MA-102
B. Tech. II Sem (2021-2022)
Tutorial sheet-04
1 Find the row-reduced echelon forms and hence find rank of following
matrices:


1 1 1
(i)  1 1 0  .
1 0 0


1 2 3 0
 2 4 3 2 

(ii) 
 3 2 1 3 
6 8 7 5


1 1 1
Solulion:-(i)Given Matrix
A =  1 1 0 ,
1 0 0
R1 → R1 − R2 ,


0 0 1
∼ 1 1 0 
1 0 0
R3 ←→ R1


1 0 0
∼ 1 1 0 
0 0 1
R2 → R2 − R1


1 0 0
∼ 0 1 0 
0 0 1
This is the Row Echelon form of A.
And
Rank of A = 3.
1
(ii) Given matrix

1
 2
B=
 3
6
2
4
2
8
3
3
1
7

0
2 

3 
5
R2 −→ R2 − 2R1
R3 → R3 − 3R1
R4 → R4 − 6R1

1 2
3
 0 0
−3
∼
 0 −4 −8
0 −4 −11

0
2 

3 
5
R2 ←→ R4

1 2
3
0
 0 −4 −11 5
∼
 0 −4 −8 3
0 0
−3 2




R3 → R3 − R2

1 2
3
0
 0 −4 −11 5 

∼
 0 0
3
−2 
0 0
−3
2

R4 −→ R4 + R3

1 2
3
0
 0 −4 −11 5
∼
 0 0
3
−2
0 0
0
0
This is Row Echelon form of B.
And
Rank of B = 3.
2




2(i) . Obtain for what values of λ and µ the equations
x+y+z =6
x + 2y + 3z = 10
x + 2y + λz = µ
have
(a)no solution.
(b) a unique solution
(c) infinitely many solutions.
Solution 2 (i) The Above system of linear equation can be written as.

  

1 1 1
x
6
 1 2 3   y  =  10 
1 2 λ
z
µ
The Augmented matrix can be

1
 1
1
R2 → R2 − R1 ,
written as

1 1 6
2 3 10 
2 λ µ
R3 → R3 − R1


6
1 1
1
 0 1
2
4 
0 1 λ−1 µ−6
R3 −→ R3 − R1


6
1 1
1
 0 1

2
4
0 0 λ − 3 µ − 10
(a) If rank A ̸= rank(A | b) then system of linear equation has no
solution.
⇒ For no solution.
rank(A) ̸= rank(A | b)
If λ = 3, &µ ̸= 10 then
rank(A) = 2,
rank(A | b) = 3.
(b)For unique solution.
Rank(A) = Rank(A | b) = 3.
⇒ λ − 3 ̸= 0 ⇒ λ ̸= 3
3
⇒ system of linear equation has unique solution when λ ̸= 3.
(c). For infinitely many solution.
Rank(A) = Rank(A | b) < 3.
⇒ λ−3=0⇒λ=3
µ − 10 = 0 ⇒ µ = 10.
2(ii). Obtain for what values of λ the equation
x+y−z =1
2x + 3y + λz = 3
x + λy + 3z = 2
have
(a)no solution
Solutions-
(b) a unique solution
(c) infinitely many solutions.

   
1 1 −1
x
1
 2 3
λ  y  =  3 
1 λ
3
z
2
The Augmented matrix

1 1 −1 1
 2 3 λ 3 
1 λ 3 2

R2 → R2 − 2R1 ,
R3 −→ R3 − R1


1
1
−1 1
 0
1
λ+2 1 
0 λ−1
4
1
R3 −→ R3 − (λ − 1)R2


1 1
−1
1
 0 1
λ+2
1 
0 0 (λ + 3)(λ − 2) 2 − λ
(a) For no solution
rank (A) ̸= rank(A | b).
⇒ λ = −3
4
(b)For Unique solution
rank(A) = rank(A | b) = 3.
⇒ λ ̸= −3, λ ̸= 2.
(c) For Infinitely many solutions
rank(A) = rank(A | b) < 3
⇒
λ = 2.
2(iii). In the following system of linear equations
ax1 + x2 + x3 = p
x1 + ax2 + x3 = q
x1 + x2 + ax3 = r
determine all values of a, p, q, r for which the resulting linear system
has
(i) unique solution (ii) infinitely many solutions (iii) no solution.
Solution-
5


a 1 1 p
 1 a 1 q 
1 1 a r
R1 ←→ R2

1 a 1 q
 a 1 1 p 
1 1 a r

R2 → R2 − aR1 ,
R3 → R3 − R1 .


