Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (Single Correct Answer)

If $$A = \left[ {\matrix{
1 & 2 & 2 \cr
2 & 1 & { - 2} \cr
a & 2 & b \cr
} } \right]$$ is a matrix satisfying the equation

$$A{A^T} = 91,$$ where $$I$$ is $$3 \times 3$$ identity matrix, then the ordered

pair $$(a, b)$$ is equal to :

$$A{A^T} = 91,$$ where $$I$$ is $$3 \times 3$$ identity matrix, then the ordered

pair $$(a, b)$$ is equal to :

A

$$(2, 1)$$

B

$$(-2, -1)$$

C

$$(2, -1)$$

D

$$(-2, 1)$$

$$\left[ {\matrix{
1 & 2 & 2 \cr
2 & 1 & { - 2} \cr
a & 2 & b \cr
} } \right]\left[ {\matrix{
1 & 2 & a \cr
2 & 1 & 2 \cr
2 & { - 2} & b \cr
} } \right] = \left[ {\matrix{
9 & 0 & 0 \cr
0 & 9 & 0 \cr
0 & 0 & 9 \cr
} } \right]$$

$$ \Rightarrow \left[ {\matrix{ {1 + 4 + 4} & {2 + 2 - 4} & {a + 4 + 2b} \cr {2 + 2 - 4} & {4 + 1 + 4} & {2a + 2 - 2b} \cr {a + 4 + 2b} & {2a + 2 - 2b} & {{a^2} + 4 + {b^2}} \cr } } \right]$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {\matrix{ 9 & 0 & 0 \cr 0 & 9 & 0 \cr 0 & 0 & 9 \cr } } \right]$$

$$ \Rightarrow a + 4 + 2b = 0$$ $$ \Rightarrow a + 2b = - 4\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

$$2a + 2 - 2b = 0 \Rightarrow 2a - 2b = - 2$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\,\,\,\,\,\,\,\,\\,$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow a - b = - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

On solving $$(i)$$ and $$(ii)$$ we get

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\,\,\,\,\,\,\,\,$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 1 + b + 2b = - 4\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$$$b=-1$$ and $$a=-2$$

$$\left( {a,b} \right) = \left( { - 2, - 1} \right)$$

$$ \Rightarrow \left[ {\matrix{ {1 + 4 + 4} & {2 + 2 - 4} & {a + 4 + 2b} \cr {2 + 2 - 4} & {4 + 1 + 4} & {2a + 2 - 2b} \cr {a + 4 + 2b} & {2a + 2 - 2b} & {{a^2} + 4 + {b^2}} \cr } } \right]$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {\matrix{ 9 & 0 & 0 \cr 0 & 9 & 0 \cr 0 & 0 & 9 \cr } } \right]$$

$$ \Rightarrow a + 4 + 2b = 0$$ $$ \Rightarrow a + 2b = - 4\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

$$2a + 2 - 2b = 0 \Rightarrow 2a - 2b = - 2$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\,\,\,\,\,\,\,\,\\,$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow a - b = - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

On solving $$(i)$$ and $$(ii)$$ we get

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\,\,\,\,\,\,\,\,$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 1 + b + 2b = - 4\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$$$b=-1$$ and $$a=-2$$

$$\left( {a,b} \right) = \left( { - 2, - 1} \right)$$

2

MCQ (Single Correct Answer)

The set of all values of $$\lambda $$ for which the system of linear equations:
$$$\matrix{
{2{x_1} - 2{x_2} + {x_3} = \lambda {x_1}} \cr
{2{x_1} - 3{x_2} + 2{x_3} = \lambda {x_2}} \cr
{ - {x_1} + 2{x_2} = \lambda {x_3}} \cr
} $$$
has a non-trivial solution

A

contains two elements

B

contains more than two elements

C

in an empty set

D

is a singleton

$$\left. {\matrix{
{2{x_1} - 2{x_2} + {x^3} = \lambda {x_1}} \cr
{2{x_1} - 3{x_2} + 2{x_3} = \lambda {x_2}} \cr
{\,\,\,\,\,\,\,\,\,\, - {x_1} + 2{x_2} = \lambda {x_3}} \cr
} } \right\}$$

$$\eqalign{ & \Rightarrow \,\,\,\,\,\,\,\left( {2 - \lambda } \right){x_1} - 2{x_2} + {x_3} = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,2{x_1} - \left( {3 + \lambda } \right){x_2} + 2{x_3} = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - {x_1} + 2{x_2} - \lambda {x_3} = 0 \cr} $$

