Detailed Solutions A-01/C-01/T-01 JUNE 2003

1. (a) C Because z = 1 is a pole for given function f and it lies outside the circle

|z| = ½ . Therefore, by Cauchy’s Theorem

(b) A Because P (x = 2) = 9 P (x = 4) + 90 P (x = 6)

=>

=>

Because m¹0, Therefore, 3m² + m⁴ – 4 = 0

=> m = 1

(c) B A vector field is solenoidal if div = 0

(d) D

=> (cos x – sin x)² (sin x + 2 cos x) = 0

Its only root which lies in .

(e) A 2y sin x + cos 2x = a

I.F.

Therefore, the solution is given as

=> 2y sin x + cos 2x = a

(f) A

where u and v are homogeneous functions of order 6 and 0 respectively. Using Euler’s theorem

= 6 u + 0 v = 6 u.

(g) D By Rodrigue’s formula,

P_n(0) = 0 if n is odd.

(h) C

=

2. (a) Let u = a³x² + b³y² + c³z²

Let where l is constant using Lagrange’s multiplier method.

For stationary values,

=> ax = by = cz = k

Then, gives a + b + c = k

Therefore,

Stationary value of u is

.

(b) The region of integration R is bounded by x = 0, y = 0, and lx + my = 1

{projection of lx + my + nz = 1 on z = 0}

.

3. (a)

This is along normal to the surface and is the maximum directional derivative. Thus is // to line .

Therefore,

Therefore, a = 4l, b= -11l, c = 10l and

Therefore, .

(b) By Divergence theorem,

Now,

Let x = 2 sin then dx = 2cos, for x = 0, = 0 and for x = 2, =

Now Surface S consists of three surfaces, the one leaving base S₁ (z = 0), second leaving top S₂ (z = 3), third the curved surface S₃ of cylinder x² + y² = 4 between z = 0, z = 3

On S₃ the outer normal is in the direction of . Therefore a unit vector along normal to the curved surface is given by

, thus

Hence divergence theorem proved.

4. (a) We know that

Putting, , Then

Comparing real and imaginary part

Let

.

Therefore, . Let b > 0, then

Hence,

Putting b = 0, we get

(b) Let U = X(x) Y(y), then equation becomes

.

Let , then

Therefore, X = A cos lx + B sin lx

Y = Ce^ly + De^-ly

U (x, y) = (A cos lx + B sin lx) (Ce^ly + De^-ly)

(iv) condition gives U (x, y) → 0 as y→ µ => C = 0

\ U = (A cos lx + B sin lx) e^-ly

(i) gives U (0, y) = 0 => A = 0

Hence, U = B sin lx . e^-ly

(iii) gives U (x, 0) = 1 => 1 = b_l sin x

5. (a) The characteristic equation of A is |A-lI| = 0

=> l² – 4l – 5 = 0

=> A² – 4A – 5I = 0, by Cayley Hamilton Theorem.

Thus A⁵ – 4A⁴ – 7A³ + 11A² – A – 10I

= (A² – 4A – 5I) (A³ – 2A + 3I) + A + 5I = A + 5I.

(b) (D² + 5D + 6) y = e^-2x sec²x (1 + 2 tan x)

A.E. : m² + 5m + 6 = 0

m = -2 , -3

C.F. = c₁e^-2x + c₂e^-3x

Thus y = c₁e^-2x + c₂ e^-3x + e^-2x tan x.

6. (a) Let and f(z) = u + iv

since u is an analytic function, thus it must

satisfies C-R equations, thus

Using Milne’s Thomson method, Let x = z , y = 0

, where c is a constant of integration.

(b) Consider the function . It has simple pole at z = 0. where C consists of the part of real axis from –R to -r and from r to R, the small semi circle C_r from –r to r with center at origin and radius r, which is small and large semi-circle C_R from R to –R as shown in Fig. f (z) is analytic inside C (z = 0, the only singularity has been deleted by indenting the origin by drawing C_r).

