Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (Single Correct Answer)

If $$\alpha ,\beta \ne 0,$$ and $$f\left( n \right) = {\alpha ^n} + {\beta ^n}$$ and
$$$\left| {\matrix{
3 & {1 + f\left( 1 \right)} & {1 + f\left( 2 \right)} \cr
{1 + f\left( 1 \right)} & {1 + f\left( 2 \right)} & {1 + f\left( 3 \right)} \cr
{1 + f\left( 2 \right)} & {1 + f\left( 3 \right)} & {1 + f\left( 4 \right)} \cr
} } \right|$$$

$$ = K{\left( {1 - \alpha } \right)^2}{\left( {1 - \beta } \right)^2}{\left( {\alpha - \beta } \right)^2},$$ then $$K$$ is equal to :

$$ = K{\left( {1 - \alpha } \right)^2}{\left( {1 - \beta } \right)^2}{\left( {\alpha - \beta } \right)^2},$$ then $$K$$ is equal to :

A

$$1$$

B

$$-1$$

C

$$\alpha \beta $$

D

$${1 \over {\alpha \beta }}$$

Consider

$$\left| {\matrix{ 3 & {1 + f\left( 1 \right)} & {1 + f\left( 2 \right)} \cr {1 + f\left( 1 \right)} & {1 + f\left( 2 \right)} & {1 + f\left( 3 \right)} \cr {1 + f\left( 2 \right)} & {1 + f\left( 3 \right)} & {1 + f\left( 4 \right)} \cr } } \right|$$

$$\,\,\,\,\,\,\,\,\,\,\, = \left| {\matrix{ {1 + 1 + 1} & {1 + \alpha + \beta } & {1 + {\alpha ^2} + {\beta ^2}} \cr {1 + \alpha + \beta } & {1 + {\alpha ^2} + {\beta ^2}} & {1 + {\alpha ^3} + {\beta ^3}} \cr {1 + {\alpha ^2} + {\beta ^2}} & {1 + {\alpha ^3} + {\beta ^3}} & {1 + {\alpha ^4} + {\beta ^4}} \cr } } \right|$$

$$\,\,\,\,\,\,\,\,\,\,\, = \left| {\matrix{ 1 & 1 & 1 \cr 1 & \alpha & \beta \cr 1 & {{\alpha ^2}} & {{\beta ^2}} \cr } } \right| \times \left| {\matrix{ 1 & 1 & 1 \cr 1 & \alpha & \beta \cr 1 & {{\alpha ^2}} & {{\beta ^2}} \cr } } \right|$$

$$\,\,\,\,\,\,\,\,\,\,\, = {\left| {\matrix{ 1 & 1 & 1 \cr 1 & \alpha & \beta \cr 1 & {{\alpha ^2}} & {{\beta ^2}} \cr } } \right|^2}$$

$$\,\,\,\,\,\,\,\,\,\,\, = {\left[ {\left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {\alpha - \beta } \right)} \right]^2}$$

So, $$k=1$$

$$\left| {\matrix{ 3 & {1 + f\left( 1 \right)} & {1 + f\left( 2 \right)} \cr {1 + f\left( 1 \right)} & {1 + f\left( 2 \right)} & {1 + f\left( 3 \right)} \cr {1 + f\left( 2 \right)} & {1 + f\left( 3 \right)} & {1 + f\left( 4 \right)} \cr } } \right|$$

$$\,\,\,\,\,\,\,\,\,\,\, = \left| {\matrix{ {1 + 1 + 1} & {1 + \alpha + \beta } & {1 + {\alpha ^2} + {\beta ^2}} \cr {1 + \alpha + \beta } & {1 + {\alpha ^2} + {\beta ^2}} & {1 + {\alpha ^3} + {\beta ^3}} \cr {1 + {\alpha ^2} + {\beta ^2}} & {1 + {\alpha ^3} + {\beta ^3}} & {1 + {\alpha ^4} + {\beta ^4}} \cr } } \right|$$

$$\,\,\,\,\,\,\,\,\,\,\, = \left| {\matrix{ 1 & 1 & 1 \cr 1 & \alpha & \beta \cr 1 & {{\alpha ^2}} & {{\beta ^2}} \cr } } \right| \times \left| {\matrix{ 1 & 1 & 1 \cr 1 & \alpha & \beta \cr 1 & {{\alpha ^2}} & {{\beta ^2}} \cr } } \right|$$

$$\,\,\,\,\,\,\,\,\,\,\, = {\left| {\matrix{ 1 & 1 & 1 \cr 1 & \alpha & \beta \cr 1 & {{\alpha ^2}} & {{\beta ^2}} \cr } } \right|^2}$$

$$\,\,\,\,\,\,\,\,\,\,\, = {\left[ {\left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {\alpha - \beta } \right)} \right]^2}$$

