Joint Entrance Examination

Graduate Aptitude Test in Engineering

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General Aptitude

1

Let A be a 3 $$ \times $$ 3 matrix such that A^{2} $$-$$ 5A + 7I = 0

**Statement - I :**

A^{$$-$$1} = $${1 \over 7}$$ (5I $$-$$ A).

**Statement - II :**

The polynomial A^{3} $$-$$ 2A^{2} $$-$$ 3A + I can be reduced to 5(A $$-$$ 4I).

Then :

A

The polynomial A

Then :

A

Statement-I is true, but Statement-II is false.

B

Statement-I is false, but Statement-II is true.

C

Both the statements are true.

D

Both the statements are false

Given,

A^{2} $$-$$ 5A + 7I = 0

$$ \Rightarrow $$ A^{2} $$-$$ 5A = $$-$$ 7I

$$ \Rightarrow $$ AAA^{$$-$$1} $$-$$ 5AA^{$$-$$1} = $$-$$ 7IA^{$$-$$1}

$$ \Rightarrow $$ AI $$-$$ 5I = $$-$$ 7A^{$$-$$1}

$$ \Rightarrow $$ A $$-$$ 5I = $$-$$ 7A^{$$-$$1}

$$ \Rightarrow $$ A^{$$-$$1} = $${1 \over 7}$$(5I $$-$$ A)

Hence, statement 1 is true.

Now A^{3} $$-$$ 2A^{2} $$-$$ 3A + I

= A(A^{2}) $$-$$ 2A^{2} $$-$$ 3A + I

= A(5A $$-$$ 7I) $$-$$ 2A^{2} $$-$$ 3A + I

= 5A^{2} $$-$$ 7A $$-$$ 2A^{2} $$-$$ 3A + I

= 3A^{2} $$-$$ 10A + I

= 3(5A $$-$$ 7I) $$-$$ 10A + I

= 15A $$-$$ 21A $$-$$ 10A + I

= 5A $$-$$ 20I

= 5(A $$-$$ 4I)

So, statement 2 is also correct.

A

$$ \Rightarrow $$ A

$$ \Rightarrow $$ AAA

$$ \Rightarrow $$ AI $$-$$ 5I = $$-$$ 7A

$$ \Rightarrow $$ A $$-$$ 5I = $$-$$ 7A

$$ \Rightarrow $$ A

Hence, statement 1 is true.

Now A

= A(A

= A(5A $$-$$ 7I) $$-$$ 2A

= 5A

= 3A

= 3(5A $$-$$ 7I) $$-$$ 10A + I

= 15A $$-$$ 21A $$-$$ 10A + I

= 5A $$-$$ 20I

= 5(A $$-$$ 4I)

So, statement 2 is also correct.

2

If A = $$\left[ {\matrix{
{ - 4} & { - 1} \cr
3 & 1 \cr
} } \right]$$,

then the determinant of the matrix (A^{2016} − 2A^{2015} − A^{2014}) is :

then the determinant of the matrix (A

A

2014

B

$$-$$ 175

C

2016

D

$$-$$ 25

Given,

$$A = \left[ {\matrix{ { - 4} & { - 1} \cr 3 & 1 \cr } } \right]$$

$${A^2} = \left[ {\matrix{ { - 4} & { - 1} \cr 3 & 1 \cr } } \right]\left[ {\matrix{ { - 4} & { - 1} \cr 3 & 1 \cr } } \right]$$

$$ = \left[ {\matrix{ {13} & 3 \cr { - 9} & { - 2} \cr } } \right]$$

A^{2} $$-$$ 2A $$-$$ I

$$ = \left[ {\matrix{ {13} & 3 \cr { - 9} & { - 2} \cr } } \right] - \left[ {\matrix{ { - 8} & { - 2} \cr 6 & 2 \cr } } \right] - \left[ {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right]$$

$$ = \left[ {\matrix{ {20} & 5 \cr { - 15} & { - 5} \cr } } \right]$$

$$\left| A \right| = \left| {\matrix{ { - 4} & { - 1} \cr 3 & 1 \cr } } \right|$$ $$=$$ $$-$$ 4 + 3 $$=$$ $$-$$ 1

Now,

$$\left| {{A^{2016}} - 2{A^{2015}} - {A^{2014}}} \right|$$

$$=$$ $${\left| A \right|^{2014}}\left| {{A^2} - 2A - {\rm I}} \right|$$

$$ = {\left( { - 1} \right)^{2014}}\left| {\matrix{ {20} & 5 \cr { - 15} & { - 5} \cr } } \right|$$

$$=$$ 1 $$ \times $$ ($$-$$ 100 + 75)

$$=$$ $$-$$ 25

$$A = \left[ {\matrix{ { - 4} & { - 1} \cr 3 & 1 \cr } } \right]$$

$${A^2} = \left[ {\matrix{ { - 4} & { - 1} \cr 3 & 1 \cr } } \right]\left[ {\matrix{ { - 4} & { - 1} \cr 3 & 1 \cr } } \right]$$

$$ = \left[ {\matrix{ {13} & 3 \cr { - 9} & { - 2} \cr } } \right]$$

A

$$ = \left[ {\matrix{ {13} & 3 \cr { - 9} & { - 2} \cr } } \right] - \left[ {\matrix{ { - 8} & { - 2} \cr 6 & 2 \cr } } \right] - \left[ {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right]$$

$$ = \left[ {\matrix{ {20} & 5 \cr { - 15} & { - 5} \cr } } \right]$$

$$\left| A \right| = \left| {\matrix{ { - 4} & { - 1} \cr 3 & 1 \cr } } \right|$$ $$=$$ $$-$$ 4 + 3 $$=$$ $$-$$ 1