Therefore, by Cauchy’s Theorem,

For C_R, we have z = Re^iq , 0 £ q £ p and C_r , z = re^iq , 0 £ q £ p

Since decreases from 1 to as q increases from 0 to

Putting values in (1) and applying limits r→0, R→¥, we get

7. (a) Since 31% of items are under 45. Hence 19% of items lies between and 45. Since ,

thus, .Similarly

Solving, we get s = 10, .

(b) p(A) = Probability of hitting target by A = 3/5

p(B) = Probability of hitting target by B = 2/5

p(C) = Probability of hitting target by C = 3/4

(i) p₁ = Chance A, B hit & C fails

=

p₂ = Chance B, C hit & A fails = 12/100

p₃ = Chance C, A hit & B fails = 27/100

Since all these events are mutually exclusive, therefore,

P(two shots hit the target) = p₁ + p₂ + p₃ = 0.45

(ii) In case atleast two shots may hit target, we must also consider case when all hit the target.

p₄ = Probability A, B, C hit target = 18/100.

Therefore, P(atleast two shot hit the target) = p₁ + p₂ + p₃ + p₄ = 63/100 = 0.63.

8. (a) The characteristic equation is |A – lI| = 0

i.e. l³ - 7l² + 36 = 0

=> l = -2 , 3, 6

These are eigen values of given matrix A. For eigen vectors we find X ≠ 0, such that (A - lI) X = 0

For l = -2

3x + y + 3z = 0

x + 7y + z = 0

Therefore, for l = -2, eigen vector is (-1, 0, 1)’

Similarly for l = 3, eigen vector is (1, -1, 1)’

for l = 6, eigen vector is (1, 2, 1)’

The modal matrix P is given by

The diagonal matrix D is given by

(b) We have

i.e. AX = B

(i) (A : B) =

R₃→R₃ – R₁, R₂→7R₁ – 2R₂

If l = 5, system will have no solution for those values of m, for which rank A ¹ rank (A : B). If l = 5, m ¹ 9, then rank (A) = 2 and rank (A : B) = 3. Hence no solution

(ii) The system admits unique solution iff coefficient matrix is of rank 3

\

Thus for unique solution l ¹ 5 and m may have any value.

(iii) If l = 5,m = 9, system of equation have infinitely many solution

9. (a) Let r be base radius and l be slant height of cone.

Total area A = area of base + area of curved surface

= pr² + prl = pr (r + l) = ph² tan a (tan a + sec a)

dA = 2ph (tan² a + tana seca) dh + ph² (2tana sec²a + sec³a + tan²a seca) da

The error in h will be compensated by error in a, when

dA = 0

radians = -0.33^o .

(b) - (A)

Differentiating (A) partially w.r.t. t, we get

Differentiating (A) partially w.r.t. r, we get

equating (1) and (2), we get n = -3/2.

10. (a) Curl

=

\ vector field is irrotational then $ a scalar function j s.t.

Integrating, we get

Because, field is irrotational

(b)

Therefore, solution is

=> t = 1 + cx

=> (log z)^-1 = 1 + cx or .

11. (a) We know

For n = 1, 2 , 3

11. (b)

We have to expand about z = i

Detailed Solutions A-01/C-01/T-01 DECEMBER 2003

1. (a) A Let z = e^iq then

= i p

(b) B Let

Residue = coefficient of

(c) B P(A) = , P(B) = . Probability problem will be solved i.e. P(AÈB)

P(AÈB) = P(A) + P(B) – P(AB)

Because A & B are independent, So P(AB) = P(A) P(B)

P(AÈB) =

(d) D

over x² – 2ax + y² = 0

=

(e) A a= 0.25, b = 1

Let F = ax² – byz – (a + 2) = 0

G = 4x²y + z³ – 4 = 0

These surfaces will be orthogonal if

Also since (1, -1, 2) lies on F

\ a + 2b – a – 2 = 0 b = 1 , thus a =

(f) C z = u + v

i.e. u is homogeneous function of degree 2 and v is homogeneous function of degree 0. By Euler’s Theorem,

(g) C

(h) D As

2. (a) The system of equations can be written as AX = B

where

Augmented matrix = [A : B]

[A : B] =

[A : B]