1
a
1
q
 0 1 − a2 1 − a p − aq 
0 1−a a−1 r−q
R2 ←→ R3


1
a
1
q
 0 1−a a−1 r−q 
0 1 − a2 1 − a p − aq
R3 −→ R3 − (1 + a)R2


1
a
1
q
 0 1−a

a−1
r−q
0
0
−(a + 2)(a − 1) (p + q) − r(1 + a)
(1) For no solution.
a = −2, a = 1 and (p + q) − r(1 + a) ̸= 0.
(2) For unique solution
a ̸= −2, a ̸= 1 and (p + q) − r(1 + a) ̸= 0.
(3) For infinitely many solutions
a) = 0.
a = −2, a = 1 and (p + q) − r(1 +
3. Does the system:
x+y+z =1
2x + 2y + z = 3
has a solution for z = 7? Find the general solution of system by Gauss
elimination.
Solutionput z = 7 in above equation we get
x + y = −6
2x + 2y = −4.
6
We will solve by Gauss eliminiation method.
1 1
x
−6
=
.
2 2
y
−4
Augmented matrix
1 1 −6
2 2 −4
1 1 −6
0 0 8
R2 −→ R2 − R1
⇒ x + y = −6
and 0.x + 0.y = 8
⇒ Given system of equation has no solution for z = 7.
4 Show that the rank of matrix AB is less than or equal to rank of A as
well as rank of B, Further prove that rank of AB to equal to rank of
A, if B is invertible.
Solution:- As rank of a matrix Mm×n is the dimension of the range
R(M ) of the matrix M . And range R(M ) is defined as
R(M ) = [y ∈ Rm | y = M x f or some x ∈ Rn ]
So we have
rank(AB) = dim(R(AB))
and
rank(A) = dim(R(A))
In general, If a vector space W is a subset of a vector space V ,
dim(W ) ⩽ dim(V )
Thus, it is sufficient to show that the vector space R(AB) is a subset
of the vector space R(A).
Now, consider any vector y ∈ R(AB), then there exists a vector x ∈ Rn
such that y = (AB)x by the definition of the range.
let z = Bx ∈ Rn
7
then, we have y = A(Bx) = Az
and thus the vector y is in R(A).
Thus R(AB) is a subset of R(A) and we have⇒ rank(AB) = dim(R(AB)) ≤ dim(R(A)) = rank(A)
⇒ rank(AB) ≤ rank(A) .... 1
Similarly we can prove for
rank(AB) ≤ rank(B)
Now the matrix B is nonsingular,i.e. it is invertible. Thus the inverse
matrix B −1 exists. We apply above result (equation(1)) with the
matrices AB and B −1 instead of A and B. Then we have,
rank((AB)B −1 ≤ rank(AB)
Now as
rank(A) = rank (AB)B −1
⇒ rank(A) ⩽ rank(AB)
... 2
from equn (1) & (2)
rank(AB) = rank(A)
5 Suppose that Am×n has rank k. Show that ∃Bm×k , Ck×n such that
rank (A) = rank(B) = k and A = BC.
Solution:
Given that A be an m × n matrix whose rank is k. By the definition
of rank it is dimension of column space. Therefore, there are k
linearly independent columns in A (equivalently, the dimension of the
column space of A is k).
Let {b1 , b2 , . . . , bk } be any basis for the column space of A, let us
place them together as column vectors to form the m × k matrix.


.. ..
..
 . . ··· . 

B=
b1 b2 · · · bk 
.. ..
..
. . ··· .
8
Therefore, every column vector of A is a linear combination of the
columns of C.


..
..
..
. ··· . 
.

To be precise, if A = a1 a2 · · · an 
 is an m × n matrix with aj
..
..
..
.
. ··· .
as the j-th column, then every column of A can be written as linear
combination of {b1 , b2 , . . . , bk }
a1 = c11 b1 + c21 b2 + · · · + ck1 bk
a2 = c12 b1 + c22 b2 + · · · + ck2 bk
..
.
aj = c1j b1 + c2j b2 + · · · + ckj bk
..
.
an = c1n b1 + c2n b2 + · · · + ckn bk
where cij ’s are the scalar coefficients of aj in terms of the basis {b1 , b2 , . . . , bk }.
Now we can write


 c
c1n
.. ..
..
11 c12 · · ·
c21 c22 · · · c2n 
 . . ··· . 


b
b
·
·
·
b
A=

..
.. 
1
2
k

  ...
.
. 
.. ..
.
. . · · · ..
ck1 ck2 · · · ckn
That is A = BC.
It is clear that rank of B is also k because it has k linearly independent
columns.
So now we have matrices Bm×k and Ck×n such that A = BC and
rank(A) = rank(B) = k.
6. Find row reduced Echelon form of the following matrices and hence
find column space, row space and null space of given matrices.




