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(MATH113)[2009](f)midterm~klkpang^L1_10445.pdf
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MATH 113 Introduction to Linear Algebra Date: 14-Oct-2009, 7:00pm C 8:00pm. Midterm Test
Name: Student ID: Answer all questions. Total marks: 100

Qn. 1 (20) 2 (40) 3 (30) 4 (10) Total (100)


. Show your working steps to receive credits.

Qn. 1 (20 marks) Consider the following linear system:
. x1 +2x3 +3x4 = . x1 . x2 . x3 +4x4 =2
.
x1 + x2 + kx3 +2x4 =2 Find all the values of k, . such that the above system has a solution. Solution: We perform elementary row operations on the augmented matrix:
. . . . . .
1 0 2 3 . 1 0 2 3 . 1 0 2 3 .
. 1 .1 .1 4 2 . . . 0 .1 .3 1 2 . . . . . 0 .1 .3 1 2 . . . .
1 1 k 2 2 0 1 k . 2 .1 2 . . 0 0 k . 5 0 4 . 2.

For the system to be consistent (has a solution), we need to have: k .. any no. or k =5,
=5,. =2.
Qn. 2 (40 marks) Find the inverse of the following matrix A:

..
0111
. 1011 .
..
A = .
..
1101
1110
Solution: We perform row operations on [ A | I4 ]:
.
.
0111 1000
. 1011 0100 .
.
.
1101 0010
1110 0001
.
.
3333
1111 R1+R2+R3+R4.R1 . 01
11 0100 .
.
.
.
.
1100 0010
1110 0001
.
.
1111 1/31/31/31/3
R3/3.R3
. 1011 0100 .
.
.
.
. 1100
0010 . 1110
0001
.
.
1111 1/31/31/31/3
R1.R2.R2,R1.R3.R3,R1.R4.R4 .
.
0100 1/3 .2/31/31/3
.
.
.
. 0010
1/31/3 .2/31/3 . 0001
1/31/31/3 .2/3
.
.
1000 .2/31/31/31/3
.
R1.R2.R3.R4.R1 .
0100 1/3 .2/31/31/3
.
.
.
. 0010
1/31/3 .2/31/3 . 0001
1/31/31/3 .2/3
Therefore:
..
111 33 3 3
. 2
1 11
. . 2 .
A.1 . 3 333 .
= .
11 1
. . 2 .
33 33
111
. 2
333 3
Qn. 3 (30 marks) Let T : R2 . R3 be a matrix transformation (or a linear transformation) satisfying the conditions:
.. ..
.. 1 .. .1
11
T = . 1 . ,T = . .1 . .
1 .1
1 .1
(i)
Find the standard matrix A of T .

(ii)
Is T a one-to-one transformation? Why?


(iii) Is T an onto transformation? Why?
Solution:
(i) We use the formula A =[T (e1) T (e2)]. Since it is easy to see that:
.. .... .. ....
1 1111 0 1111
e1 == + , e2 == . ,
0 212 .1 1 212 .1
so we have:
.. ....
.. .. 1 .10
11111 1
T (e1)= T + T = . 1 . + . .1 . = . 0 . ;
21 2 .12 2
1 .10
.. ....
.. .. 1 .11
11111 1
T (e2)= T . T = . 1 . . . .1 . = . 1 . .
21 2 .12 2
1 .11
Hence:
..
01 A =[T (e1) T (e2)] = . 01 . . 01
(ii) Since T (e1)=0 but e1 .
= 0, T is not one-to-one.
(iii) Since the span of the columns of A is
..
1
span{. 1 .}
1
.. ..
11
which is not R3 (for instance, . 0 . is not in the span of . 1 .). We see that T is not onto.
01

..
1
..
0
..
Qn. 4 (10 marks) Is the vector u = a linear combination of the vectors in {v1, v2, v3}? Why?
. 0 . 1
.. ....
100 . 1 .. 1 .. 0 .
.. ....
v1 = , v2 = , v3 = .
.. ....
101
100
Solution: We note that: u = v1 . v2 . v3.
So u is a linear combination of the vectors in {v1, v2, v3}.
END