Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geomatics Engineering Or Surveying

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General Aptitude

1

If the system of linear equations

x $$-$$ 4y + 7z = g

3y $$-$$ 5z = h

$$-$$2x + 5y $$-$$ 9z = k

is consistent, then :

x $$-$$ 4y + 7z = g

3y $$-$$ 5z = h

$$-$$2x + 5y $$-$$ 9z = k

is consistent, then :

A

g + 2h + k = 0

B

g + h + 2k = 0

C

2g + h + k = 0

D

g + h + k = 0

x $$-$$ $$4y + 7z = g$$

$$3y$$ $$-$$ $$5z = h$$

$$-$$$$2x + 5y$$ $$-$$ $$9z = k$$

$$D = \left| {\matrix{ 1 & { - 4} & 7 \cr 0 & 3 & { - 5} \cr { - 2} & 5 & { - 9} \cr } } \right|$$

$$D = 1\left( { - 27 + 25} \right) - 2\left( {20 - 21} \right)$$

$$D = - 2 + 2 = 0$$

If system is consistent then $${D_1} = {D_2} = {D_3} = 0$$

$$\left| {\matrix{ 1 & { - 4} & g \cr 0 & 3 & h \cr { - 2} & 5 & k \cr } } \right| = 0$$

$$1\left( {3k - 5h} \right) - 2\left( { - 4h - 3g} \right) = 0$$

$$3k - 5h + 8h + 6g = 0$$

$$6g + 3h + 3k = 0$$

$$2g + h + k = 0$$

$$3y$$ $$-$$ $$5z = h$$

$$-$$$$2x + 5y$$ $$-$$ $$9z = k$$

$$D = \left| {\matrix{ 1 & { - 4} & 7 \cr 0 & 3 & { - 5} \cr { - 2} & 5 & { - 9} \cr } } \right|$$

$$D = 1\left( { - 27 + 25} \right) - 2\left( {20 - 21} \right)$$

$$D = - 2 + 2 = 0$$

If system is consistent then $${D_1} = {D_2} = {D_3} = 0$$

$$\left| {\matrix{ 1 & { - 4} & g \cr 0 & 3 & h \cr { - 2} & 5 & k \cr } } \right| = 0$$

$$1\left( {3k - 5h} \right) - 2\left( { - 4h - 3g} \right) = 0$$

$$3k - 5h + 8h + 6g = 0$$

$$6g + 3h + 3k = 0$$

$$2g + h + k = 0$$

2

If $$A = \left[ {\matrix{
{{e^t}} & {{e^{ - t}}\cos t} & {{e^{ - t}}\sin t} \cr
{{e^t}} & { - {e^{ - t}}\cos t - {e^{ - t}}\sin t} & { - {e^{ - t}}\sin t + {e^{ - t}}co{\mathop{\rm s}\nolimits} t} \cr
{{e^t}} & {2{e^{ - t}}\sin t} & { - 2{e^{ - t}}\cos t} \cr
} } \right]$$

then A is :

then A is :

A

invertible for all t$$ \in $$**R**.

B

invertible only if t $$=$$ $$\pi $$

C

not invertible for any t$$ \in $$**R**

D

invertible only if t $$=$$ $${\pi \over 2}$$.

$$A = \left[ {\matrix{
{{e^t}} & {{e^{ - t}}\cos t} & {{e^{ - t}}\sin t} \cr
{{e^t}} & { - {e^{ - t}}\cos t - {e^{ - t}}\sin t} & { - {e^{ - t}}\sin t + {e^{ - t}}co{\mathop{\rm s}\nolimits} t} \cr
{{e^t}} & {2{e^{ - t}}\sin t} & { - 2{e^{ - t}}\cos t} \cr
} } \right]$$

$$\left| A \right| = {e^t}.\,{e^{ - t}}.{e^{ - t}}\left| {\matrix{ 1 & {\cos t} & {\sin t} \cr 1 & { - \cos t - \sin t} & { - \sin t + \cos t} \cr 1 & {2\sin t} & { - 2\cos t} \cr } } \right|$$

Apply operations R_{2} < R_{2} $$-$$R_{1}, R_{3} < R_{3} $$-$$ R_{1}, R_{1} < R_{1}

$$\left| A \right| = {e^{ - t}}\left| {\matrix{ 1 & {\cos t} & {\sin t} \cr 0 & { - \sin t - 2\cos t} & { - 2\sin t + \cos t} \cr 0 & {2\sin t - \cos t} & { - 2\cos t - \sin t} \cr } } \right|$$

Open the determinant by R_{1}

$$\left| A \right| = 5{e^{ - t}}$$

Invertible for all t $$ \in $$ R

$$\left| A \right| = {e^t}.\,{e^{ - t}}.{e^{ - t}}\left| {\matrix{ 1 & {\cos t} & {\sin t} \cr 1 & { - \cos t - \sin t} & { - \sin t + \cos t} \cr 1 & {2\sin t} & { - 2\cos t} \cr } } \right|$$

Apply operations R

$$\left| A \right| = {e^{ - t}}\left| {\matrix{ 1 & {\cos t} & {\sin t} \cr 0 & { - \sin t - 2\cos t} & { - 2\sin t + \cos t} \cr 0 & {2\sin t - \cos t} & { - 2\cos t - \sin t} \cr } } \right|$$

