Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geomatics Engineering Or Surveying

Engineering Mechanics

Hydrology

Transportation Engineering

Strength of Materials Or Solid Mechanics

Reinforced Cement Concrete

Steel Structures

Irrigation

Environmental Engineering

Engineering Mathematics

Structural Analysis

Geotechnical Engineering

Fluid Mechanics and Hydraulic Machines

General Aptitude

1

The number of values of k for which the system of linear equations,

(k + 2)x + 10y = k

kx + (k +3)y = k -1

has**no solution,** is :

(k + 2)x + 10y = k

kx + (k +3)y = k -1

has

A

1

B

2

C

3

D

infinitely many

System of linear equation have no solution,

$$\therefore\,\,\,$$ determinant of coefficient = 0

$$\left| {\matrix{ {k + 2} & {10} \cr k & {k + 3} \cr } } \right| = 0$$

$$ \Rightarrow $$ $$\,\,\,\,$$ (k + 2) (k + 3) $$-$$ 10 K = 0

$$ \Rightarrow $$ $$\,\,\,\,$$ k^{2} $$-$$ 5k + 6 = 0

$$\therefore\,\,\,\,$$ k = 2, 3

When, k = 2 then equations become,

4x + 10y = 2

and 2x + 5y = 1

It has in finite number of solutions.

When k = 3, equations becomes

5x + 10y = 3

3x + 6y = 2

Those equation has no solutions.

$$\therefore\,\,\,\,$$ When k = 3, then system of equations have no solutions.

$$\therefore\,\,\,$$ determinant of coefficient = 0

$$\left| {\matrix{ {k + 2} & {10} \cr k & {k + 3} \cr } } \right| = 0$$

$$ \Rightarrow $$ $$\,\,\,\,$$ (k + 2) (k + 3) $$-$$ 10 K = 0

$$ \Rightarrow $$ $$\,\,\,\,$$ k

$$\therefore\,\,\,\,$$ k = 2, 3

When, k = 2 then equations become,

4x + 10y = 2

and 2x + 5y = 1

It has in finite number of solutions.

When k = 3, equations becomes

5x + 10y = 3

3x + 6y = 2

Those equation has no solutions.

$$\therefore\,\,\,\,$$ When k = 3, then system of equations have no solutions.

2

Let A = $$\left[ {\matrix{
1 & 0 & 0 \cr
1 & 1 & 0 \cr
1 & 1 & 1 \cr
} } \right]$$ and B = A^{20}. Then the sum of the elements of the first column of B is :

A

210

B

211

C

231

D

251

=

A = $$\left[ {\matrix{
1 & 0 & 0 \cr
1 & 1 & 0 \cr
1 & 1 & 1 \cr
} } \right]$$

A^{2} = A.A = $$\left[ {\matrix{
1 & 0 & 0 \cr
1 & 1 & 0 \cr
1 & 1 & 1 \cr
} } \right] \times \left[ {\matrix{
1 & 0 & 0 \cr
1 & 1 & 0 \cr
1 & 1 & 1 \cr
} } \right]$$

= $$\left[ {\matrix{ 1 & 0 & 0 \cr 2 & 1 & 0 \cr 3 & 2 & 1 \cr } } \right]$$

A^{3} = A^{2}.A = $$\left[ {\matrix{
1 & 0 & 0 \cr
2 & 1 & 0 \cr
3 & 2 & 1 \cr
} } \right] \times \left[ {\matrix{
1 & 0 & 0 \cr
1 & 1 & 0 \cr
1 & 1 & 1 \cr
} } \right]$$

= $$\left[ {\matrix{ 1 & 0 & 0 \cr 3 & 1 & 0 \cr 6 & 3 & 1 \cr } } \right]$$

Similarly

A^{4} = $$\left[ {\matrix{
1 & 0 & 0 \cr
4 & 1 & 0 \cr
{10} & 4 & 1 \cr
} } \right]$$

