FIRS T SEM E

STER M. Tee
H, INiERNA
t., L EXAMINATION
'"OVe n,b
er 2023

Depa rt
ment of El 8
ctrlca1 E
nglne ering
contr ol Syste ms Guid ance N .
• av,gation
Powe r Ele
ctronics & p and Control p
ower Syste~s ower Systems,

22 1TEEtoo
LINEAR ALG
EBRAA
ND LINEAR SYSTEMS

Time: 2 hours
Max . Marks: 40
Part A
LAns wer all guest ionsj

Q. Marks co KL
no

1 Find a basis for the null space of

l 4 1 K3

l
2 2 -1 0 1
A= -1-1 2 - 3 1
1 1-2 0-1
0 0 1 1 1

3 2
d by 4 2 K3
2 Check wheth er the following transformation T: R to R define

T[n~ rx ~ 2:.z: Jz]is linear ornot. Also, find the matrix representation.

th e K3
Find a matrix P so that by performing P· AP will diagonalize
1 2
4
J
o o - 2]
A matr ix A= 1 2 1
[1 0 3
 1

vabilitY of th e sy
st em de sc rib ed b
·1·rv an d o bser
Ch ec k for controlla 11 4
4 b ; y

0 l) x

fl lr t~
~n sw er any four
qu es tio ns }

a.no. Marks
C()
Find the rank, nulli
ty, row space, and 6
5 co lu m n sp ac e of
-1 2 0 4 5 -
31
A==
l 3 -7 2 0 1
z -5 2 4 6 41
4 -9 z - 4 - 4
7

For the following trans

6 formation T: R3 10 R3 6
defined by
2 'f·
'
x] [-2x + 3zl
I
T( : ) = ; : ;

Find the matrix T w.r.

t the ordered basis

a) B =([I I 0], [IO

O] , [O O l]} for R3
b)E=([I 00], [11 O]
, [00 I]} fo rR J
 3 K4

, tmi q1 K' solu1io11 b. 110 s o 1llt1.on c

•1 . inftnite solut ·ton

by . ._
ror t11c sys tem des cribed x - Ax + Bu and y ::::: 6 4 K3
~
-
w1I ere A -.

" === f-01 -

osl•13= [1I] ' c· r' 21, X(O) ::: [il
===
,,
l re
fin d the solution. and the tota sponse wh en the
input is a step sig nal

sh ow that 6 1 K4
o vector 1 .
a) A subset containing a zer a one is a subspace
or not.
b) A set of rea.l numbers ov
er a field of complex
not.
numbers IS a vector space or
c) A set of co~plex numb
ers over a field of real
or not.
numbers IS a vector space

Definition

r spaces.
Apply the concepts of vecto
on s in linear systems
Apply linear transformati

Solve LTI and LTV Systems

vers
An alyse linear systems an
d design controllers and obser
 FIRST SEMESTER M. TECH. INTERNAL EXAMINATION

November 2023
De

1 Systems, Guidance Navigation and Control, Power Systems, Power

contro Electronics & Power S stems
221TEEIO0
LINEAR ALGEBRA AND LINEAR SYSTEMS

Max. Marks: 40
r;me,. 2 hours
Part A
ANSWER KEY Answer all uestions

Ma rks co KL

l 1
RREF= [~
1
0
0
0
0
1
0
0
0
0
1
0
1
1
0
0
~l 2 marks
4 1 K3

Basis for the null space are v1 -

1
_ -1
0 v2=
0
-1
~ l l-l~
2 marks

K3
T (x2) check an d ven·ry . I mark 4 2
( I +x2) = T(x I)+
2 Property! -T x k dverify. l mark
T (C (xi)= CT(xl) chec an
Property2 -

Matrix wrt stan dard basis

=

T (v2) T (v3)]
Tu= [T (v1)

2 MARKS

4 2

3
 ( \ - 1) ( ,\ _ ) 2 I &
2 () \

l
cigcn va lues I ,11:.1 rk
.. 2
\ I.
I• l

.
- (J1 '
I
v,-m ,\ = 1 1'3 =
l ~1 \\
I
l·.igc n vc c.:·t.rm,- 2 \
ma rk I

p - [ - I
(J

I
()
I
CJ -:1
Writi n g I' I mark

1 l 2+ J • 4mark."

_ _ ·i
-- -- 4 5
4 K3
Steps and showing Rank(Qc)=2 1,5 mark

Not controllable - 0.5 mark

Steps and showing Rank(Qo)=2 1.5 mark

s
Not observable - 0.5 mark

2 +2 = 4marks

Part B

(Answer any four que stio ns)

.