[A : B]

This is row Echelon Form of A. Since the number of non-zero rows in the row-echelon form is 3. So,

Hence system of equations has unique solution, and is given by solving

(b) Characteristic equation is | A – lI | = 0

Eigen values of matrix A are 1, 1, 5

For l = 5, eigen vector is obtained by solving the system of equations

i.e. –3x + 2y + z = 0

x – 2y + z = 0

x + 2y –3z = 0

Solving, we get

i.e. eigen vector is (1, 1, 1)’

For l = 1, eigen vector is obtained by solving the system of equations

i.e. x + 2y + z = 0

There are two linearly independent eigen vectors for l = 1 . These are obtained by putting x=0 and y =0 respectively in the equation.

For x = 0, 2y + z = 0

Eigen vector is (0, -1, 2)’

For y = 0, Eigen vector is (-1, 0, 1)’

\ Modal Matrix = P =

\ Diagonal Matrix = D = P^-1AP

3. (a)

Then

By C.R. equations,

Subtracting (2) from (1)

Adding (1) and (2)

Using Milne’s Thomson method, putting x = z, y = 0 in (3) and (4), we get

3(b) Since,

Thus image of real axis in the z plane (means y = 0) is given by

4(u²+v²) + 7v – 2 = 0

i.e.

which is an equation of circle with centre at and radius

For u = 0, v = , we get x = 0, y = . Thus centre of circle in w plane is image of point in z plane.

4. (a) x = 0 is a regular singular point.

Let

Substituting in the given differential equation

The lowest power of x is x^r-1. Its coefficient equated to zero gives C₀r(r+3) = 0. Because C₀ ¹ 0

=> r = 0, r = –3

The coefficient of x^m+r is equated to zero gives

When r = 0

, for r = 0. Thus one solution is

When r = -3, C₁ = -5C₀ , C₂ = 10C₀ , C₃ = -10C₀ , C₄ = 5C₀ ------Thus is another solution.

is general solution.

(b)

5. (a)

except for x = -2.

Therefore, given vector field is not solenoidal.

= 0

\ Given vector field is irrotational. Thus it can be expressed as , where f is scalar function.

, ,

Integrating w.r.t. x, y, z we get

Since these three must be equal

+c

(b)

We know that

5. (c)

6. (a) Given

Adding & subtracting respectively, we get

Let

Similarly

Similarly

Adding, we get

(b) L = aM (1 – b) , 1 – b = bb⁴M²

Eliminating from (1) and (2), we get

is the required expression of percent change in L. Since 0 < β < 1, we have

7(a) (i)

Let

I.F.

(ii)

Integrating,

7. (b)

This integration is first w.r.t. y and then w.r.t. x. On changing the order of integration, we first integrate w.r.t. x and then w.r.t. y. This divided into three regions.

Region I: The strip extends from parabola y² = 2ax i.e. to x = 2a.

Then y = a to y = 2a

Region II: The strip extends from to circle .

Then y = 0 to a

Region III: The strip extends from circle, to x = 2a

Then y = 0 to y = a

8. (a) The vibration of the string is given by

- (i)

As the end points of the strings are fixed, for all time

y (0, t) = 0 = y (L, t) - (ii)

Since initial transverse velocity at any point of the string is zero.

- (iii)

Also, - (iv)

Using method of separation of variables and since the vibration of the string is periodic, therefore, the solution of (i) is of the form

- (v)

By (ii),

This should be true if c₁ = 0

Hence - (vi)

By (iii)

If c₂ = 0, then (vi) will be y (x, t) = 0

\ C₄ = 0

Thus (vi) becomes

Hence (vi) reduces to

From (iv)

\ Solution is

(b) Let the equation for conduction of heat be

Prior to temperature change at end B, when t = 0, the heat flow was independent of time (steady state condition), when u depends only on x i.e.

Since u = 20 for x = 0 and u = 80 for x = L

\ b = 20 , a =

Thus initial condition is expressed as

The boundary conditions are

The temperature function u (x, t) can be written as

where u_s(x) is a solution of (i) involving x only and satisfying boundary conditions (iii), which is steady state solution, u_t (x, t) is a transient part of the solution which decreases with increase of t.

