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

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

2

If the system of linear equations

x + ky + 3z = 0

3x + ky - 2z = 0

2x + 4y - 3z = 0

has a non-zero solution (x, y, z), then $${{xz} \over {{y^2}}}$$ is equal to

x + ky + 3z = 0

3x + ky - 2z = 0

2x + 4y - 3z = 0

has a non-zero solution (x, y, z), then $${{xz} \over {{y^2}}}$$ is equal to

A

30

B

-10

C

10

D

-30

System of equations has non-zero solution when determinant of coefficient = 0.

So, in this questions,

$$\left| {\matrix{ 1 & K & 3 \cr 3 & K & { - 2} \cr 2 & 4 & { - 3} \cr } } \right| = 0$$

$$ \Rightarrow \,\,\,\,$$ ($$-$$ 3K + 8) $$-$$ K ($$-$$9 + 4) + 3(12 $$-$$ 2K) = 0

$$ \Rightarrow \,\,\,\,$$ $$-$$ 3K + 8 + 9K $$-$$ 4K + 36 $$-$$ 6K = 0

$$ \Rightarrow \,\,\,\,$$ $$-$$ 4K + 44 = 0

$$ \Rightarrow \,\,\,\,$$ K = 11

Now the equations become

x + 11y + 3z = 0 . . . (1)

3x + 11y $$-$$ 2z = 0 . . . (2)

2x + 4y $$-$$ 3z = 0 . . . (3)

By adding equation (1) and (3) we get,

3x + 15y = 0

$$ \Rightarrow \,\,\,\,$$ x = $$-$$ 5y

Putting x = $$-$$ 5y in equation (1) we get

$$-$$ 5y + 11y + 3z = 0

$$ \Rightarrow \,\,\,\,$$ 6y + 3z = 0

$$ \Rightarrow \,\,\,\,$$ z = $$-$$ 2y

$$\therefore\,\,\,\,$$ $${{xz} \over {{y^2}}}$$

$$ = {{\left( { - 5y} \right)\left( { - 2y} \right)} \over {{y^2}}}$$

$$ = {{10{y^2}} \over {{y^2}}}$$

$$ = 10$$

So, in this questions,

$$\left| {\matrix{ 1 & K & 3 \cr 3 & K & { - 2} \cr 2 & 4 & { - 3} \cr } } \right| = 0$$

$$ \Rightarrow \,\,\,\,$$ ($$-$$ 3K + 8) $$-$$ K ($$-$$9 + 4) + 3(12 $$-$$ 2K) = 0

$$ \Rightarrow \,\,\,\,$$ $$-$$ 3K + 8 + 9K $$-$$ 4K + 36 $$-$$ 6K = 0

$$ \Rightarrow \,\,\,\,$$ $$-$$ 4K + 44 = 0

$$ \Rightarrow \,\,\,\,$$ K = 11

Now the equations become

x + 11y + 3z = 0 . . . (1)

3x + 11y $$-$$ 2z = 0 . . . (2)

2x + 4y $$-$$ 3z = 0 . . . (3)

By adding equation (1) and (3) we get,

3x + 15y = 0

$$ \Rightarrow \,\,\,\,$$ x = $$-$$ 5y

Putting x = $$-$$ 5y in equation (1) we get

$$-$$ 5y + 11y + 3z = 0

$$ \Rightarrow \,\,\,\,$$ 6y + 3z = 0

$$ \Rightarrow \,\,\,\,$$ z = $$-$$ 2y

$$\therefore\,\,\,\,$$ $${{xz} \over {{y^2}}}$$

$$ = {{\left( { - 5y} \right)\left( { - 2y} \right)} \over {{y^2}}}$$

$$ = {{10{y^2}} \over {{y^2}}}$$

$$ = 10$$

3

If $$\left| {\matrix{
{x - 4} & {2x} & {2x} \cr
{2x} & {x - 4} & {2x} \cr
{2x} & {2x} & {x - 4} \cr
} } \right| = \left( {A + Bx} \right){\left( {x - A} \right)^2}$$

then the ordered pair (A, B) is equal to

then the ordered pair (A, B) is equal to

A

(4, 5)

B

(-4, -5)

C

(-4, 3)

D

(-4, 5)

$$\left| {\matrix{
{x - 4} & {2x} & {2x} \cr
{2x} & {x - 4} & {2x} \cr
{2x} & {2x} & {x - 4} \cr
} } \right|$$

Applying c_{1} $$ \to $$ c_{1} + c_{2} + c_{3}

$$ = \,\,\,\,\left| {\matrix{ {5x - 4} & {2x} & {2x} \cr {5x - 4} & {x - 4} & {2x} \cr {5x - 4} & {2x} & {x - 4} \cr } } \right|$$