Q.no .
Rank =2 3 marks
Marks co Kl
5
6 1 K3
Nullity= 4 1 marks

Dim (Rowspace) =2 1 marks

Dim (Rowspace) =2
1 marks

I -
 •' I' (I I I OJ. II OO j. [00 II } fn rR ' 6 2 KJ
ti
, \) i11
Steps nnd nns\\ cr upto ti nt.lin g 1 (v i ) l'( , 2) I'( I
standnrd bnsis - I .5 lllnrk

~teps nnd answer to rewrite [(T( v I ))11 ( 1'( v2))1

1 ( f(" J ))11 I \
11, tcnn s o f B = 1.5 marl,.
I
E =(( I O0). [I I OJ. [O o 11} for R3
t,)

) in
Steps und answer upto findin g T(v l) T(v2 ) T(v3
I standard basis - t .5 lllark
(T(v3))E]
Steps and answer l o rewrite [(T( v I ))E (T(v 2))E \
in tenn s of E= 1.5 mark

·-
.... K4
6 3
7 Row reductio n - 3 marks
r combinations of b I .
I Unique solution - not possible for any linea
I
r
b2. b3 ,b4 1 marks

BI +b2 +b3+b4 =0 infinite no of solutions 1 murks

BI +b2+b3+b4 not equal to 0 inconsistent 1 murks

6 4 K3
8 Finding the solution for x(t) 3 mark
a step sign al 3 mark
Finding the total response when the input is
6 1 K4
9 Sho w that
tor alone is a
a) Stat ing subset containing a zero vec
subspace)0.5 mark)
cation
( vec tor addition 1 mark and scalar multipli
O.Smark)
will not result in a real
b) Sho win g scalar multipli cation
space 1 ma rk
num ber 1 mar k and stating not a vec tor \
'· ks for 8 axioms I
c) Sho win g a vec tor space 0.25 mar

co Definition

co 1 Apply the concep ts of vector spaces.

sys te ms
co 2 Apply lin ear tran sfor m atio ns in line a r

co 4 olvc I.Tl and LTV syste ms

 FIRST IIDMJDI

D1
0
aa1TJDB1

Timar 2 hr■

Max, Markis 40

Marlul CO Kl,
4 COl K:J

4 CO2 K2

4 003 K2

4 004 K3

0 001 1(2
 ' /
' (, I
:.;
I; I
IV
Marks
6

G. Let Ube· t Ile subspace of JR" geuerated by

l -2, 9)}
2 3 '
{ (1, 3, -2, 2, 3), (1, 4, -3, 4, 2), ( , , -
by
and let W be the subspace generated
2 5 3, 2, 1)}
,l), (1, 5,- 6,6 , 3), ( ' '
{(1 ,3, 0,2
. .
Find a basis and dimension of U + vV
6 CO2 l\3
rna tnx is
sys tern
st f'orma-
7- Analy(se the cliagonalisabilty of a sy ern whose
. t l1e trans
obtaLn ~
A~ i
- 4 -6 -7 )
~ i
if diagonalisable,
6 C03 I\2
erations
tion matrix . . tar" row op
. en J
8. Solve the given system usmg elem
-3x1 + 2x2 + 4xs == 12

X1 - 2X3 == -4
-3
2x 1 - 3x2 + 4x3 --
6 C04 K2
.b I by x(t) = A(t)x(t)
st clescn ec · ·

n
9. Solve the linear time varying sy em
·where
A( t)- u

Given, x 0 = ( X10)
X20

Course Outcomes:
COI: Explain the concepts of vec
tor spaces .
in line ar systems.
CO2: Apply linear transformations
ns and inte rpre t their results.
C03: Solve systems of linear equatio
C04: Solve LTI and LTV Systems.
 •. l II l· . hil l !

1 A n s,ver a ll q u estio n s
I. P rov(' that l-F 1 U H ·2 is a subspace if a nd only if W ;; \\''2 m it·_~ :t
wh e re l\ '1 a nd lF2 are subspace. ·
( COl ( 0 .5 marK

•) Pro\·\., t hat e\·e11 when B 1 and B2 are two differem bas.5 :or · he 5ame
\·error :--pace t he number of elements in both basis will ~e · he ... =~"'
( C Ol ) (0 . 25 ma:K )

J. For t he giYen set of simulta neous equation

x 1 + '.b:2 + 3x3 , 5I4 = b1

2x 1 + -1x2 + 8x3 + 12x .i = b-i
3x 1 + 6.r2 + 7x3 - 13x4 = b3
,L F ind the condi tion o n b1 • 02 and b-J so that the ~t ui equatJ ' J

ha,·e no solution and infini te solutions.