Since, u_s(0) = 40 , u_s(L) = 60 \ using (ii), we get

From (iv), we get

General solution of (i) is given as

where

9. (a)

(b)

Let

The pole of ¦(z) enclosed by contour C, which consists of semicircle C_R and real axis segment from –R to R taken in counter clockwise directions, are z = i, z = 3i each of order 1.

Residue at

Residue at

Let R → ¥

, since second integral on the left hand side tends to 0.

10. (a) :Let

Let Using Lagrange’s method of multiplier, for extreme values, , f = g = 0.

Multiplying (1), (2), (3) by x, y, z respectively and adding, we get

Substituting in (4), we get

is satisfied by stationary points.

(b)

11. (a) Let us call a forward step “a success” and a backward step “a failure”.

Let X = no. of forward steps. Then X has binomial distribution with n = 11 and probability of success p = 0.4.

Required probability =P(X=6)+P(X=5)

(b) , n = 10

\ m = np = 0.02, e^-0.2 = 0.9802

(i) Probability of no defective blade = P (X=0)

= e^–m = 0.9802 (approximately)

\ Mean number of packets containing no defective blade is

= 10000 x 0.9802 = 9802

(ii) The mean number of packets containing one defective blade

= 10000 x me^–m = 196

(iii) The mean number of packets containing two defective blades

= 10000 x

= 2

____________________________________________

Detailed Solution A-01/C-01/T-01 June 2004

1. (a) A The matrix is singular if its determinant is zero. Solving determinant, we

get equations x(x-3)²=0.

(b) B Because

(c) D As

Since it is an odd function.

(d) B As

(e) A Dividing by z, we get

,

Let 1/(log z)=u, then above differential equation becomes

(f) C The given function has a pole at z=1, which lies outside the circle C. So

by Cauchy’s theorem integral is zero.

(g) D By orthogonal property of Legendre’s polynomial.

(h) C Probability of getting head=1/2= probability of getting tail.

If A starts the game, then in first chance either A wins the game, in second case A fails, B fails and A won the match and so on, we get an infinite series. Let H_A, H_B, T_A, T_B, denotes the getting of head and tails by A and B respectively.

P(wining of A)=P(H_A)+P(T_AT_BH_A)+ P(T_AT_BT_AT_BH_A)+…….

=

This is an infinite G.P. series with common ratio 1/4. Thus

P(winning of A) = .

2. (a) It is given that x^xy^yz^z=c, taking log we get

x log x+ y log y +z log z=log c

Differentiating the above equation w.r.t. y respectively, we get

, now differentiating w.r.t. x we get

At the point x=y=z, we get

2(b) Let edges of parallelopipid be 2x, 2y, 2z parallel to the coordinate axes. The volume V is given by V = 8xyz.

Let

Using Lagrange’s multiplier method, for maxima and minima, let

F = V + λ, where λ is a constant. For stationary values,

, = 0

3(a) On changing order of integration, the elementary strip has to be taken parallel to x

axis for which the region of integration has to be divided into two regions R₁and R₂. The region R₁: 0 ≤ x ≤ y, 0 ≤ y ≤ 1 and the region R₂: 0 ≤ x ≤ 2-y, 1 ≤ y ≤ 2.

= 2log2-1.

(b) The volume V is given as

4(i) The auxiliary equation is m² - 4m + 4 = 0, which gives m = - 2, -2.

Thus C.F. is given as (C₁ + C₂ x) e^-2x.

, since 2 is a root of order 2.

Therefore, general solution is

.

Let Y = y + k, X = x + h,

where h, k are so chosen so that

-7h+3k+7=0

-3h+7k+3=0

Solving, we get h = 1, k = 0. Let Y = VX,

Integrating, we get

5(a) The augmented matrix is given as

(A : B) =

Applying R₃ àR₃ – 2R₁,

If l ¹ 7, rank (A:B) = 3 ¹ rank A = 2. Hence system is inconsistent.

