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

Let A be any 3 $$ \times $$ 3 invertible matrix. Then which one of the following is **not** always true ?

A

adj (A) = $$\left| \right.$$A$$\left| \right.$$.A^{$$-$$1}

B

adj (adj(A)) = $$\left| \right.$$A$$\left| \right.$$.A

C

adj (adj(A)) = $$\left| \right.$$A$$\left| \right.$$^{2}.(adj(A))^{$$-$$1}

D

adj (adj(A)) = $$\left| \, \right.$$A $$\left| \, \right.$$.(adj(A))^{$$-$$1}

We know, the formula

A^{-1} = $${{adj\left( A \right)} \over {\left| A \right|}}$$

$$ \therefore $$ adj (A) = $$\left| \right.$$A$$\left| \right.$$.A^{$$-$$1}

**So, Option (A) is true.**

We know, the formula

adj (adj (A)) = $${\left| A \right|^{n - 2}}.A$$

Now if we put n = 3 as given that A is a 3 $$ \times $$ 3 matrix, we get

adj (adj (A)) = $${\left| A \right|^{3 - 2}}.A$$ = $$\left| A \right|.A$$

**So, Option (B) is also true.**

We know, the formula

adj (adj (A)) = $${\left| A \right|^{n - 1}}{\left( {adj\left( A \right)} \right)^{ - 1}}$$

Now if we put n = 3 as given that A is a 3 $$ \times $$ 3 matrix, we get

adj (adj (A)) = $${\left| A \right|^{3 - 1}}{\left( {adj\left( A \right)} \right)^{ - 1}}$$ = $${\left| A \right|^{2}}{\left( {adj\left( A \right)} \right)^{ - 1}}$$

**So, Option (C) is also true.**

Now in this formula

adj (adj (A)) = $${\left| A \right|^{n - 1}}{\left( {adj\left( A \right)} \right)^{ - 1}}$$

if we put n = 2, we get

adj (adj (A)) = $${\left| A \right|^{2 - 1}}{\left( {adj\left( A \right)} \right)^{ - 1}}$$ = $${\left| A \right|}{\left( {adj\left( A \right)} \right)^{ - 1}}$$

But as A is a 3 $$ \times $$ 3 matrix so we can not take n = 2, so we can say for a 3 $$ \times $$ 3 matrix option (D) is not true.

**So, Option (D) is false.**

A

$$ \therefore $$ adj (A) = $$\left| \right.$$A$$\left| \right.$$.A

We know, the formula

adj (adj (A)) = $${\left| A \right|^{n - 2}}.A$$

Now if we put n = 3 as given that A is a 3 $$ \times $$ 3 matrix, we get

adj (adj (A)) = $${\left| A \right|^{3 - 2}}.A$$ = $$\left| A \right|.A$$

We know, the formula

adj (adj (A)) = $${\left| A \right|^{n - 1}}{\left( {adj\left( A \right)} \right)^{ - 1}}$$

Now if we put n = 3 as given that A is a 3 $$ \times $$ 3 matrix, we get

adj (adj (A)) = $${\left| A \right|^{3 - 1}}{\left( {adj\left( A \right)} \right)^{ - 1}}$$ = $${\left| A \right|^{2}}{\left( {adj\left( A \right)} \right)^{ - 1}}$$

Now in this formula

adj (adj (A)) = $${\left| A \right|^{n - 1}}{\left( {adj\left( A \right)} \right)^{ - 1}}$$

if we put n = 2, we get

adj (adj (A)) = $${\left| A \right|^{2 - 1}}{\left( {adj\left( A \right)} \right)^{ - 1}}$$ = $${\left| A \right|}{\left( {adj\left( A \right)} \right)^{ - 1}}$$

But as A is a 3 $$ \times $$ 3 matrix so we can not take n = 2, so we can say for a 3 $$ \times $$ 3 matrix option (D) is not true.

2

The number of real values of $$\lambda $$ for which the system of linear equations

2x + 4y $$-$$ $$\lambda $$z = 0

4x + $$\lambda $$y + 2z = 0

$$\lambda $$x + 2y + 2z = 0

has infinitely many solutions, is :

2x + 4y $$-$$ $$\lambda $$z = 0

4x + $$\lambda $$y + 2z = 0

$$\lambda $$x + 2y + 2z = 0

has infinitely many solutions, is :

A

0

B

1

C

2

D

3

3

If

$$S = \left\{ {x \in \left[ {0,2\pi } \right]:\left| {\matrix{ 0 & {\cos x} & { - \sin x} \cr {\sin x} & 0 & {\cos x} \cr {\cos x} & {\sin x} & 0 \cr } } \right| = 0} \right\},$$

then $$\sum\limits_{x \in S} {\tan \left( {{\pi \over 3} + x} \right)} $$ is equal to :

$$S = \left\{ {x \in \left[ {0,2\pi } \right]:\left| {\matrix{ 0 & {\cos x} & { - \sin x} \cr {\sin x} & 0 & {\cos x} \cr {\cos x} & {\sin x} & 0 \cr } } \right| = 0} \right\},$$

then $$\sum\limits_{x \in S} {\tan \left( {{\pi \over 3} + x} \right)} $$ is equal to :

