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Krash (Sample)

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Manual for Krash

Why Krash?

Towards the end of preparation, a student has lost the time to revise all the chapters from

his / her class notes / standard text books. This is the reason why K-Notes is specifically

intended for Quick Revision and should not be considered as comprehensive study material.

It’s very overwhelming for a student to even think about finishing 300-400 questions per

subject when the clock is ticking at the last moment. This is the reason why Kuestion serves

the purpose of being the bare minimum set of questions to be solved from each chapter

during revision.

What is Krash?

Krash is a combination of K-Notes and Kuestion which effectively becomes a great tool for

revision. A 50 page or less notebook for each subject which contains all concepts covered in

GATE Curriculum in a concise manner to aid a student in final stages of his/her preparation.

A set of 100 questions or less for each subject covering almost every type which has been

previously asked in GATE. Along with the Solved examples to refer from, a student can try

similar unsolved questions to improve his/her problem solving skills.

When do I start using Krash?

It is highly recommended to use Krash in the last 2-3 months before GATE Exam.

How do I use Krash?

Once you finish the entire K-Notes for a particular subject, you should practice the respective

Subject Test / Mixed Question Bag containing questions from all the Chapters to make best

use of it. Kuestion should be used as a tool to improve your speed and accuracy subject-

wise. It should be treated as a supplement to our K-Notes and should be attempted once

you are comfortable with the concepts and basic problem solving ability of the subject. You

should refer K-Notes before solving any “Type” problems from Kuestion.

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in any form, or by means, electronic, mechanical, photocopying, recording, or otherwise, without

the prior written consent of Kreatryx, is a violation of copyright law. Any sale/resale of this material

is punishable under law. Subject to Ghaziabad Jurisdiction only.

Krash (Sample)


Krash - Sample

Engineering Maths

(Linear Algebra &

Differential Equation)

Krash (Sample)


Krash (Sample)


LINEAR ALGEBRA

MATRICES

A matrix is a rectangular array of numbers (or functions) enclosed in brackets. These

numbers (or function) are called entries of elements of the matrix.

Example:

2 0.4 8

5 -32 0 order = 2 x 3, 2 = number of rows, 3 = number of columns

Special Type of Matrices

1. Square Matrix

A m x n matrix is called as a square matrix if m = n i.e. number of rows = number of columns

The elements ij

a when i = j 11 22a a ......... are called diagonal elements

Example:

1 2

4 5

2. Diagonal Matrix

A square matrix in which all non-diagonal elements are zero and diagonal elements may or

may not be zero.

Example:

1 0

0 5

Properties

a. diag [x, y, z] + diag [p, q, r] = diag [x + p, y + q, z + r]

b. diag [x, y, z] × diag [p, q, r] = diag [xp, yq, zr]

c.

11 1 1diag x, y, z diag , ,

x y z

d.

tdiag x, y, z = diag [x, y, z]

e.

n n n ndiag x, y, z diag x , y , z

f. Eigen value of diag [x, y, z] = x, y & z

g. Determinant of diag [x, y, z] = xyz

3. Scalar Matrix

A diagonal matrix in which all diagonal elements are equal.

Krash (Sample)


4. Identity Matrix

A diagonal matrix whose all diagonal elements are 1. Denoted by I

Properties:

a. AI = IA = A

b. nI I

c. 1I I

d. det(I) = 1

5. Null matrix

An m x n matrix whose all elements are zero. Denoted by O.

Properties:

a. A + O = O + A = A

b. A + (- A) = O

6. Upper Triangular Matrix

A square matrix whose lower off diagonal elements are zero.

Example:

3 4 5

0 6 7

0 0 9

7. Lower Triangular Matrix

A square matrix whose upper off diagonal elements are zero.

Example:

3 0 0

4 6 0

5 7 9

8. Idempotent Matrix

A matrix is called Idempotent if 2A A

Example:

1 0

0 1

9. Involutary Matrix

A matrix is called Involutary if 2A I .

Krash (Sample)


Matrix Equality

Two matrices m nA and p qB are equal if

m = p ; n = q i.e., both have same size

ija =

ijb for all values of i & j.

Addition of Matrices

For addition to be performed, the size of both matrices should be same.