If l = 7, rank (A:B) = 2 = rank A . Hence system is consistent. The solution is obtained as follows :

Set x₃ = t₁, x₄ = t₂, t₁, t₂, arbitrary, then

x₂ = 4t₁- t₂+ 1, x₁ = 3t₁+ t₂+ 3.

5(b) The characteristic equation of A is |A-lI| = 0

ð l³ - 5l² + 7l – 3 = 0, is the characteristic equation.

By Cayley Hamilton theorem, A must satisfy this equation i.e.

A³ – 5A² + 7A – 3I = 0

Thus A⁸ – 5A⁷ +7A⁶ -3A⁵ +A⁴ – 5A³ + 8A² – 2A +I

= (A⁵ +A) (A³ –5 A² + 7A - 3I) + A² +A + I = A² + A + I

=

vector is solenoidal.

Therefore is irrotational. Thus , where is a scalar function.

6(b)

Unit normal to the plane 2x + 2y + z = 6 is along the vector , is given as .

Thus projection of given plane z=6-2x-2y on z=0 is region bounded by x=0, y=0,

y=3-x.

7(a) Let p₁ , p₂ , p₃ be the probabilities that thesis is approved by examiner A,B,C

p₁ = 5/7, p₂ = 4/7, p₃ = 3/7. A majority approves thesis if atleast two examiners are favorable.

P = p₁p₂q₃ + p₁p₃q₂ + p₂p₃q₁+ p₁p₂p₃

= =209/343.

7(b)

=0.0126,

We know that probable error is given as (2/3) =0.0084.

8(a) Let Z=X(x)Y(y), then given differential equation becomes

X’’Y-2X’Y+XY’=0, where X’’,Y’, X’,Y’’ are first and second order derivatives.

8(b) Let U=X(x)T(t), then given differential equation becomes

. Condition (i) is satisfied.

If k² =0, X=ax+b, T=c, thus by condition (ii), we get a=0

Thus U = bc.

Now U= (A cos kx + B sin kx) C

By (ii) condition, we get B=0, kl=nπ. Thus

C is solution for all n.

Since x-x² =

9 (a) Since x = 0 is a regular singular point

Let

Now the given differential equation becomes

The terms with lowest power of x is x^r-1. Its coefficient equated to zero gives C₀r(r-2) = 0. Because C₀ ¹ 0

=> r = 0, r = 2

The coefficient of x^m+r is equated to zero gives

,………

9(b) We know

For n = 1, 2 , 3

10(a) If f(x) =∑c_nP_n(x), then

where we have used the fact that

10(b) Let , and f(z) = u + iv

since u is an analytic function, thus it must satisfies

C-R equations, thus

Using Milne’s Thomson method, Let x = z , y = 0

\ where c is an arbitrary constant.

11(a)

For the real axis in the z plane y=0, i.e. =1, Also <1 implies y>0. hence the result.

11(b) , since 3.5 is a point which lies outside C, thus F(3.5) = 0 by Cauchy theorem.

Also -1lies within C, by Cauchy Integral Formula

.

Detailed Solutions A-01/C-01/T-01 December 2004

1. (a) D Let y= mx² be equation of curve. As x→0, y also tends to zero.

=

, which depends on m.

Thus it does not exist.

(b) A

Eliminating we get

(c) B f (x, y) = y² – x³

f_x = - 3x² = 0 , f_y = 2y = 0

gives (0,0) is a critical point.

∆f (x, y) = f(∆x,∆y)= (∆y)² –(∆ x)³

> 0 , if (∆y)² >(∆ x)³

< 0 , if (∆y)² < (∆ x)³

This means in the neighborhood of (0,0) f changes sign. Thus (0,0) is neither a point of maximum nor minimum.

(d) A y = (x – k)² Diff. w.r.t. x

y₁ = 2(x – k) => y₁ = 2

For orthogonal trajectories y₁ is replaced by -1/y₁.