A

$$4 + 2\sqrt 3 $$

B

$$ - 2 + \sqrt 3 $$

C

$$ - 2 - \sqrt 3 $$

D

$$-\,\,4 - 2\sqrt 3 $$

Given,

$$\left| {\matrix{ 0 & {\cos x} & { - \sin x} \cr {\sin x} & 0 & {\cos x} \cr {\cos x} & {\sin x} & 0 \cr } } \right|$$ = 0

$$ \Rightarrow $$$$\,\,\,$$ 0 (0 $$-$$ cosx sinx) $$-$$ cosx (0 $$-$$ cos^{2}x) $$-$$ sinx(sin^{2}x) = 0

$$ \Rightarrow $$$$\,\,\,$$ cos^{3}x $$-$$ sin^{3} x = 0

$$ \Rightarrow $$$$\,\,\,$$ tan^{3}x = 1

$$ \Rightarrow $$$$\,\,\,$$ tanx = 1

$$ \therefore $$ $$\,\,\,$$ $$\sum\limits_{x\, \in \,\,S} {\,\tan \left( {{\pi \over 3} + x} \right)} $$

= $${{\tan {\pi \over 3} + \tan x} \over {1 - \tan {\pi \over 3}\tan x}}$$

= $${{\sqrt 3 + 1} \over {1 - \sqrt 3 }}$$

= $${{\left( {\sqrt 3 + 1} \right)} \over {\left( {1 - \sqrt 3 } \right)}} \times {{1 + \sqrt 3 } \over {1 + \sqrt 3 }}$$

= $${{1 + 3 + 2\sqrt 3 } \over { - 2}}$$

= $$-$$ 2 $$-$$ $$\sqrt 3 $$

$$\left| {\matrix{ 0 & {\cos x} & { - \sin x} \cr {\sin x} & 0 & {\cos x} \cr {\cos x} & {\sin x} & 0 \cr } } \right|$$ = 0

$$ \Rightarrow $$$$\,\,\,$$ 0 (0 $$-$$ cosx sinx) $$-$$ cosx (0 $$-$$ cos

$$ \Rightarrow $$$$\,\,\,$$ cos

$$ \Rightarrow $$$$\,\,\,$$ tan

$$ \Rightarrow $$$$\,\,\,$$ tanx = 1

$$ \therefore $$ $$\,\,\,$$ $$\sum\limits_{x\, \in \,\,S} {\,\tan \left( {{\pi \over 3} + x} \right)} $$

= $${{\tan {\pi \over 3} + \tan x} \over {1 - \tan {\pi \over 3}\tan x}}$$

= $${{\sqrt 3 + 1} \over {1 - \sqrt 3 }}$$

= $${{\left( {\sqrt 3 + 1} \right)} \over {\left( {1 - \sqrt 3 } \right)}} \times {{1 + \sqrt 3 } \over {1 + \sqrt 3 }}$$

= $${{1 + 3 + 2\sqrt 3 } \over { - 2}}$$

= $$-$$ 2 $$-$$ $$\sqrt 3 $$

4

For two 3 × 3 matrices A and B, let A + B = 2B^{T} and 3A + 2B = I_{3}, where B^{T} is
the transpose of B and I_{3} is 3 × 3 identity matrix. Then :

A

5A + 10B = 2I_{3}

B

10A + 5B = 3I_{3}

C

B + 2A = I_{3}

D

3A + 6B = 2I_{3}

Given, A + B = 2B^{T} .......(1)

$$ \Rightarrow $$ (A + B)^{T} = (2B^{T})^{T}

$$ \Rightarrow $$ A^{T} + B^{T} = 2B

$$ \Rightarrow $$ B = $${{{A^T} + {B^T}} \over 2}$$

Now put this in equation (1)

So, A + $${{{A^T} + {B^T}} \over 2}$$ = 2B^{T}

$$ \Rightarrow $$2A + A^{T} = 3B^{T}

$$ \Rightarrow $$ A = $${{3{B^T} - {A^T}} \over 2}$$

Also, 3A + 2B = I_{3} .......(2)

$$ \Rightarrow $$ $$3\left( {{{3{B^T} - {A^T}} \over 2}} \right) + 2\left( {{{{A^T} + {B^T}} \over 2}} \right)$$ = I_{3}

$$ \Rightarrow $$ 11B^{T} - A^{T} = 2I_{3}

$$ \Rightarrow $$ (11B^{T} - A^{T})^{T} = (2I_{3})^{T}

$$ \Rightarrow $$ 11B - A = 2I_{3} ........(3)

Multiply (3) by 3 and then adding (2) and (3) we get,

35B = 7I_{3}

$$ \Rightarrow $$ B = $${{{I_3}} \over 5}$$

From (3), 11$${{{I_3}} \over 5}$$ - A = 2I_{3}

$$ \Rightarrow $$ A = $${{{I_3}} \over 5}$$

$$ \therefore $$ 5A = 5B = I_{3}

$$ \Rightarrow $$ 10A + 5B = 3I_{3}

$$ \Rightarrow $$ (A + B)

$$ \Rightarrow $$ A

$$ \Rightarrow $$ B = $${{{A^T} + {B^T}} \over 2}$$

Now put this in equation (1)

So, A + $${{{A^T} + {B^T}} \over 2}$$ = 2B

$$ \Rightarrow $$2A + A

$$ \Rightarrow $$ A = $${{3{B^T} - {A^T}} \over 2}$$

Also, 3A + 2B = I

$$ \Rightarrow $$ $$3\left( {{{3{B^T} - {A^T}} \over 2}} \right) + 2\left( {{{{A^T} + {B^T}} \over 2}} \right)$$ = I

$$ \Rightarrow $$ 11B

$$ \Rightarrow $$ (11B

$$ \Rightarrow $$ 11B - A = 2I

Multiply (3) by 3 and then adding (2) and (3) we get,

35B = 7I

$$ \Rightarrow $$ B = $${{{I_3}} \over 5}$$

From (3), 11$${{{I_3}} \over 5}$$ - A = 2I

$$ \Rightarrow $$ A = $${{{I_3}} \over 5}$$

$$ \therefore $$ 5A = 5B = I

$$ \Rightarrow $$ 10A + 5B = 3I

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