If [C] = [A] + [B]

Then ij ij ij

c a b

i.e., elements in same position in the two matrices are added.

Subtraction of Matrices

[C] = [A] – [B] = [A] + [–B]

Difference is obtained by subtraction of all elements of B from elements of A.

Hence here also, same size matrices should be there.

Scalar Multiplication

The product of any m × n matrix A jka and any scalar c, written as cA, is the m × n matrix

cA = jkca obtained by multiplying each entry in A by c.

Multiplication of two matrices

Let m nA and p qB be two matrices and C = AB, then for multiplication, [n = p] should

hold. Then, n

jkj 1

ik ijc a b

Properties:

If AB exists then BA does not necessarily exists.

Example: 3 4A , 4 5

B , then AB exits but BA does not exists as 5 ≠ 3

So, matrix multiplication is not commutative.

Matrix multiplication is not associative.

A(BC) ≠ (AB)C .

Matrix Multiplication is distributive with respect to matrix addition

A(B + C) = AB +AC

Krash (Sample)


If AB = AC B = C (if A is non-singular)

BA = CA B = C (if A is non-singular)

Transpose of a matrix

If we interchange the rows by columns of a matrix and vice versa we obtain transpose of a

matrix.

eg., A =

1 3

2 4

6 5

;

T 1 2 6A

3 4 5

Conjugate of a matrix

The matrix obtained by replacing each element of matrix by its complex conjugate.

Properties

a. A A

b. A B A B

c. KA K A

d. AB AB

Transposed conjugate of a matrix

The transpose of conjugate of a matrix is called transposed conjugate. It is represented by

A .

a.

A A

b. A B A B

c. KA KA

d. AB B A

Trace of matrix

Trace of a matrix is sum of all diagonal elements of the matrix.

Classification of Real Matrix

a. Symmetric Matrix : T

A A

b. Skew symmetric matrix : T

A A

c. Orthogonal Matrix : T 1 TA A ; AA =I

Krash (Sample)


Note:

a. If A & B are symmetric, then (A + B) & (A – B) are also symmetric

b. For any matrix TAA is always symmetric.

c. For any matrix,

TA + A

2 is symmetric &

TA A

2 is skew symmetric.

d. For orthogonal matrices, A 1

Classification of complex Matrices

a. Hermitian matrix : A A

b. Skew – Hermitian matrix : A A

c. Unitary Matrix : 1A A ;AA 1

Determinants

Determinants are only defined for square matrices.

For a 2 × 2 matrix

11 12

21 22

a a

a a

= 11 22 12 21

a a a a

Minors & co-factor

If

11 12 13

21 22 23

31 32 33

a a a

a a a

a a a

Minor of element 12 13

21 2132 33

a aa : M

a a

Co-factor of an element i j

ij ija 1 M

To design cofactor matrix, we replace each element by its co-factor

Determinant

Suppose, we need to calculate a 3 × 3 determinant

3 3 3

1j 1 j 2 j 2 j 3j 3jj 1 j 1 j 1

a cof a a cof a a cof a

We can calculate determinant along any row or column of the matrix.

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Properties

Value of determinant is invariant under row & column interchange i.e., TA A

If any row or column is completely zero, then A 0

If two rows or columns are interchanged, then value of determinant is multiplied by -1.

If one row or column of a matrix is multiplied by ‘k’, then determinant also becomes k

times.

If A is a matrix of order n × n , then

nKA K A

Value of determinant is invariant under row or column transformation

AB A * B

nnA A

1 1

AA

Adjoint of a Square Matrix

Adj(A) = T

cof A

Inverse of a matrix

Inverse of a matrix only exists for square matrices

1

Adj AA

A

Properties

a. 1 1AA A A I

b. 1 1 1AB B A

c. 1 1 1 1ABC C B A

d. T1

T 1A A

e. The inverse of a 2 × 2 matrix should be remembered

1a b d b1

c d c aad bc

I. Divide by determinant.

II. Interchange diagonal element.

III. Take negative of off-diagonal element.

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Rank of a Matrix

a. Rank is defined for all matrices, not necessarily a square matrix.

b. If A is a matrix of order m × n,

then Rank (A) ≤ min (m, n)

c. A number r is said to be rank of matrix A, if and only if

There is at least one square sub-matrix of A of order ‘r’ whose determinant is non-zero.