Now,

$$\left| {{A^{2016}} - 2{A^{2015}} - {A^{2014}}} \right|$$

$$=$$ $${\left| A \right|^{2014}}\left| {{A^2} - 2A - {\rm I}} \right|$$

$$ = {\left( { - 1} \right)^{2014}}\left| {\matrix{ {20} & 5 \cr { - 15} & { - 5} \cr } } \right|$$

$$=$$ 1 $$ \times $$ ($$-$$ 100 + 75)

$$=$$ $$-$$ 25

3

If $$A = \left[ {\matrix{
2 & { - 3} \cr
{ - 4} & 1 \cr
} } \right]$$,

then adj(3A^{2} + 12A) is equal to

then adj(3A

A

$$\left[ {\matrix{
{51} & {63} \cr
{84} & {72} \cr
} } \right]$$

B

$$\left[ {\matrix{
{51} & {84} \cr
{63} & {72} \cr
} } \right]$$

C

$$\left[ {\matrix{
{72} & {-63} \cr
{-84} & {51} \cr
} } \right]$$

D

$$\left[ {\matrix{
{72} & {-84} \cr
{-63} & {51} \cr
} } \right]$$

We have, $$A = \left[ {\matrix{
2 & { - 3} \cr
{ - 4} & 1 \cr
} } \right]$$

$$ \therefore $$ A^{2} = A.A = $$\left[ {\matrix{
2 & { - 3} \cr
{ - 4} & 1 \cr
} } \right]\left[ {\matrix{
2 & { - 3} \cr
{ - 4} & 1 \cr
} } \right]$$

= $$\left[ {\matrix{ {4 + 12} & { - 6 - 3} \cr { - 8 - 4} & {12 + 1} \cr } } \right]$$

= $$\left[ {\matrix{ {16} & { - 9} \cr { - 12} & {13} \cr } } \right]$$

Now, 3A^{2} + 12A

= $$3\left[ {\matrix{ {16} & { - 9} \cr { - 12} & {13} \cr } } \right] + 12\left[ {\matrix{ 2 & { - 3} \cr { - 4} & 1 \cr } } \right]$$

= $$\left[ {\matrix{ {48} & { - 27} \cr { - 36} & {39} \cr } } \right] + \left[ {\matrix{ {24} & { - 36} \cr { - 48} & {12} \cr } } \right]$$

= $$\left[ {\matrix{ {72} & { - 63} \cr { - 84} & {51} \cr } } \right]$$

$$ \therefore $$ adj(3A^{2} + 12A) = $$\left[ {\matrix{
{51} & {63} \cr
{84} & {72} \cr
} } \right]$$

$$ \therefore $$ A

= $$\left[ {\matrix{ {4 + 12} & { - 6 - 3} \cr { - 8 - 4} & {12 + 1} \cr } } \right]$$

= $$\left[ {\matrix{ {16} & { - 9} \cr { - 12} & {13} \cr } } \right]$$

Now, 3A

= $$3\left[ {\matrix{ {16} & { - 9} \cr { - 12} & {13} \cr } } \right] + 12\left[ {\matrix{ 2 & { - 3} \cr { - 4} & 1 \cr } } \right]$$

= $$\left[ {\matrix{ {48} & { - 27} \cr { - 36} & {39} \cr } } \right] + \left[ {\matrix{ {24} & { - 36} \cr { - 48} & {12} \cr } } \right]$$

= $$\left[ {\matrix{ {72} & { - 63} \cr { - 84} & {51} \cr } } \right]$$

$$ \therefore $$ adj(3A

4

If S is the set of distinct values of 'b' for which the following system of linear equations

x + y + z = 1

x + ay + z = 1

ax + by + z = 0

has no solution, then S is :

x + y + z = 1

x + ay + z = 1

ax + by + z = 0

has no solution, then S is :

A

an empty set

B

an infinite set

C

a finite set containing two or more elements

D

a singleton

$$\left| {\matrix{
1 & 1 & 1 \cr
1 & a & 1 \cr
a & b & 1 \cr
} } \right| = 0$$

$$ \Rightarrow $$ 1 [a – b] – 1 [1 – a] + 1 [b – a^{2}] = 0

$$ \Rightarrow $$ (a - 1)^{2} = 0

$$ \Rightarrow $$ a = 1

For a = 1, the equations become

x + y + z = 1

x + y + z = 1

x + by + z = 0

These equations give no solution for b = 1

$$ \Rightarrow $$ S is singleton set.

$$ \Rightarrow $$ 1 [a – b] – 1 [1 – a] + 1 [b – a

$$ \Rightarrow $$ (a - 1)

$$ \Rightarrow $$ a = 1

For a = 1, the equations become

x + y + z = 1

x + y + z = 1

x + by + z = 0

These equations give no solution for b = 1

$$ \Rightarrow $$ S is singleton set.

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