Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

The system of linear equations

$$\matrix{ {x + \lambda y - z = 0} \cr {\lambda x - y - z = 0} \cr {x + y - \lambda z = 0} \cr } $$

has a non-trivial solution for:A

exactly two values of $$\lambda .$$

B

exactly three values of $$\lambda .$$

C

infinitely many values of $$\lambda .$$

D

exactly one value of $$\lambda .$$

For trivial solution, $$\left| {\matrix{
1 & \lambda & { - 1} \cr
\lambda & { - 1} & { - 1} \cr
1 & 1 & { - \lambda } \cr
} } \right| = 0$$

$$ \Rightarrow - \lambda \left( {\lambda + 1} \right)\left( {\lambda + 1} \right) = 0$$

$$ \Rightarrow \lambda = 0, + 1, - 1$$

$$ \Rightarrow - \lambda \left( {\lambda + 1} \right)\left( {\lambda + 1} \right) = 0$$

$$ \Rightarrow \lambda = 0, + 1, - 1$$

2

MCQ (Single Correct Answer)

If $$A = \left[ {\matrix{
{5a} & { - b} \cr
3 & 2 \cr
} } \right]$$ and $$A$$ adj $$A=A$$ $${A^T},$$ then $$5a+b$$ is equal to :

A

$$4$$

B

$$13$$

C

$$-1$$

D

$$5$$

$$A\left( {Adj\,\,A} \right) = A\,{A^T}$$

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

$$Adj\,\,A = {A^T}$$

$$ \Rightarrow \left[ {\matrix{ 2 & b \cr { - 3} & {5a} \cr } } \right] = \left[ {\matrix{ {5a} & 3 \cr { - b} & 2 \cr } } \right]$$

$$ \Rightarrow a = {2 \over 5}\,\,$$ and $$\,\,b = 3$$

$$ \Rightarrow 5a + b = 5$$

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

$$Adj\,\,A = {A^T}$$

$$ \Rightarrow \left[ {\matrix{ 2 & b \cr { - 3} & {5a} \cr } } \right] = \left[ {\matrix{ {5a} & 3 \cr { - b} & 2 \cr } } \right]$$

$$ \Rightarrow a = {2 \over 5}\,\,$$ and $$\,\,b = 3$$

$$ \Rightarrow 5a + b = 5$$

3

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 $$a=-2$$

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

4

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.

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