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Exam (elaborations)

MAT3701 EXAM PACK 2025 {DETAILED QUESTIONS AND ANSWERS}

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MAT3701 EXAM PACK 2025 {DETAILED QUESTIONS AND ANSWERS}

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lOMoAR cPSD| 51600623




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QUESTION 1

Given that
B= 5 3t + 2t2; 2 + t2; 1 + 2t; 2t + 7t2; 2 + t + 10t2

is a generating set for P2 (C); find a basis for P2 (C) from among the vectors of B:

SOLUTION

Write the coefficients of the polynomials in B as the columns of a matrix and reduce to echelon
form:

5 2 1 0 23 1 4 1 14 18 R 2R3
3 0 2 2 1 1 1 2 9 11 R + R3
24 2 1 0 7 10 5 ! 421 22 1 0 7 10 3 5 21


! 24 0 3 1 5 7 35 R3 + R 2 ! 24 1 1 2 9 11 35 3R2
1 1 2 9 11 0 3 1 5 7 R1
0 3 4 25 32 R + 2R 0 0 5 20 25 R + R1

Since the leading coefficients occur in columns 1; 2 and 3; it follows that the first three polynomials
in B is a basis for P2 (C); namely,

:
QUESTION 2

Let f0 (x); f1 (x); f2 (x) denote the Lagrange polynomials over R associated with 0; 1 and 2;
respectively.

(a) Calculate f0 (x); f1 (x) and f2 (x):

(b) Use the Lagrange interpolation formula to express, 1; x; and x2 as linear combinations of f0
(x); f1 (x); and f2 (x):

(c) Without any further calculations, explain why

= ff0; f1; f2g

is a basis for P2 (R):

(d) Let
= 1; x; x2

and write down the change of coordinate matrix Q which changes coordinates into –
coordinates.

(e) Use Q to express
p(x) = 1 + x + x2




2


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MAT3701/201
as a linear combination of :

(f) Check your answer in (e) by using the Lagrange interpolation formula to express p(x) as a
linear combination of :


SOLUTION


(a)




(b)

1 = 1 f0 (x) + 1 f1 (x) + 1 f2 (x)
x = 0 f0 (x) + 1 f1 (x) + 2 f2 (x)
x2 = 0 f0 (x) + 1 f1 (x) + 4 f2 (x)
(c) forIt followsP2 (R). from (b) that generates P2 (R), and since j j = 3 =
dim(P2 (R)); it also is a basis


(d) It follows from (b) that 24 1 0 0 35
Q= 1 1 1 :
1 2 4

(e) 24 1 0 0 3524 1 35 24 1 35
[p(x)] = Q[p(x)] = 1 1 1 1 = 3 ;
1 2 4 1 7
so
p(x) = 1 f0 (x) + 3 f1 (x) + 7 f2 (x):

(f) According to the Lagrange interpolation formula,

p(x) = p(0) f0 (x) + p(1)f1 (x) + p(2)f2 (x) =
1 f0 (x) + 3 f1 (x) + 7 f2 (x)



QUESTION 3

Let and let T : M2 2 (C) ! M2 2 (C) be defined by T (X) = XA:




3

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(a) Show that T is a linear transformation over C:

(b) Find a basis for R(T):

(c) Find a basis for N (T):

(d) Determine whether or not M2 2 (C) = R(T) N (T):

SOLUTION

(a) For X; Y 2 M2 2 (C) and z 2 C;

T (X + Y ) = (X + Y )A
= XA + Y A
= T (X) + T (Y )
and
T (zX) = (zX)A
= z (XA)
= zT (X):
Thus, T is a linear transformation since it satisfies both conditions for a linear transformation.
(b) Since




is a basis for M2 2 (C);
R(T) span fT (A11); T (A12); T (A21); T (A22)g span
=
fA11A; A12A; A21A; A22Ag
=
1 i i 1 0 0
span ; ; ; 0 0
=
0 0 0 0 1 i i 1
= 1 i 0 0
span ;
0 0 1 i
since
and :

Thus,
1 i 0 0
= ;
0 0 1 i

is a basis for R(T) since it is clearly linearly independent.

(c) Let where z1;z2;z3; and z4 are complex numbers.




4


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