From this we can say,

A^{n} = $$\left[ {\matrix{
1 & 0 & 0 \cr
n & 1 & 0 \cr
{{{n\left( {n + 1} \right)} \over 2}} & n & 1 \cr
} } \right]$$

$$\therefore\,\,\,$$ A^{20} = $$\left[ {\matrix{
1 & 0 & 0 \cr
{20} & 1 & 0 \cr
{210} & {20} & 1 \cr
} } \right]$$

$$\therefore\,\,\,$$ Sum of the first column

= 1 + 20 + 210

= 231

A

= $$\left[ {\matrix{ 1 & 0 & 0 \cr 2 & 1 & 0 \cr 3 & 2 & 1 \cr } } \right]$$

A

= $$\left[ {\matrix{ 1 & 0 & 0 \cr 3 & 1 & 0 \cr 6 & 3 & 1 \cr } } \right]$$

Similarly

A

From this we can say,

A

$$\therefore\,\,\,$$ A

$$\therefore\,\,\,$$ Sum of the first column

= 1 + 20 + 210

= 231

3

If $$A = \left[ {\matrix{
{\cos \theta } & { - \sin \theta } \cr
{\sin \theta } & {\cos \theta } \cr
} } \right]$$, then the matrix A^{–50} when $$\theta $$ = $$\pi \over 12$$, is equal to :

A

$$\left[ {\matrix{
{ {{\sqrt 3 } \over 2}} & { - {1 \over 2}} \cr
{{{ 1} \over 2}} & {{{\sqrt 3 } \over 2}} \cr
} } \right]$$

B

$$\left[ {\matrix{
{{1 \over 2}} & -{{{\sqrt 3 } \over 2}} \cr
{{{\sqrt 3 } \over 2}} & {{{ - 1} \over 2}} \cr
} } \right]$$

C

$$\left[ {\matrix{
{{{\sqrt 3 } \over 2}} & {{1 \over 2}} \cr
-{{1 \over 2}} & {{{\sqrt 3 } \over 2}} \cr
} } \right]$$

D

$$\left[ {\matrix{
{{1 \over 2}} & {{{\sqrt 3 } \over 2}} \cr
{-{{\sqrt 3 } \over 2}} & {{{ 1} \over 2}} \cr
} } \right]$$

(A^{$$-$$50}) = (A^{$$-$$1})^{50}

We know,

A^{$$-$$1} = $${{adjA} \over {\left| A \right|}}$$

$$\left| A \right|$$ = cos^{2}$$\theta $$ + sin^{2}$$\theta $$ = 1

cofactor of A = $$\left[ {\matrix{ {\cos \theta } & { - \sin \theta } \cr {\sin \theta } & {\cos \theta } \cr } } \right]$$

Adjoint of A = Transpose of cofactor matrix

$$ \therefore $$ Adj A = $$\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$$

$$ \therefore $$ A^{$$-$$1} = $$\left[ {\matrix{
{\cos \theta } & {\sin \theta } \cr
{ - \sin \theta } & {\cos \theta } \cr
} } \right]$$

$$ \therefore $$ (A^{$$-$$1})^{2} = $$\left[ {\matrix{
{\cos \theta } & {\sin \theta } \cr
{ - \sin \theta } & {\cos \theta } \cr
} } \right]\left[ {\matrix{
{\cos \theta } & {\sin \theta } \cr
{ - \sin \theta } & {\cos \theta } \cr
} } \right]$$

= $$\left[ {\matrix{ {\cos 2\theta } & {\sin 2\theta } \cr { - \sin 2\theta } & {\cos 2\theta } \cr } } \right]$$

Similarly,

(A^{$$-$$1})^{3} = $$\left[ {\matrix{
{\cos 3\theta } & {\sin 2\theta } \cr
{ - \sin 3\theta } & {\cos 3\theta } \cr
} } \right]$$

:

:

:

(A^{$$-$$1})^{50} = $$\left[ {\matrix{
{\cos 50\theta } & {\sin 50\theta } \cr
{ - \sin 50\theta } & {\cos 50\theta } \cr
} } \right]$$

when $$\theta $$ = $${\pi \over {12}}$$ then

$${A^{ - 50}}$$ = $$\left[ {\matrix{ {\cos {{25\pi } \over 6}} & {\sin {{25\pi } \over 6}} \cr { - \sin {{25\pi } \over 6}} & {\cos {{25\pi } \over 6}} \cr } } \right]$$

= $$\left[ {\matrix{ {{{\sqrt 3 } \over 2}} & {{1 \over 2}} \cr { - {1 \over 2}} & {{{\sqrt 3 } \over 2}} \cr } } \right]$$

__Note:__

$$\cos {{25\pi } \over 6} = \cos \left( {4\pi + {\pi \over 6}} \right) = \cos {\pi \over 6} = {{\sqrt 3 } \over 2}$$

We know,

A

$$\left| A \right|$$ = cos

cofactor of A = $$\left[ {\matrix{ {\cos \theta } & { - \sin \theta } \cr {\sin \theta } & {\cos \theta } \cr } } \right]$$