So, $$k=1$$

2

MCQ (Single Correct Answer)

If $$P = \left[ {\matrix{
1 & \alpha & 3 \cr
1 & 3 & 3 \cr
2 & 4 & 4 \cr
} } \right]$$ is the adjoint of a $$3 \times 3$$ matrix $$A$$ and

$$\left| A \right| = 4,$$ then $$\alpha $$ is equal to :

$$\left| A \right| = 4,$$ then $$\alpha $$ is equal to :

A

$$4$$

B

$$11$$

C

$$5$$

D

$$0$$

$$\left| P \right| = 1\left( {12 - 12} \right) - \alpha \left( {4 - 6} \right) + $$

$$\,\,\,\,\,\,\,\,\,\,\,3\left( {4 - 6} \right) = 2\alpha - 6$$

Now, $$adj\,\,A = P\,$$ $$\,\,\,\,\,\,\,\, \Rightarrow \left| {adj\,A} \right| = \left| P \right|$$

$$ \Rightarrow {\left| A \right|^2} = \left| P \right|$$

$$ \Rightarrow \left| P \right| = 16$$

$$ \Rightarrow 2\alpha - 6 = 16$$

$$ \Rightarrow \alpha = 11$$

$$\,\,\,\,\,\,\,\,\,\,\,3\left( {4 - 6} \right) = 2\alpha - 6$$

Now, $$adj\,\,A = P\,$$ $$\,\,\,\,\,\,\,\, \Rightarrow \left| {adj\,A} \right| = \left| P \right|$$

$$ \Rightarrow {\left| A \right|^2} = \left| P \right|$$

$$ \Rightarrow \left| P \right| = 16$$

$$ \Rightarrow 2\alpha - 6 = 16$$

$$ \Rightarrow \alpha = 11$$

3

MCQ (Single Correct Answer)

Let $$P$$ and $$Q$$ be $$3 \times 3$$ matrices $$P \ne Q.$$ If $${P^3} = {Q^3}$$ and

$${P^2}Q = {Q^2}P$$ then determinant of $$\left( {{P^2} + {Q^2}} \right)$$ is equal to :

$${P^2}Q = {Q^2}P$$ then determinant of $$\left( {{P^2} + {Q^2}} \right)$$ is equal to :

A

$$-2$$

B

$$1$$

C

$$0$$

D

$$-1$$

Given

$${P^3} = {q^3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

$${P^2}Q = {Q^2}p\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$

Subtracting $$(1)$$ and $$(2)$$, we get

$${P^3} - {P^2}Q = {Q^3} - {Q^2}P$$

$$ \Rightarrow {P^2}\left( {P - Q} \right) + {Q^2}\left( {P - Q} \right) = 0$$

$$ \Rightarrow \left( {{P^2} + {Q^2}} \right)\left( {P - Q} \right) = 0$$

$$ \Rightarrow \left| {{p^2} + {Q^2}} \right| = 0$$

as $$P \ne Q$$

$${P^3} = {q^3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

$${P^2}Q = {Q^2}p\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$

Subtracting $$(1)$$ and $$(2)$$, we get

$${P^3} - {P^2}Q = {Q^3} - {Q^2}P$$

$$ \Rightarrow {P^2}\left( {P - Q} \right) + {Q^2}\left( {P - Q} \right) = 0$$

$$ \Rightarrow \left( {{P^2} + {Q^2}} \right)\left( {P - Q} \right) = 0$$

$$ \Rightarrow \left| {{p^2} + {Q^2}} \right| = 0$$

as $$P \ne Q$$

4

MCQ (Single Correct Answer)