If there is a sub-matrix of order (r + 1), then determinant of such sub-matrix should be 0.

Linearly Independent and Dependent

Let X1 and X2 be the non-zero vectors

If X1=kX2 or X2=kX1 then X1,X2 are said to be Linearly Dependent vectors.

If X 1 kX2 or X2 kX1 then X1,X2 are said to be Linearly Independent vectors.

Note:

Let X1,X2 ……………. Xn be n vectors of matrix A

If rank(A)=no of vectors then vector X1,X2………. Xn are Linearly Independent

If rank(A)<no of vectors then vector X1,X2………. Xn are Linearly Dependent

System of Linear Equations

There are two type of linear equations

Homogenous equation

n11 1 12 2 1na x a x .......... a x 0

n21 1 22 2 2na x a x .......... a x 0

------------------------------------

------------------------------------

mn nm1 1 m2 2a x a x .......... a x 0

This is a system of ‘m’ homogenous equations in ‘n’ variables

Let

11 12 13 1n 1

21 22 2n 2

mn nm1 m2 m 1n 1m n

a a a a x 0

a a a x 0

-A ; x ; 0=

-

a a a x 0

This system can be represented as

AX = 0

Krash (Sample)


Important Facts

Inconsistent system

Not possible for homogenous system as the trivial solution

T T

n1 2x ,x ....., x 0,0,......,0 always exists.

Consistent unique solution

If rank of A = r and r = n |A| ≠ 0, so A is non-singular. Thus trivial solution exists.

Consistent infinite solution

If r < n, number of independent equation < (number of variables) so, value of (n – r)

variables can be assumed to compute rest of r variables.

This solution set has the following additional properties:

1. If u and v are two vectors representing solutions to a homogeneous system, then the

vector sum u + v is also a solution to the system.

2. If u is a vector representing a solution to a homogeneous system, and r is any scalar,

then ru is also a solution to the system.

Non-Homogenous Equation

n11 1 12 2 1n 1a x a x .......... a x b

n21 1 22 2 2n 2a x a x .......... a x b

-------------------------------------

-------------------------------------

mn n nm1 1 m2 2a x a x .......... a x b

This is a system of ‘m’ non-homogenous equation for n variables.

11

2

12 1n 1 1

21 22 2n 2

mn n mm1 m2 m 1m n

a a a x b

a a a x b

A ; X ; B= -

-

a a a x b

Augmented matrix = [A | B] =

11 n12 1

21 22 2

mn mm1 m2

a a a b

a a b

a a a b

https://en.wikipedia.org/wiki/Vector_(mathematics)

https://en.wikipedia.org/wiki/Scalar_(mathematics)

Krash (Sample)


Conditions

Inconsistency

If r(A) ≠ r(A | B), system is inconsistent

Consistent unique solution

If r(A) = r(A | B) = n, we have consistent unique solution.

Consistent Infinite solution

If r(A) = r (A | B) = r & r < n, we have infinite solution

Specifically, if p is any solution of the system Ax=B, then entire solution set can be described

as {p+v: v is any solution to Ax=O}.

The solution of system of equations can be obtained by using Gauss elimination Method.

(Not required for GATE)

Note:

Let Anxn and rank(A)=r, then the number of Linearly Independent solutions of Ax = 0 is “n-

r”.

Cramer’s Rule

Suppose we have the following system of equations,

1 1 1 1

2 2 2 2

3 3 3 3

a x b y c z d

a x b y c z d

a x b y c z d

We define the following matrices,

1 1 1 1 1 1

1 2 2 2 2 2 2 2

3 3 3 3 3 3

1 1 1 1 1 1

3 2 2 2 2 2 2

3 3 3 3 3 3

d b c a d c

d b c a d c

d b c a d c

a b d a b c

a b d a b c

a b d a b c

Then, the solution is given by,

31 2x y z

This rule can only be applied in case of unique solution.

Krash (Sample)


Eigen values & Eigen Vectors

If A is n × n square matrix, then the equation

Ax = λ x is called Eigen value problem.

Where λ is called as Eigen value of A.

x is called as Eigen vector of A.

Characteristic polynomial

11 12 1n

21 22 2n

mnm1 m2

a a a

a a aA I

a a a

Characteristic equation A I 0

The roots of characteristic equation are called as characteristic roots or the Eigen values.