For non-trivial solution, $$\Delta = 0$$

i.e. $$\,\,\,\left| {\matrix{ {2 - \lambda } & { - 2} & 1 \cr 2 & { - \left( {3 + \lambda } \right)} & 2 \cr { - 1} & 2 & { - \lambda } \cr } } \right| = 0$$

$$ \Rightarrow \left( {2 - \lambda } \right)\left[ {\lambda \left( {3 + \lambda } \right) - 4} \right] + $$

$$\,\,\,\,\,\,\,\,\,2\left[ { - 2\lambda + 2} \right] + 1\left[ {4 - \left( {3 + \lambda } \right)} \right] = 0$$

$$ \Rightarrow {\lambda ^3} + {\lambda ^2} - 5\lambda + 3 = 0$$

$$ \Rightarrow \lambda = 1,1,3$$

Hence, $$\lambda $$ has $$2$$ values.

$$\eqalign{ & \Rightarrow \,\,\,\,\,\,\,\left( {2 - \lambda } \right){x_1} - 2{x_2} + {x_3} = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,2{x_1} - \left( {3 + \lambda } \right){x_2} + 2{x_3} = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - {x_1} + 2{x_2} - \lambda {x_3} = 0 \cr} $$

For non-trivial solution, $$\Delta = 0$$

i.e. $$\,\,\,\left| {\matrix{ {2 - \lambda } & { - 2} & 1 \cr 2 & { - \left( {3 + \lambda } \right)} & 2 \cr { - 1} & 2 & { - \lambda } \cr } } \right| = 0$$

$$ \Rightarrow \left( {2 - \lambda } \right)\left[ {\lambda \left( {3 + \lambda } \right) - 4} \right] + $$

$$\,\,\,\,\,\,\,\,\,2\left[ { - 2\lambda + 2} \right] + 1\left[ {4 - \left( {3 + \lambda } \right)} \right] = 0$$

$$ \Rightarrow {\lambda ^3} + {\lambda ^2} - 5\lambda + 3 = 0$$

$$ \Rightarrow \lambda = 1,1,3$$

Hence, $$\lambda $$ has $$2$$ values.

3

MCQ (Single Correct Answer)

If $$A$$ is a $$3 \times 3$$ non-singular matrix such that $$AA'=A'A$$ and

$$B = {A^{ - 1}}A',$$ then $$BB'$$ equals:

$$B = {A^{ - 1}}A',$$ then $$BB'$$ equals:

A

$${B^{ - 1}}$$

B

$$\left( {{B^{ - 1}}} \right)'$$

C

$$I+B$$

D

$$I$$

$$BB' = B\left( {{A^{ - 1}}A'} \right)' = B\left( {A'} \right)'\left( {{A^{ - 1}}} \right)' = BA\left( {{A^{ - 1}}} \right)'$$

$$ = \left( {{A^{ - 1}}A'} \right)\left( {A\left( {{A^{ - 1}}} \right)'} \right)$$

$$ = {A^{ - 1}}A.A'.\left( {{A^{ - 1}}} \right)'\,\,\,\,\,\,$$ $$\left\{ {} \right.$$ as $$\,\,\,\,\,\,$$ $$AA' = A'A$$ $$\left. \, \right\}$$

$$ = I\left( {{A^{ - 1}}A} \right)'$$

$$ = I.I = {I^2} = I$$

$$ = \left( {{A^{ - 1}}A'} \right)\left( {A\left( {{A^{ - 1}}} \right)'} \right)$$

$$ = {A^{ - 1}}A.A'.\left( {{A^{ - 1}}} \right)'\,\,\,\,\,\,$$ $$\left\{ {} \right.$$ as $$\,\,\,\,\,\,$$ $$AA' = A'A$$ $$\left. \, \right\}$$

$$ = I\left( {{A^{ - 1}}A} \right)'$$

$$ = I.I = {I^2} = I$$

4

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, $$k=1$$

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