Let $$A = \left( {\matrix{
1 & 0 & 0 \cr
2 & 1 & 0 \cr
3 & 2 & 1 \cr
} } \right)$$. If $${u_1}$$ and $${u_2}$$ are column matrices such

that $$A{u_1} = \left( {\matrix{ 1 \cr 0 \cr 0 \cr } } \right)$$ and $$A{u_2} = \left( {\matrix{ 0 \cr 1 \cr 0 \cr } } \right),$$ then $${u_1} + {u_2}$$ is equal to :

that $$A{u_1} = \left( {\matrix{ 1 \cr 0 \cr 0 \cr } } \right)$$ and $$A{u_2} = \left( {\matrix{ 0 \cr 1 \cr 0 \cr } } \right),$$ then $${u_1} + {u_2}$$ is equal to :

A

$$\left( {\matrix{
-1 \cr
1 \cr
0 \cr
} } \right)$$

B

$$\left( {\matrix{
-1 \cr
1 \cr
-1 \cr
} } \right)$$

C

$$\left( {\matrix{
-1 \cr
-1 \cr
0 \cr
} } \right)$$

D

$$\left( {\matrix{
1 \cr
-1 \cr
-1 \cr
} } \right)$$

Let $$A{u_1} = \left( {\matrix{
1 \cr
0 \cr
0 \cr
} } \right)\,\,\,\,\,\,A{u_2} = \left( {\matrix{
0 \cr
1 \cr
0 \cr
} } \right)$$

Then, $$A{u_1} + A{u_2} = \left( {\matrix{ 1 \cr 0 \cr 0 \cr } } \right) + \left( {\matrix{ 0 \cr 1 \cr 0 \cr } } \right)$$

$$ \Rightarrow A\left( {{u_1} + {u_2}} \right) = \left( {\matrix{ 1 \cr 1 \cr 0 \cr } } \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

Also, $$A = \left( {\matrix{ 1 & 0 & 0 \cr 2 & 1 & 0 \cr 3 & 2 & 1 \cr } } \right)$$

$$ \Rightarrow \left| A \right| = 1\left( 1 \right) - 0\left( 2 \right) + 0\left( {4 - 3} \right) = 1$$

We know,

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

( as $$\left| A \right| = 1$$ )

Now, from equation $$(1)$$, we have

$${u_1} + {u_2} = {A^{ - 1}}\left( {\matrix{ 1 \cr 1 \cr 0 \cr } } \right)$$

$$ = \left[ {\matrix{ 1 & 0 & 0 \cr { - 2} & 1 & 0 \cr 1 & { - 2} & 1 \cr } } \right]\left( {\matrix{ 1 \cr 1 \cr 0 \cr } } \right)$$

$$ = \left[ {\matrix{ 1 \cr { - 1} \cr { - 1} \cr } } \right]$$

Then, $$A{u_1} + A{u_2} = \left( {\matrix{ 1 \cr 0 \cr 0 \cr } } \right) + \left( {\matrix{ 0 \cr 1 \cr 0 \cr } } \right)$$

$$ \Rightarrow A\left( {{u_1} + {u_2}} \right) = \left( {\matrix{ 1 \cr 1 \cr 0 \cr } } \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

Also, $$A = \left( {\matrix{ 1 & 0 & 0 \cr 2 & 1 & 0 \cr 3 & 2 & 1 \cr } } \right)$$

$$ \Rightarrow \left| A \right| = 1\left( 1 \right) - 0\left( 2 \right) + 0\left( {4 - 3} \right) = 1$$

We know,

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

( as $$\left| A \right| = 1$$ )

Now, from equation $$(1)$$, we have

$${u_1} + {u_2} = {A^{ - 1}}\left( {\matrix{ 1 \cr 1 \cr 0 \cr } } \right)$$

$$ = \left[ {\matrix{ 1 & 0 & 0 \cr { - 2} & 1 & 0 \cr 1 & { - 2} & 1 \cr } } \right]\left( {\matrix{ 1 \cr 1 \cr 0 \cr } } \right)$$

$$ = \left[ {\matrix{ 1 \cr { - 1} \cr { - 1} \cr } } \right]$$

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