To find the Eigen vector, we need to solve

A I x 0

This is a system of homogenous linear equation.

We substitute each value of λ one by one & calculate Eigen vector corresponding to each

Eigen value.

Important Facts

a. If x is an eigenvector of A corresponding to λ, the KX is also an Eigenvector where K is a

constant.

b. If a n × n matrix has ‘n’ distinct Eigen values, we have ‘n’ linearly independent Eigen

vectors.

c. Eigen Value of Hermitian/Symmetric matrix are real.

d. Eigen value of Skew - Hermitian / Skew – Symmetric matrix are purely imaginary or zero.

e. Eigen Value of unitary or orthogonal matrix are such that | λ | = 1.

f. If n1 2, ......., are Eigen value of A, n1 2

k ,k .......,k are Eigen values of kA.

g. Eigen Value of 1A are reciprocal of Eigen value of A.

h. If n1 2, ...., are Eigen values of A,

n1 2

A A A, , ........., are Eigen values of Adj(A).

i. Sum of Eigen values = Trace (A)

j. Product of Eigen values = |A|

k. In triangular or diagonal matrix, Eigen values are diagonal elements.

Krash (Sample)


Cayley - Hamiltonian Theorem

Every matrix satisfies its own Characteristic equation.

e.g., If characteristic equation is

n n 1n1 2

C C ...... C 0

Then, n n 1

n1 2C A C A ...... C I O

Where I is identity matrix

O is null matrix

DIFFERENTIAL EQUATION The order of a deferential equation is the order of highest derivative appearing in it.

The degree of a differential equation is the degree of the highest derivative occurring in it,

after the differential equation is expressed in a form free from radicals & fractions.

For equations of first order & first degree

Variable Separation method

Collect all function of x & dx on one side.

Collect all function of y & dy on other side.

like f(x) dx = g(y) dy

Solution: f x dx g y dy c

Example: dz

1 xtanz 1dx

dzxtanz

dx

dzxdx

tanz

cotzdz xdx

2x

log sinz c2

Krash (Sample)


Linear Differential Equations

An equation is linear if it can be written as:

y' P x y r x

If r(x) = 0 ; equation is homogenous

else r(x) ≠ 0 ; equation is non-homogeneous

y(x) = p x dx

ce is the solution for homogenous form

for non-homogenous form, h = P x dx

hhy x e e rdx c

Example: dx

t x tdt

dx xDivide by t, 1

dt t

1/t logtIF e dt e t

Solution is 2t

x t 1 t dt c2

Exact differential equation

An equation of the form

M(x, y) dx + N (x, y) dy = 0

For equation to be exact.

M N

y x

; then only this method can be applied.

The solution can be given by any one of the following two equations,

Mdx terms of N not containing 'x' dy = c (y constant)

Ndy terms of M not containing 'y' dx = c (treat x as constant)

Else, some special terms can be rearranged as given below,

xdy ydx

d log xyxy

2

x 1 ydx xdy ydx xdyd log

y x / y xyy

y xdy ydxd log

x xy

1

2 2

x ydx xdyd tan

y x y

Krash (Sample)


1

2 2

y xdy ydxd tan

x x y

2 2

2 2

2 xdx ydyd log x y

x y

x x x

2

e ye dx e dyd

y y

y y y

2

e xe dy e dxd

x x

2 2

2

x 2xydx x dyd

y y

2 2

2

y 2xydy y dxd

x x

2 2 2

2 4

x 2xy dx 2x ydyd

y y

2 2 2

2 4

y 2x ydy 2xy dxd

x x

Example: 1

y cosy dx x xsiny 1 dy 02 x

1

ydx xdy cosydx xsinydy dx dy 02 x

1

d xy d xcos y dx dy 02 x

xy xcosy x y 0

Integrating factors


P(x, y) dx + Q (x, y) dy = 0

This can be reduced to exact form by multiplying both sides by IF.

Rule I

If the given equation Pdx + Qdy = 0 (1) isn’t an exact equation but it is a homogenous

equation and Px + Qy 0, then 1

(2)Px Qy

is an integrating factor of (1).