Open the determinant by R

$$\left| A \right| = 5{e^{ - t}}$$

Invertible for all t $$ \in $$ R

3

If the system of equations

x + y + z = 5

x + 2y + 3z = 9

x + 3y + az = $$\beta $$

has infinitely many solutions, then $$\beta $$ $$-$$ $$\alpha $$ equals -

x + y + z = 5

x + 2y + 3z = 9

x + 3y + az = $$\beta $$

has infinitely many solutions, then $$\beta $$ $$-$$ $$\alpha $$ equals -

A

8

B

21

C

18

D

5

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

$$ = \left( {\alpha - 1} \right) - 4 = \left( {\alpha - 5} \right)$$

for infinite solutions $$D = 0 \Rightarrow \alpha = 5$$

$${D_x} = 0 \Rightarrow \left| {\matrix{ 5 & 1 & 1 \cr 9 & 2 & 3 \cr \beta & 3 & 5 \cr } } \right| = 0$$

$$ \Rightarrow \left| {\matrix{ 0 & 0 & 1 \cr { - 1} & { - 1} & 3 \cr {\beta - 15} & { - 2} & 5 \cr } } \right| = 0$$

$$ \Rightarrow 2 + \beta - 15 = 0 \Rightarrow \beta - 13 = 0$$

on $$\beta = 13$$ we get $${D_y} = {D_z} = 0$$

$$\alpha = 5,\beta = 13$$

$$ = \left( {\alpha - 1} \right) - 4 = \left( {\alpha - 5} \right)$$

for infinite solutions $$D = 0 \Rightarrow \alpha = 5$$

$${D_x} = 0 \Rightarrow \left| {\matrix{ 5 & 1 & 1 \cr 9 & 2 & 3 \cr \beta & 3 & 5 \cr } } \right| = 0$$

$$ \Rightarrow \left| {\matrix{ 0 & 0 & 1 \cr { - 1} & { - 1} & 3 \cr {\beta - 15} & { - 2} & 5 \cr } } \right| = 0$$

$$ \Rightarrow 2 + \beta - 15 = 0 \Rightarrow \beta - 13 = 0$$

on $$\beta = 13$$ we get $${D_y} = {D_z} = 0$$

$$\alpha = 5,\beta = 13$$

4

Let d $$ \in $$ R, and

$$A = \left[ {\matrix{ { - 2} & {4 + d} & {\left( {\sin \theta } \right) - 2} \cr 1 & {\left( {\sin \theta } \right) + 2} & d \cr 5 & {\left( {2\sin \theta } \right) - d} & {\left( { - \sin \theta } \right) + 2 + 2d} \cr } } \right],$$

$$\theta \in \left[ {0,2\pi } \right]$$ If the minimum value of det(A) is 8, then a value of d is -

$$A = \left[ {\matrix{ { - 2} & {4 + d} & {\left( {\sin \theta } \right) - 2} \cr 1 & {\left( {\sin \theta } \right) + 2} & d \cr 5 & {\left( {2\sin \theta } \right) - d} & {\left( { - \sin \theta } \right) + 2 + 2d} \cr } } \right],$$

$$\theta \in \left[ {0,2\pi } \right]$$ If the minimum value of det(A) is 8, then a value of d is -

A

$$-$$ 7

B

$$2\left( {\sqrt 2 + 2} \right)$$

C

$$-$$ 5

D

$$2\left( {\sqrt 2 + 1} \right)$$

$$\det A = \left| {\matrix{
{ - 2} & {4 + d} & {\sin \theta - 2} \cr
1 & {\sin \theta + 2} & d \cr
5 & {2\sin \theta - d} & { - \sin \theta + 2 + 2d} \cr
} } \right|$$

(R_{1} $$ \to $$ R_{1} + R_{3} $$-$$ 2R_{2})

$$ = \left| {\matrix{ 1 & 0 & 0 \cr 1 & {\sin \theta + 2} & d \cr 5 & {2\sin \theta - d} & {2 + 2d - \sin \theta } \cr } } \right|$$

= (2 + sin $$\theta $$) ( 2 + 2d $$-$$ sin$$\theta $$) $$-$$ d(2sin$$\theta $$ $$-$$ d)

= 4 + 4d $$-$$ 2sin$$\theta $$ + 2sin$$\theta $$ + 2dsin$$\theta $$ $$-$$ sin^{2}$$\theta $$ $$-$$ 2dsin$$\theta $$ + d^{2}

d^{2} + 4d + 4 $$-$$ sin^{2}$$\theta $$

= (d + 2)^{2} $$-$$ sin^{2}$$\theta $$

For a given d, minimum value of

det(A) = (d + 2)^{2} $$-$$ 1 = 8

$$ \Rightarrow $$ d = 1 or $$-$$ 5

(R

$$ = \left| {\matrix{ 1 & 0 & 0 \cr 1 & {\sin \theta + 2} & d \cr 5 & {2\sin \theta - d} & {2 + 2d - \sin \theta } \cr } } \right|$$

= (2 + sin $$\theta $$) ( 2 + 2d $$-$$ sin$$\theta $$) $$-$$ d(2sin$$\theta $$ $$-$$ d)

= 4 + 4d $$-$$ 2sin$$\theta $$ + 2sin$$\theta $$ + 2dsin$$\theta $$ $$-$$ sin

d

= (d + 2)

For a given d, minimum value of

det(A) = (d + 2)

$$ \Rightarrow $$ d = 1 or $$-$$ 5

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