b. Find the basis fo r col umnspace( ..\) and nulbpace\ A

C03 )1 l mark }

A E nm xn . Prove tli ut N ( A) is a s ubspace

O~ -1. Let
tC0l )(0.25 mark

,.---...------:-
-----
1- ,-;-----:=-;t:::_~o:,f;-;v:e~c;to;:;-;:-r s paces. - -
COl : .-\pp y t ie conce p ~ . d interpret their results.
C03 : Solve systems of linear equations _a_ n _ __:_:.~ - - - - -
 :221 TEE J . Quiz 2
OO Line ar Algebra and Liuca r Systc 111s

T11rn• 1 110 111

ToLal 2 mark.,

1 Answ er a ll ques tions

1 F ind u no n-singular mat.rix P such t.hat p - 1 AP is diagonal. Given

A=[ ~ ~ ~]
-8 - 14 - 7

(C02 ,K2 )( 1 mark )

2. Find a . eigen space and b. dimensi on of eigen space assocai ted with each
cigen va lues of

0
Also. state with reaso11 whether the ubove matrix is cliagontllisahll'
(C02 ,K4 )(l mark)

CO2 : Apply li11car transfo rmatio ns in liucm systems. ]

• t L .•
 Quiz 3
22 1TE El 00 Lin ear Alg ebr a and Lin ear Sy ste ms

Tim e: 1 ho ur Totnl: 2 marks

1 An sw er all qu est ion s

, x2, X3) =
T : R ~ R be a linea r transform ation given by T(x1for T w.r. t
3 2
1. Let
tatio n
l- 2x 1 + 3x3, X1 + 2x2 - x3 J. Find the matr ix repr esen
the ord ered basi s B = {[1, -3, 2] , [-4, 13,- 3]
, [2, - 3,20]} for R3 and
2
C = {(- 2, 1), (5,3)} for R
(C0 2)(1 mar k ,
K3)
given by T(x 1, x 2 ,x3 , x 3) =
2. Let T: R 4 ~ R be a linea r trans form ation
3

lx 1 - x 2 + x 3 + x 4 , x1 + 2x3 - X 4 , x, + x2 - 3x4 + 3x3J

. Find the mat rix
confirm
Q,1':> repr esentation of T w.r. t the stan dard basis. Also find K er(T ) and
the mapping is 1-1 or not
( C02 )( 1 mar k ,
K3 )

I CO2 : I Apply linear tran sformations in li nearsyste ms. I

: .:t d '· ~ r.: :

- l ,,,: I_ :.. ] • :.. , <~ l '· .:.
 Quiz 4
22 1T EE 100 L. in ear Al bra a nd Lin ear Systems
ge

T im e: I ho ur To ta l: 2 ,narks

io ns
1 A n sw e r a ll q u e st
r m desc ri bed by
I. Evuluu t.r t he sol u1 io n ,o r l 1l C syste
•

(0 )x 2( 0) J' = ll OI'
(1)
+ x2 ,y = x, /o r !x 1
.i:1 = .c, . :i-2 = x,
)
i11 pu1 re sp on se (C 04 ,K3 )( l m a rk
A lso liu d I he ze ro
described by
ar e fe ed back co nt ro ller for lh e sy st em
'J. . Dcsiv;n II full st (2)
tLling
syijt cm exh ib its o response with o se
su t hu t t hi: resulti
ng closed lo op o percentage
l I sec (w ith 2% tolerance) and
eq ua lO
ti me less than or
k)
ov ers ho ot < 16 % {C05 ,K 3) ( 1 m ar

Systems.
Solve LT I and LTV 1., uml design t·ontroller::, an d obSt•rvcr
s.
C 04 : ea r ;._\-slen
Analyse lin
C OS :

• - L tl · •- .. - .-
I
--
'
 Qui:l 5
221 TEE l00 Line ar Algebra and Linear Systems

Time: l hour
ToLal · 2 mar.r.s

1 Ans wer all questio ns

1. Evalu ate the observ ability gramm ian for the system when t he system
out-
put is meas ured for 2 second s and hence comm ent on the observability o:
the system

±1 = - 0.5x1 + u,
0 (C05 ,K4)( 1 mark
2. Desig n a full order observer for the system descri bed by

~,i 'J :i:1 = -X3,

V so t hat the obser ver poles are at -10, - 10. -10
(C05, K3){1 mark )

A nulyse linear system s and design contro llers and observers.

COS :
 Quiz 6
221 T EEl0 0 L· A .
1near lgebra and Linear System s

Time: l hour Total: 2 marks

1 Answ er a ll quest ions

l. Check the controlla bility and stabillzabilty of the system described by

O,fS (C05,K 2)(1 mark)

2. Check the detectab ility and observability of the system described by

(2)
±1 =:t1, ±2=2x2 +2:r3+u , :r3=:t3 + u, y=x1

(C05,K2 }(1 mark)

I
Ana yse
r r systems and design controllers and observers.
C OS:

l._ ..,_) t. -~ .·d I

. L .-
' -" -' .:.t ! - !....! - - .. t