Rule II

In the given equation Pdx Qdy 0 (1) is not exact equation but it is in the form

f xy ydx g xy xdy 0 and Px Qy 0 . Then 1

IF ...............(2)Px Qy

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Rule III

If

1 P Q

Q y x is a function of x, then R(x) =

1 P Q

Q y x

. Then, Integrating Factor

IF = exp R x dx

Rule IV

Otherwise, if Q P1

P x y

is a function of y then, S(y) =

Q P1P x y

. Then, Integrating

Factor, IF = exp S y dy

Bernoulli’s equation

The equation ndyPy Qy

dx

Where P & Q are function of x

Divide both sides of the equation by ny & put 1 n

y z

dz

P 1 n z Q 1 ndx

This is a linear equation & can be solved easily.

Clairaut’s equation

An equation of the form y = Px + f (P), is known as Clairaut’s equation where P = dydx

The solution of this equation is y = cx + f (c) where c = constant

Linear Differential Equation of Higher Order

Constant coefficient differential equation

n

n n 1

n 1

n1

d y d yk .............. k y X

dx dx

Where X is a function of x only

a. If n1 2y , y ,.........., y are n independent solution, then

n n1 1 2 2c y c y .......... c y x is complete solution

where 1 2 nc ,c ,..........,c are arbitrary constants.

b. The procedure of finding solution of nth order differential equation involves computing

complementary function (C. F) and particular Integral (P. I).

Krash (Sample)


c. Complementary function is solution of

n

n n 1

n 1

n1

d y d yk ............. k y 0

dx dx

d. Particular integral is particular solution of

n

n n 1

n 1

n1

d y d yk ............ k y x

dx dx

e. y = CF + PI is complete solution

Finding complementary function

Method of differential operator

Replace d

dx by D →

dyDy

dx

Similarly, n

n

d

dx by

nD → n

nn

d yD y

dx

n

n n 1

n 1

n1

d y d yk ............ k y 0

dx dx

becomes

n n 1n1

D k D ........... k y 0

Let 1 2 nm ,m ,............,m be roots of

n1

n 1nD k D ................ K 0 ………….(i)

Case I: All roots are real & distinct

n1 2D m D m ............ D m 0 is equivalent to (i)

y = 1 2 nm xm x m x

n1 2c e c e ........... c e is solution of differential equation

Case II: If two roots are real & equal i.e., 1 2m m m

y = 23 n

m x m xmxn1 3

c c x e c e .......... c e

Case III: If two roots are complex conjugate

1m j ; 2

m j

y = 1 2nm x

nxe c 'cos x c 'sin x .......... c e

Finding particular integral

Suppose differential equation is

n n 1

nn 1 n 1

d y d yk .......... k y X

dx dx

Particular Integral,

Krash (Sample)


PI =

n1 2n1 2

W x W x W xy Xdx y Xdx .......... y Xdx

W x W x W x

Where n1 2y , y ,............y are solutions of Homogenous from of differential equations.

n1 2

n1 2

n n nn1 2

y y y

y ' y ' y 'W x

y y y

n1 2 i 1

n1 2 i 1

i

n n n nn1 2 i 1

y y y 0 y

y ' y ' y '0 y 'W x

0

y y y 1 y

iW x is obtained from W(x) by replacing ith column by all zeroes & last 1.

Example: 3x

2

2

eD 6D 9 y

x

3x

2

2

eD 3 y

x

Finding Complimentary Function,

D 3,3

1 2

3x 3x 3x

1 2 1 2

y y

y c c x e c e c xe

Wronskian is given by, 3x 3x

6x

3x 3x 3x

e xe W e

3e 3xe e

3x3x

1 3x 3x

0 xe W xe

1 3xe e

3x3x

2 3x

e 0 W e

3e 1

Coefficients for Particular integral can be computed as, 3x 3x

2 6x

e xeA dx logx

x e

3x 3x

2 6x

e eB dx

x e

1logx

x

P 1 2y Ay By

Euler-Cauchy Equation


n n 1n n 1

nn 1 n 1

d y d yx k x .......... k y 0

dx dx

is called as Euler-Cauchy theorem

Krash (Sample)