Adjoint of A = Transpose of cofactor matrix

$$ \therefore $$ Adj A = $$\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$$

$$ \therefore $$ A

$$ \therefore $$ (A

= $$\left[ {\matrix{ {\cos 2\theta } & {\sin 2\theta } \cr { - \sin 2\theta } & {\cos 2\theta } \cr } } \right]$$

Similarly,

(A

:

:

:

(A

when $$\theta $$ = $${\pi \over {12}}$$ then

$${A^{ - 50}}$$ = $$\left[ {\matrix{ {\cos {{25\pi } \over 6}} & {\sin {{25\pi } \over 6}} \cr { - \sin {{25\pi } \over 6}} & {\cos {{25\pi } \over 6}} \cr } } \right]$$

= $$\left[ {\matrix{ {{{\sqrt 3 } \over 2}} & {{1 \over 2}} \cr { - {1 \over 2}} & {{{\sqrt 3 } \over 2}} \cr } } \right]$$

$$\cos {{25\pi } \over 6} = \cos \left( {4\pi + {\pi \over 6}} \right) = \cos {\pi \over 6} = {{\sqrt 3 } \over 2}$$

4

The system of linear equations

x + y + z = 2

2x + 3y + 2z = 5

2x + 3y + (a^{2} – 1) z = a + 1 then

x + y + z = 2

2x + 3y + 2z = 5

2x + 3y + (a

A

has infinitely many solutions for a = 4

B

has a unique solution for |a| = $$\sqrt3$$

C

is inconsistent when |a| = $$\sqrt3$$

D

is inconsistent when a = 4

$$D = \left| {\matrix{
1 & 1 & 1 \cr
2 & 3 & 2 \cr
2 & 3 & {{\alpha ^2} - 1} \cr
} } \right|$$

D = 3$$a$$^{2} $$-$$ 3 $$-$$ 6 $$-$$ 2$$a$$^{2} + 2 + 4 + 2$$a$$^{2} $$-$$ 2 $$-$$ 4

D = ($$a$$^{2} $$-$$ 3)

When D $$ \ne $$ 0 then system of equiation has unique solution.

$$ \therefore $$ 3($$a$$^{2} $$-$$ 3) $$ \ne $$ 0

$$ \Rightarrow $$ $$\left| a \right|$$ $$ \ne \sqrt 3 $$

When $$3({a^2} - 3) = 0$$

$$ \Rightarrow $$ $$\left| a \right| = \sqrt 3 $$ then D = 0

If D = 0 then two cases possible

(1) System of equation has infinite many solution.

(2) System of equation has no solution and inconsistent.

Here D_{1} = $$\left| {\matrix{
2 & 1 & 1 \cr
5 & 3 & 2 \cr
{a + 1} & 3 & {{a^2} - 1} \cr
} } \right| = {a^2} - a + 1$$

D_{2} = $$\left| {\matrix{
1 & 2 & 1 \cr
2 & 5 & 2 \cr
2 & {a + 1} & {{a^2} - 1} \cr
} } \right| = {a^2} - 3$$

D_{3} = $$\left| {\matrix{
1 & 1 & 2 \cr
2 & 3 & 5 \cr
2 & 3 & {a + 1} \cr
} } \right| = a - 4$$

System of equation will have infinite solution if D_{1} = D_{2} = D_{3} = 0.

And system of equation will have no solution if at last one of D_{1}, D_{2}, D_{2} is non zero.

At $$\left| a \right| = \sqrt 3 $$ we get D = 0

But D_{1} = 3 $$ \pm $$ $$\sqrt 3 + 1$$ $$ \ne $$ 0

and D_{3} = $$ \pm $$ $$\sqrt 3 - 4$$ $$ \ne $$ 0

So, system of equations has no solution at $$\left| a \right| = \sqrt 3 $$ then system is in consistent

D = 3$$a$$

D = ($$a$$

When D $$ \ne $$ 0 then system of equiation has unique solution.

$$ \therefore $$ 3($$a$$

$$ \Rightarrow $$ $$\left| a \right|$$ $$ \ne \sqrt 3 $$

When $$3({a^2} - 3) = 0$$

$$ \Rightarrow $$ $$\left| a \right| = \sqrt 3 $$ then D = 0

If D = 0 then two cases possible

(1) System of equation has infinite many solution.

(2) System of equation has no solution and inconsistent.

Here D

D

D

System of equation will have infinite solution if D

And system of equation will have no solution if at last one of D

At $$\left| a \right| = \sqrt 3 $$ we get D = 0

But D

and D

So, system of equations has no solution at $$\left| a \right| = \sqrt 3 $$ then system is in consistent

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