Therefore, -1/y₁ = 2

=> 2 dy + dx = 0

Integrating, we get y^3/2 = ¾ (c-x)

(e) B yy” – (y’)² = 0

Because, y₁, y₂ are solutions

Therefore, y₁y₁” – (y₁’)² = 0

y₂y₂” – (y₂’)² = 0

Now (c₁y₁ + c₂y₂) (c₁y₁ + c₂y₂)” – ((c₁y₁ + c₂y₂)’)²

= (c₁y₁ + c₂y₂) (c₁y₁” + c₂y₂”) – (c₁y₁’² + c₂y₂’²) - 2 c₁y₁’c₂y₂’

= c₁²(y₁y₁” – (y₁’)²) + c₂² (y₂y₂” – (y₂’)²) + c₁ c₂ (y₁ y₂”+y₂y₁” - 2y₁’y₂’)

= 0, if c₁c₂ = 0.

(f) B Since A, B are square matrix of order n s.t. AB = 0, then rank of both A and B is less than n.

(g) D Because i is one eigen value so another eigen value must be – i.

(h) C Let I =

Let cosq = t. –sinqdq = dt

= 0

2. (b) Because x + y = 2e^qcosj, x – y = 2ie^qsinj

Adding & Subtracting, we get

2x = 2e^q (cosj + isinj) = 2e^{q + ij}

=> x = e^{q + ij}

Similarly y = e^{q - ij}

Let v = f (x, y), x = g (q, j), y = h (q, j)

3. (a) f(x, y) = x^y , f(1, 1) = 1

f_x (x, y) = yx^y-1 , f_x(1, 1) = 1

f_y (x, y) = x^ylog x , f_y(1, 1) = 0

(x, y) = y (y-1) x^y-2 , (1, 1) = 0

(x, y) = x^y (log x)² , (1, 1) = 0

f_xy (x, y) = x^y-1 + yx^y-1 log x , f_xy (1, 1) = 1

(x, y) = y (y-1) (y-2) x^y-3 , (1, 1) = 0

(x, y) = (2y-1)x^y-2 + y (y-1) x^y-2 log x , (1, 1) = 1

(x, y) = yx^y-1 (log x)² + 2 log x x^y-1 , (1, 1) = 0

By Taylor’s Theorem

3. (b) Let f = x² + y² + z² + xy + xz + zy

g = x + y + z – 1 = 0

h = x + 2y + 3z – 3 = 0

Let l₁, l₂ be two constants. Using Lagrange’s multiplier method, we get

F = f + l₁g + l₂h OR

F = x² + y² + z² + xy + xz + zy + l₁(x + y + z – 1) + l₂(x + 2y + 3z – 3)

For extreme values,

, x + y + z = 1, x + 2y + 3z = 3.

=> 2x + y + z + l₁ + l₂ = 0 => x + l₁ + l₂ + 1 = 0

2y + x + z + l₁ + 2l₂ = 0 y + l₁ + 2l₂ + 1 = 0 (A)

2z + x + y + l₁ + 3l₂ = 0 z + l₁ + 3l₂ + 1 = 0

Adding (A) and using x + y + z = 1, we get

3l₁ + 6l₂ + 4 = 0

Multiplying equation (ii) of ‘A’ by 2 and (iii) by 3 and adding all and using

x +2 y +3 z = 1, we get 6l₁ + 14l₂ + 9 = 0

Solving, 3l₁ + 6l₂ + 4 = 0

6l₁ + 14l₂ + 9 = 0, we get

l₁ = –1 /3, l₂ = –1 /2

From (A), we get

x = –1/6, y = 1/3, z = 5/6

Therefore, (–1/6, 1/3, 5/6) is a point of extremum, with extreme value

F(–1/6, 1/3, 5/6)= (-1/6)² + (1/3)² + (5/6)² – 1/6*1/3 – 1/6 * 5/6 + 1/3 * 5/6 =11/12

4. (a) Applying R₁ R₃, R₂ R₄

R₃ R₄

R₃ àR₃ – 3R₁, R₄ à R₄ – 9R₁

R₄ àR₄ – R₂, R₃ à R₂ – 2R₃

R₄ à 9R₃ + 5R₄

Thus |A| 0 Hence, rank of A = 4.