Substitute y = xm

The equation becomes

mn1

m m 1 ........ m n k m(m 1)......... m n 1 ............. k x 0

The roots of equation are

Case I: All roots are real & distinct

1 2 nm m m

n1 2y c x c x ........... c x

Case II: Two roots are real & equal

1 2m m m

3m mm

n1 2 3ny c c nx x c x ........ c x

Case III: Two roots are complex conjugate of each other

1m j ; 2

m j

y =

3m nmn3

x Acos nx Bsin nx c x ........... c x

Krash (Sample)


Type 1: Linear Algebra

For Concepts, please refer to Engineering Mathematics K-Notes, Linear Algebra

Problems:

01. If for a matrix, rank equals both the number of rows and number of columns, then the

matrix is called

(A) Non-singular (B) singular

(C) transpose (D) minor

02. The equation 2

2 1 1

1 1 1 0

y x x

represents a parabola passing through the points.

(A) (0, 1), (0, 2), (0, -1) (B) (0, 0), (-1, 1), (1, 2)

(C) (1, 1), (0, 0), (2, 2) (D) (1, 2), (2, 1), (0, 0)

03. If matrix 2

a 1X

a a 1 1 a

and 2X X I 0 then the inverse of X is

(A) 2

1 a 1

a a

(B) 2

1 a 1

a a 1 a

(C) 2

a 1

a a 1 a 1

(D) 2a a 1 a

1 1 a

04. Consider the matrices 4 3 4 3 2 3X , Y andP . The order of

T1T TP X Y P

will be

(A) 2x2 (B) 3x3

(C) 4x3 (D) 3x4

05. The rank of the matrix

1 1 1

1 1 0

1 1 1

is

(A) 0 (B) 1

(C) 2 (D) 3

Krash (Sample)


06. The matrix

2 2 3

M 2 1 6

1 2 0

has given values -3, -3, 5. An eigen vector corresponding

to the eigen value 5 is T

1 2 1 . One of the eigen vector of the matrix 3M is

(A) T

1 8 1 (B) T

1 2 1

(C) T

31 2 1 (D) T

1 1 1

07. The system of equations x + y + z = 6, x + 4y + 6z = 20, x + 4y + z has no solutions

for values of and given by

(A) 6, 20 (B) 6, 20

(C) 6, 20 (D) 6, 20

08. Let P be 2x2 real orthogonal matrix and x is a real vector T

1 2x x with length

1 2

2 21 2

x x x . Then which one of the following statement is correct?

(A) px x where at least one vector satisfies px x

(B) px x for all vectors x

(C) px x where at least one vector satisfies px x

(D) No relationship can be established between x and px

09.The characteristics equation of a 3 x 3 matrix P is defined as

3 2I P 2 1 0 . If I denotes identity matrix then the inverse of P will be

(A) 2P P 2I (B)

2P P I

(C) 2P P I (D) 2P P 2I

10. A set of linear equations is represented by the matrix equations Ax = b. The necessary

condition for the existence of a solution for the system is

(A) A must be invertible

(B) b must be linearly dependent on the columns of A

(C) b must be linearly independent on the columns of A

(D) None.

Krash (Sample)


11. Consider a non-homogeneous system of linear equations represents mathematically an

over determined system. Such a system will be

(A) Consistent having a unique solution.

(B) Consistent having many solution.

(C) Inconsistent having a unique solution.

(D) Inconsistent having no solution.

12. The trace and determinant of a 2x2 matrix are shown to be -2 and -35 respectively. Its

eigen values are

(A) -30, -5 (B) -37, -1

(C) -7, 5 (D) 17.5, -2

Krash (Sample)


Type 2: Differential Equation For Concepts, please refer to Engineering Mathematics K-Notes, Differential Equations.

Problems:

01. For the differential equation dy

f x, y g x, y 0dx

to be exact is

(A) gf

y x

(B)

gf

x y

(C) f = g (D)

2

2

2

2

gf

x y

02. The differential equation dy

py Qdx

, is a linear equation of first order only if,

(A) P is a constant but Q is a function of y

(B) P and Q are functions of y (or) constants

(C) P is function of y but Q is a constant

(D) P and Q are function of x (or) constants

03. Biotransformation on of an organic compound having concentration (x) can be modelled

using an ordinary differential equation 2dx

kx 0dt

, where k is reaction rate constant. If

x = a at t = 0 then solution of the equation is

(A) kxx a e (B)