Taking common (5x $$-$$ 4) from c_{1}

$$ = \,\,\,\,\left( {5x - 4} \right)\left| {\matrix{ 1 & {2x} & {2x} \cr 1 & {x - 4} & {2x} \cr 1 & {2x} & {x - 4} \cr } } \right|$$

Apply R_{2} $$ \to $$R_{2} $$-$$ R_{1} and R_{3} $$ \to $$R_{3} $$-$$ R_{1}

$$ = \,\,\,\,\left( {5x - 4} \right)\left| {\matrix{ 1 & {2x} & {2x} \cr 0 & { - \left( {x + 4} \right)} & 0 \cr 0 & 0 & { - \left( {x + 4} \right)} \cr } } \right|$$

$$ = \,\,\,\,\left( {5x - 4} \right){\left( {x + 4} \right)^2}$$

So, (A + Bx) (x $$-$$ A)^{2} = (5x $$-$$ 4) (x + 4)^{2}

By comparing both sides we get, A = $$-$$ 4 and B = 5

Applying c

$$ = \,\,\,\,\left| {\matrix{ {5x - 4} & {2x} & {2x} \cr {5x - 4} & {x - 4} & {2x} \cr {5x - 4} & {2x} & {x - 4} \cr } } \right|$$

Taking common (5x $$-$$ 4) from c

$$ = \,\,\,\,\left( {5x - 4} \right)\left| {\matrix{ 1 & {2x} & {2x} \cr 1 & {x - 4} & {2x} \cr 1 & {2x} & {x - 4} \cr } } \right|$$

Apply R

$$ = \,\,\,\,\left( {5x - 4} \right)\left| {\matrix{ 1 & {2x} & {2x} \cr 0 & { - \left( {x + 4} \right)} & 0 \cr 0 & 0 & { - \left( {x + 4} \right)} \cr } } \right|$$

$$ = \,\,\,\,\left( {5x - 4} \right){\left( {x + 4} \right)^2}$$

So, (A + Bx) (x $$-$$ A)

By comparing both sides we get, A = $$-$$ 4 and B = 5

4

Let S be the set of all real values of k for which the systemof linear equations

x + y + z = 2

2x + y $$-$$ z = 3

3x + 2y + kz = 4

has a unique solution. Then S is :

x + y + z = 2

2x + y $$-$$ z = 3

3x + 2y + kz = 4

has a unique solution. Then S is :

A

an empty set

B

equal to {0}

C

equal to **R**

D

equal to **R** $$-$$ {0}

As system of linear equations have unique solutions so, determinant of coefficient $$ \ne $$ 0

$$ \therefore $$ $$\left| {\matrix{ 1 & 1 & 1 \cr 2 & 1 & { - 1} \cr 3 & 2 & k \cr } } \right|$$ $$ \ne $$ 0

$$ \Rightarrow $$ k + 2 - (2k + 3) + 1 $$ \ne $$ 0

$$ \Rightarrow $$ k $$ \ne $$ 0

$$ \therefore $$ k $$ \in $$ R - {0}

$$ \therefore $$ $$\left| {\matrix{ 1 & 1 & 1 \cr 2 & 1 & { - 1} \cr 3 & 2 & k \cr } } \right|$$ $$ \ne $$ 0

$$ \Rightarrow $$ k + 2 - (2k + 3) + 1 $$ \ne $$ 0

$$ \Rightarrow $$ k $$ \ne $$ 0

$$ \therefore $$ k $$ \in $$ R - {0}

Number in Brackets after Paper Name Indicates No of Questions

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Trigonometric Functions & Equations *keyboard_arrow_right*

Properties of Triangle *keyboard_arrow_right*

Inverse Trigonometric Functions *keyboard_arrow_right*

Complex Numbers *keyboard_arrow_right*

Quadratic Equation and Inequalities *keyboard_arrow_right*

Permutations and Combinations *keyboard_arrow_right*

Mathematical Induction and Binomial Theorem *keyboard_arrow_right*

Sequences and Series *keyboard_arrow_right*

Matrices and Determinants *keyboard_arrow_right*

Vector Algebra and 3D Geometry *keyboard_arrow_right*

Probability *keyboard_arrow_right*

Statistics *keyboard_arrow_right*

Mathematical Reasoning *keyboard_arrow_right*

Functions *keyboard_arrow_right*

Limits, Continuity and Differentiability *keyboard_arrow_right*

Differentiation *keyboard_arrow_right*

Application of Derivatives *keyboard_arrow_right*

Indefinite Integrals *keyboard_arrow_right*

Definite Integrals and Applications of Integrals *keyboard_arrow_right*

Differential Equations *keyboard_arrow_right*

Straight Lines and Pair of Straight Lines *keyboard_arrow_right*

Circle *keyboard_arrow_right*

Conic Sections *keyboard_arrow_right*