 and x4 = -s.
Thus N(A) = { (-2r + s, r , 0, -s, s) : r, s are real numbers} = Span{(-2, 1, 0, 0, 0), (1, 0, 0, -1, 1) }. 1
Since dimN(A) = Nullity(A) = 5 – 3 = 2, then {(-2, 1, 0, 0, 0), (1, 0, 0, -1, 1) } is a basis of N(A). 1
b) If c1 , c2 , c3 , c4 , c5 are the column vectors of A, write c2 and c5 as linear combination of c1 , c3 , c4.
Solution:
Let c1‘, c2‘, c3‘, c4’, c5’ be the column vectors of A''.
Then c2’ = 2 c1’ + 0 c3‘ + 0 c4’ and c5’ = -1 c1’ + 0 c3‘ + 1 c4’. 2
It follows that c2 = 2c1 + 0c3 + 0c4 and c5 = -1 c1 + 0 c3 + 1 c4 . 2