1 1kt

x a

(C) ktx a 1 e (D) x = a + k t

04. Transformation to linear form by substituting v=1 ny

of the equation

ndyp t y q t y

dt , n > 0 will be

(A) dv

1 n pv 1 n qdt

(B) dv

1 n pv 1 n qdt

(C) dv

1 n pv 1 n qdt

(D) dv

1 n pv 1 n qdt

Krash (Sample)


05. For

22x

2

d y dy4 3y 3e

dxdx , the particular integral is

(A) 2x1

e15

(B) 2x1

e5

(C) 2x3e (D)

3x1 2

xC e C e

06. Solution of the differential equation dy

3y 2x 0dx

represents a family of

(A) ellipses (B) circles

(C) parabolas (D) hyperbolas

07. Which one of the following differential equation has a solution given by the function

y 5sin 3x3

(A) dy 5

cos 3x 0dx 3

(B) dy 5

cos3x 0dx 3

(C)

2

2

d y9y 0

d x (D)

2

2

d y9y 0

dx

08. The order and degree of a differential equation

3

3

32d y dy

4 y 0dxdx

are

respectively

(A) 3 and 2 (B) 2 and 3

(C) 3 and 3 (D) 3 and 1

09. The maximum value of the solution y (t) of the differential equation y(t)+ y..

(t) = 0 with

initial conditions y 0 1.

and y(0) = 1, for t ≥ 0 is

(A) 1 (B) 2

(C) (D) 2

10. It is given that y 2y y 0 , y(0) = 0 & y(1) = 0. What is y(0.5) ?

(A) 0 (B) 0.37

(C) 0.82 (D) 1.13

Krash (Sample)


11. A body originally at 60° cools down to 400 in 15 minutes when kept in air at a

temperature of 25°c. What will be the temperature of the body at the end of 30 minutes?

(A) 35.2° C (B) 31.5°C

(C) 28.7°C (D) 15°C

12. The solution

2

2

d y dy2 17y 0

dxdx ;

x4

dyy 0 1, 0

dx

in the range 0 x

4

is given by

(A) x 1

e cos4x sin4x4

(B) x 1

e cos4x sin4x4

(C) 4x 1e cos4x sin x

4

(D) 4x 1e cos4x sin4x

4

Krash (Sample)


Solutions Type 1: Linear Algebra

01. Ans: (A)

Solution: Rank=no. of rows=no. of columns

A 0 As all rows & columns are linearly independent

A=Non-Singular matrix

02. Ans: (B)

Solution: 2

2 1 1

I 1 1 1 0

y x x

By just looking at matrix I we can conclude that:-

One solution of above matrix is (0,0) as row3’s all elements are zero hence |I|=0

Second solution is (1,2) as R1=R3Linearly dependent rows

Third solution is (-1,1) as R2=R3Linearly dependent rows

03. Ans: (B)

Solution: 2X X I Multiplying by 1X

1X I X I

Or

1

2

1 0 a 1X I X

0 1 a a 1 1 a

2

1 a 1

a a 1 a

04. Ans: (A)

Solution: 4 3 4 3 2 3

X ,Y & P

and T

1T TP X y P

T1

1 T TPY X P

Matrix operation Order TX 3 4

Y 4 3 TX Y 3 3

1

TX Y

3 3

T 1P X Y 2 3

T 1 TP X Y P 2 2

Krash (Sample)


05. Ans: (C)

Solution: Replace

R1 R1 R3 ,

We get

B 2

0 0 0

1 1 0A11 1

A 0

Rank=2

06. Ans: (B)

Solution: Eigen values=-3, 3, 5

Eigen vector corresponding to 5 is

1

2

1

For 3M Eigen values-27, 27, 125

But Eigen vector will remains the same T

1 2 1

07. Ans: (B)

Solution:

1 1 1 6

A |B 1 4 6 20

1 4 u

For no solution

A A |B n

A n

A 0

1 4 24 1 6 1 4 4 0

4 24 6 0

3 18 6

But A |B 0

1 4u 80 1 20 u 6 4 4 0 => u 20

Krash (Sample)


08. Ans: (B)

Solution: Orthogonal matrix, T 1P P .............. 1

Note: Orthogonal matrix presence the length of vector [Euclidean Norm]

1 2X X X

T 1

2

XX

X

2 T

PX PX PX T TX P PX

From (1)

2 T 1PX X P PX T T 2X IX X X X

09. Ans: (D)

Solution: Characteristic equation= 2 22 1 0 . Every matrix satisfies its characteristic

equation [Cayley Hamiltonian theorem]

3 2P P 2P I

xP Multiplying by 1P 2 1P P 2I P

10. Ans: (C)

Solution: This equation will be solvable even for non-square matrix A of B vector is linearly

independent of columns of A.