(b) Let

y₁ = 1.25 z₁ + 3 z₂ – 2.3 z₃

y₂ = 0.75 z₁ + 2 z₂ – 2.3 z₃

y₃ = 0.5 z₁ - z₂ + z₃

5. (a) Let A be a square matrix of order n.

Then |A – lI | = (-1)ⁿlⁿ + k₁l^n-2 + ----- + k_n = 0

where k’s are expressible in terms of elements a_ij of matrix A. The roots of this equation are eigen values of matrix A. Since this is n^th polynomial in l which has n distinct roots which are either real or complex conjugates.

Hence, eigen values of matrix are either real or complex conjugates.

If l is an eigen value of orthogonal matrix then 1/l is an eigen value of A^-1. Because A is an orthogonal matrix. Therefore A^-1 is same as A’.

Therefore 1/l is eigen value of A’. But A and A’ have same eigen values.

Hence, 1/l is also an eigen value of A. The product of eigen value of orthogonal matrix = 1 and hence if the order of A is odd it must have 1 as eigen value. Since product of eigen value of matrix A is equal to its determinant. Therefore |A| = ±1.

(b) |A – lI | = 0

=> l³ + l² – 21l - 45 = 0

=> l = 5, -3, - 3

Eigen values are 5, -3, -3

For l = 5, eigen vectors are obtained from

=> -7x + 2y - 3z = 0

2x – 4y – 6z = 0

-x –2y – 5z = 0

Solving we get,

i.e. (1, 2, -1)’ is an eigen vector

For l = -3, eigen vectors are obtained from

i.e. x +2y – 3z = 0

There are two linearly independent eigen vectors for l = -3. These are obtained by putting x = 0 and y = 0 respectively in the equation.

Let x = 0 then 2y – 3z = 0

i.e. is an eigen vector

Let y = 0, then x – 3z = 0

is an eigen vector.

Eigen vectors corresponding to 5, -3 , -3 are

[1, 2, -1]’ , [0, 3, 2]’ and [3, 0, 1]’.

6. (a) A =

| A - lI | = 0

=> l² - 7l + 6 = 0

=> l = 1, 6

For l = 1,

Therefore, x + y = 0

Eigen vector is (1, -1)’

For l = 6,

Therefore, x - 4y = 0

Eigen vector is (4, 1)’

Therefore, modal matrix is X = and

X^-1 =

D = X^-1AX = which is diagonal matrix

Also A = XDX^-1

A⁵⁰ = XD⁵⁰X^-1

X^-1

(b) The system of equation can be written as AX = B

Rank(A) = Rank(A,B) = 3. Thus system is consistent. Homogeneous system AX=0, has 5 – 3 = 2 linearly independent solutions. Clearly ( , -1, 3, 1, 0),

( , 1, -2, 0, 1) are linearly independent and satisfy the homogeneous system

AX = 0. Also ( , 2, 0, 0, 0) is a particular solution of non-homogeneous system AX = B. Thus general solution of non homogeneous system is

(x₁, x₂, x₃, x₄, x₅) = ( , 2, 0, 0, 0) + a ( , -1, 3, 1, 0) + b ( , 1, -2, 0, 1), where a, b are arbitrary.

For I₁, region of integration is bonded by the lines x = 1, x = 2, y=0 y = 1 i.e. region R₁ in figure. For I₂, region of integration is bonded by the lines x = y,

x = 2, y = 1, y = 2 i.e. region R₂ in figure.

Now the region R of integration i.e. union of R₁ and R₂ is bonded by the lines

y = 0, y = x, x = 1, x = 2

.

(b) Let V be the volume of solid. The two surfaces intersect at z = 1. Therefore

Let . Then dydx = rdrdθ, r varies from 0 to and θ varies from 0 to 2π. Then

8. (a) Since m be an integrating factor for differential equation Mdx + Ndy = 0. Thus m(Mdx + Ndy) = 0 is an exact differential equation.

Also dj = m (Mdx + Ndy) (given)

Because, j = constant, is a solution.

Let G(j) be any function of j

Therefore G(j)dj = m G(j) (Mdx + Ndy).

Let _<