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 Topics: 1)Matrices
2) Differential calculus.
 Submitted by:
Name: Sharwan Solanki
Roll No.: 18951A05G3
Class: CSE-C
 Submitted to:
Mrs. V Subba Laxmi, Associate
Professor
LAC ASSIGNMENT
Math’s Assignment
 RANK OF A MATRIX
1) Reduce the matrix A to normal form and hence find its rank where
A=












61152
5732
1430
4312
.
Solution:
A=












61152
5732
1430
4312
Applying R3→R3-R1, R4→R4-R1
∼












2840
1420
1430
4312
Applying C1→C1(
1
2
)
∼











 
2840
1420
1430
4311
Applying C2→C2-C1, C3→C3-3C1, C4→C4-4C1
∼












2840
1420
1430
0001
Applying R2→R2-R3, R4→R4-2R3
∼












0000
1420
0010
0001
Applying R3→R3-2R2
∼












0000
1400
0010
0001
Applying C3↔C4
∼












0000
4100
0010
0001
Applying C4→C4-4C3
∼












0000
0100
0010
0001
this is in normal form.
∴ the rank of A is 3.
 ROLLE’S THEOREM
Theorem: let f(x) be a function such that
1. It is continuous in closed interval ⦋a, b⦌;
2. It is differentiable in open interval (a, b) and
3. f(a)=f(b)
Then there exists at least one point c ϵ (a, b) such that f ’(c) =0
1) Verify Rolle’s theorem for f(x)= 2x3 + x2 -4x-2 in [−√2, √2].
Solution: f(x) is a polynomial and hence is continuous and differentiable
for all x. Hence the first two conditions of Rolle ’s Theorem are satisfied.
Now f (√2 ) = 2(√2 )3 + (√2 )2 - 4√2 -2 = 4√2 +2 - 4√2 - 2 = 0
And f (-√2 ) = 2(−√2 )3 + (−√2 )2 - 4(−√2 ) -2 = -4√2 +2 + 4√2 - 2 = 0
Hence f (−√2 ) = f (√2 ) =0. Thus condition (3) of Rolle ’s Theorem is also
satisfied.
Hence by Rolle ’s Theorem, there exists a point c in (−√2 , √2 ) such that
f ’(c)
Verification: Consider f ’(c) =0
i.e. , 6c2 +2c -4 =0
⇒ 3c2 +c -2 =0
⇒ (3c -2)(c +1)=0
⇒ c=
2
3
or c= -1
Both c =
2
3
, -1 are in between (−√2 , √2 ).
Hence Rolle ’s Theorem is verified.
Name: Sharwan Solanki
Roll.no: 18951A05G3

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18951A05G3 - Sharwan Solanki ( LAC ASSIGNMENT)

  • 1.  Topics: 1)Matrices 2) Differential calculus.  Submitted by: Name: Sharwan Solanki Roll No.: 18951A05G3 Class: CSE-C  Submitted to: Mrs. V Subba Laxmi, Associate Professor LAC ASSIGNMENT
  • 2. Math’s Assignment  RANK OF A MATRIX 1) Reduce the matrix A to normal form and hence find its rank where A=             61152 5732 1430 4312 . Solution: A=             61152 5732 1430 4312 Applying R3→R3-R1, R4→R4-R1 ∼             2840 1420 1430 4312 Applying C1→C1( 1 2 ) ∼              2840 1420 1430 4311 Applying C2→C2-C1, C3→C3-3C1, C4→C4-4C1 ∼             2840 1420 1430 0001 Applying R2→R2-R3, R4→R4-2R3 ∼             0000 1420 0010 0001 Applying R3→R3-2R2
  • 4.  ROLLE’S THEOREM Theorem: let f(x) be a function such that 1. It is continuous in closed interval ⦋a, b⦌; 2. It is differentiable in open interval (a, b) and 3. f(a)=f(b) Then there exists at least one point c ϵ (a, b) such that f ’(c) =0 1) Verify Rolle’s theorem for f(x)= 2x3 + x2 -4x-2 in [−√2, √2]. Solution: f(x) is a polynomial and hence is continuous and differentiable for all x. Hence the first two conditions of Rolle ’s Theorem are satisfied. Now f (√2 ) = 2(√2 )3 + (√2 )2 - 4√2 -2 = 4√2 +2 - 4√2 - 2 = 0 And f (-√2 ) = 2(−√2 )3 + (−√2 )2 - 4(−√2 ) -2 = -4√2 +2 + 4√2 - 2 = 0 Hence f (−√2 ) = f (√2 ) =0. Thus condition (3) of Rolle ’s Theorem is also satisfied. Hence by Rolle ’s Theorem, there exists a point c in (−√2 , √2 ) such that f ’(c) Verification: Consider f ’(c) =0 i.e. , 6c2 +2c -4 =0 ⇒ 3c2 +c -2 =0 ⇒ (3c -2)(c +1)=0 ⇒ c= 2 3 or c= -1 Both c = 2 3 , -1 are in between (−√2 , √2 ). Hence Rolle ’s Theorem is verified. Name: Sharwan Solanki Roll.no: 18951A05G3