This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Kuwait University Department of Mathematics and Computer Science
Math. 111, Linear Algebra Date: 25/5/2000 Final Exam Time Allowed: 2 Hrs.
M Answer all questions. Calculators and Mobile Phones are not allowed.
1. [14 pts‘] Answer each of the following as True or False (justify your answer): a. The angle between the two Vectors X 2 (1,0,0,0) and Y 2 (1,0. l, 1) is b. If X1, X2 and X3 are solutions of AX : B, then EEXI + ng — X3 is
a solution to AX = O. c. L: R3 —v R2, given by L(a:,y, z) = (:5 + y.y2), is a. linear transformation. d. The set of all 2 x 2 singular matrices is a. subspace of szg (the space of
all 2 x 2 matrices). e. If A is a 3 X 3 skewsymmetric matrix, then iAI = 0‘
 f. If A and B are n x n matrices such that A = BT, then I2A‘lBi 2 2.
g. If A and B are similar n x n matrices, then Al = 2. [5 pts.] Let X1 :: (1,0,—2), X2 = (l, —1,0) and X3 = (0,2,4). (a) Show that (1, 5, —4) is in spam {X1,X2,X3}.
(b) Let L : R3 —» P2 be a linear transformation such that L(X1) = t2 + 3,
L(X2) = t2 + z — 2, LfXa) :z+4. Find L(1,5, —4) 10—123 I 2 1 —2 2 1
3 [5 pts.iLetA»— 0 1 H5 8 U
1 1 —1 4. 3 (3.} Find a basis for the null space of A. (b) Find the rank of A.
4. {6 pts.] Let L : P2 + P2 be a linear transformation deﬁned by
Lfot2 + bt + c} z (b  a)t2 + (a — cjt.
(a) Is it? +3t — 2 in the range of L ?
(b) Ist2+t+1 in the kernel OH)?
(0) Find a basis for Ker L.
(d) Is L onetoone ?
(a) (e) Is L onto?. Justify your answers in ail parts of this question. 2 3 0
5.[6pts.]LetA= 10 0
5 5 ——1 (a) Find the eigenvalues of A. I
(b) Find, ifpossible, a. nonsingular matrix P and a diagonal matrix D such that D = P“1AP.
6. {4 pts.] Let A and B be 71 x 7; matrices. Prove that (a) If A and B are nonsinguiar, then AB is nonsingular and (AB)‘1 = B'IA‘I.
(b) If A is an upper triangular matrix, then the eigenvalues of A are the elements on the main diagonal of A.
m\
9 O . Math 111, Linear Algebra 25 May, 2000, Final Exam. Solutions
W
Al. a. F, (0.5 pi“), c059 : ﬁ implies 0 yé ’3—', (1.5 pt).
b. F, (0.5 pt.).A(%X1+§Xi —X3) = B, (1.5 m).
c. F, (0.5 1313.), LOG + X2) 96 L(Xi) + L(X2), (1.5 pt). 3
d. F, (0.5 pt.), Let W be the subset in the question. Then A : 0 0 L B :—= [i 2 0 I " 0 1 0 0 i u
l
W implies A + BI = 2 v7]: 0. Hence A + B is not in W, (1.5 pt).
3. T, (0.5 m), A 2 —AT. (0.5 pt), implies (A; :  sATl implies ;A = —A, (0.5 m). Hence IA] : 0, (0.5 pt.).
f. F, (0.5 pt.), I2A“1B : 2“[A‘1A = 2“. (1.5 pt).
g. T, (0.5 pt), B = P‘JAP, (0.5pt.) implies A = BI, (1 pt). A2. (a) Let c1(1,0,—2)+02(1,ﬁ1,0)+c;~,(0,2,4)=(1,5,—4), (0.5 pt.) 1 1 0 Cl 1
Form and solve the equivalent system 0 ‘—1 2 C2 = 5 , (2 pt.).
*3 U 4 C3 —4 The solution of the system is c1 = 4, 02 =
span, (0.5 pt.) (‘0) L(1,5, —4) : L(1,0, v—2) + L(1, 1,0) + L(0,2,4),(1pt.).
L(i,5,—4) : 4(i2+3) u3(t2+t—2)+t+4=t2~—2t+22, (1 p0). —3,ca : 1. Hence the given vector lies in the 1 O —l 2 3 1 0 0 0 2 0m 3 i .0. s .5 35?: Hit)
0 —1 0 2 5 0 0 0 0 0
The reduced system 021+ 2m5 = 0, :02 — 2:04 H 5035 = 0,033 — 234 ~ 035 = 0, set 55 = t and
0:4 :T,(1pt.). '
—2 0
5 2
Hence, X :t 1 +1“ 2 ,(1 pt.)
0 1
1 O
—2 0
5 2
The set 1 , 2 is a basis for the null space of A, (1 pt).
0 1
1 0 (b) Rank of A = 5 — nullity of A m 5 i 2 2 3, (1pt.) (or any other way).
A4. (a) No, (0.5 pt), since t2 + 3t — 2 = (b i a)i2 + (a — c)t implies i2 = 0, (0.5 pt).
(b) Yes, (0.5 pt), since L(t2+t+ 1) 2 O, (0.5 pt). (0) To find a. basis for kerL, we seek the solution of (b — a):2 + (a — c)t : O which is
b: a,c = a, (1 pt). Hence kerL = {a(t2 + t+ 1) : a E R}. Hence, {t2 + t + l} is a basis for kerL, (ipt).
(cl) No, (0.5pt), since liter}: yé {0,3,} (0.5pt).
(e) No, (0.5pt), since dim(rangeL) % 3 (0.5pt). OO 99 ...
View
Full Document
 Spring '10
 Linear
 Linear Algebra, Determinant, Matrices, Diagonal matrix

Click to edit the document details