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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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Krash (Sample)
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Krash - Sample
Engineering Maths
(Linear Algebra &
Differential Equation)
Krash (Sample)
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Krash (Sample)
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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.
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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 .
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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
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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
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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
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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
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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.
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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.
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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
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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
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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
An equation of the form
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).
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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,
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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
An equation of the form
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
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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
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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
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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)
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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)
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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)
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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)
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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)
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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)
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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)
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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)
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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)
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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)
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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)
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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