11. Ans: (D)

Solution: Over determined system will have no solution as number of equations are more

than number of unknowns.

12. Ans: (C)

Solution: Eigen 2 & Eigen 35

1 22 &

1 235

1 12 35

2

1 12 35 0

1 27 or 5

Krash (Sample)


Type 2: Differential Equation

01. Ans: (B)

Solution: N M

f x, y dy g x, y dx 0

For exact differential equation

M N

y x

gf

x y

02. Ans: (D)

Solution: For linear differential equation:-

dyPy Q

dx

P & Q must be function of x Or constant 0

k f x

03. Ans: (B)

Solution: 2

t 0

dxkx 0 , x a

dt

2dxkx

dt Variable separate type

2

dxk dt

x

1kt c

x

At t=0

1c

a

1 1kt

x a

04. Ans: (A)

Solution: ndy

Py q.ydt

[ Bernoulli’s equation]………….(i)

Let 1 n

V y

n dydv1 n y

dt dt

Krash (Sample)


ndy y dv................... 2

dt dt1 n

Equation (1) can be written as

1 nn dyy Py q

dt

dv

Pv q1 n dt

dv

1 n pv 1 n qdt

05. Ans: (B)

Solution: Let 2dD, f D D 4D 3 0

dx

2x 2x 2x3e 3e eP.I

15 5f 2

06. Ans: (A)

Solution: 3ydy 2xdx 0 [Variable separable]

On integration

223y

x K2

[Ellipse,2

2 yx

2k

3

]

For circle, coefficient of 2x coefficient of 2y

07. Ans: (C)

Solution: y 5sin 3x ............. 13

On differentiating both sides

dy

5 3 cos 3xdx 3

Again differentiating both

2

2

d y5 3 3 sin 3x

3dx

9 5 sin 3x

3

From (1) 2

2

d y9y 0

dx

Krash (Sample)


08. Ans: (A)

Solution:

332

3

d y dy4 y

dxdx

On squaring both sides

2 332

3

d y dy16 y

dxdx

Order=3, Degree=2

09. Ans: (D)

Solution: 2m 1 0

m i

Solution

1 2y C cosx C sinx

1y 0 1 => C 1 and 2

y 0 1 => C 1

0y cosx sinx 2 sin x 45

maxy 2 at

4

10. Ans: (A)

Solution: y 2y y 0

y 0 0 & y 1 0

2m 2m 1 0

2

m 1 0 m=-1 (Replacing Roots)

Solution

x

1 2y e C xC

1y 0 0 C

x

2

2

y e xC

y 1 0 C 0

Solution of difference equation

y 0

y 0.5 0

Krash (Sample)


11. Ans: (B)

Solution: By Newton’s law of cooling (0=Temp)

0Q Secondary temperature

1Q Initial temperature

0

dk Q Q

dt

0

dQdt

k Q Q

kt kc

0 0ln Q Q k t c Q Q e .e

kt

0Q Pe Q

1 0At t 0 P Q Q

0 0 0P 60 25 35

ktQ 35e 25

After, 15 minutes k 15

40 25 35e

: K .0565

After, 30minutes 0.565 30 0Q 35e 25 31.48 31.5 C

12. Ans: (A)

Solution: Let d

mydx

Then, 2m 2m 17 0

m 1 4i

Solution,

x

1 2y e C cos4x C sin4x

y 0 1 => 1

1 C

x

2y e cos4x C sin4x

On differentiating both sides

x x

2 2y e cos4x C sin4x e 4sin4x 4C cos4x

At y 04

:

20 4C 1

21C

4

Solution= x sin4xy e cos 4x

4

manual for krash · adjoint of a square matrix adj(a) = a t inverse of a matrix inverse of a matrix...

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