1 2 2
1 2
1 1 1
(a) A1 =
[(b)] A2 =  1 3 4  [(c)] A3 =  1 3 
2 3 2
1 2 3
1 2


1 2 1
 1 3 1 

(d) A4 = 
 1 2 1 
2 5 2
9
Solution:
1 1 1
(a) A1 =
2 3 2
1 1 1
A1 =
2 3 2
Applying R2 → R2 − 2R1 , we get
1 1 1
→
0 1 0
1 1 1
RREF of A1 is
0 1 0
1
1
}
,
Column Space of A1 = Span{
1
0
Row Space of A1 = Span {(1, 1, 1), (0, 1, 0)}  
x1
Let W = {X ∈ R3 | A1 X = O} where X = x2 .
x3
Now,
 
x1
1 1 1  
0
x2 =
0 1 0
0
x3
Therefore,
 we
 have, x1 + x2 + x3 = 0 and x2 = 0. Thus we have
1
X = x1  0 
−1
 
1
Therefore, The null space of A1 = Span{ 0 }
−1
(b)


1 2 2
A2 = 1 3 4
1 2 3
Applying R2 → R2 − R1 and R3

1
0
0
10
→ R3 − R1 , we get

2 2
1 2
0 1
     
1
2
2





Column Space of A2 = Span{ 0 , 1 , 2}
0
0
1
Row Space of A2 = Span{(1, 2, 2), (0, 1, 2), (0,
1)}
 0, 
x1
Let W = {X ∈ R3 | A2 X = O} where X = x2 .
x3
Now,

   
1 2 2
x1
0
0 1 2 x2  = 0
0 0 1
x3
0
Therefore, we have,
 x1 + x2 + x3 = 0,x2 + 2x3 = 0 and x3 = 0. Thus
0
we have X = 0
0
 
0

Therefore, The null space of A2 = { 0}
0
(c)


1 2
A3 = 1 3
1 2
Applying R2 → R2 − R1 and R3 → R3 − R1 , we get


1 2
0 1
0 0
   
1
2
Column Space of A3 = Span{0 , 1}
0
0
Row Space of A3 = Span{(1, 2), (0, 1)} x
Let W = {X ∈ R2 |A3 X = O} where X = 1 .
x2
Now,


 
1 2 0
0 1 x1 = 0
x2
0 0
0
11
Therefore,
we have, x1 + 2x2 = 0,x2 = 0 and x1 = 0. Thus we have
0
X=
0
0
Therefore, The null space of A3 = {
}
0
(d)


1 2 1
1 3 1

A4 = 
1 2 1
2 5 2
Applying R2 → R2 − R1 , R3 → R3 − R1 and R4 → R4 − 2R1 , we get


1 2 1
0 1 0


0 0 0
0 1 0
Applying R4 → R4 − R2


1 2 1
0 1 0


0 0 0
0 0 0
   
2
1
0 1
  
Column Space of A4 = Span{
0 , 0}
0
0
Row Space of A4 = Span{(1, 2, 1), (0, 1, 0)} 
x1
Let W = {X ∈ R3 |A4 X = O} where X = x2 .
x3
Now,


 
1 2 1  
0
0
0 1 0 x1

   
0 0 0 x2 = 0
x3
0
0 0 0
Therefore,
 we
 have, x1 + 2x2 + x3 = 0 and x2 = 0. Thus we have
1
X = x1  0 
−1
12


1
Therefore, The null space of A4 = Span{ 0 }
−1
7. Use Gauss elimination method to find all polynomials f ∈ P2 : f (1) =
2 and f (−1) = 6.
Solution:
Let us take f (x) ∈ P2 as
f (x) = a0 + a1 x + a2 x2 ,
a0 , a1 , a2 ∈ R
Then,
f (1) = 2 ⇒a0 + a1 + a2 = 2 and
f (−1) = 6 ⇒a0 − a1 + a2 = 6
Now we can write these equations in matrix form as,
 
a0
1 1 1  
2
a1 =
1 −1 1
6
a2
So Augmented matrix is,
R2 → R2 − R1
1 1 1 2
1 −1 1 6
1 1 1 2
0 −2 0 4
Now, we have equations
a0 + a1 + a2 = 2 and
−2a1 = 4
⇒ a1 = −2 and a0 + a2 = 4
Put a2 = c then we have
(a0 , a1 , a2 ) = (4 − c, −2, c)
So, all f (x) = (4−c)−2x+cx2 ∈ P2 where c ∈ R are such polynomials
in P2 which satisfies f (1) = 2 and f (−1) = 6.
13