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1

JEE Main 2016 (Offline)

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 $$\lambda .$$

Explanation

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

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

$$\Rightarrow \lambda = 0, + 1, - 1$$
2

JEE Main 2016 (Offline)

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
$$5$$

Explanation

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

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

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

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

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

$$\Rightarrow 5a + b = 5$$
3

JEE Main 2015 (Offline)

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
$$(2, 1)$$
B
$$(-2, -1)$$
C
$$(2, -1)$$
D
$$(-2, 1)$$

Explanation

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

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

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

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

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

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

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

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

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

$$\left( {a,b} \right) = \left( { - 2, - 1} \right)$$
4

JEE Main 2015 (Offline)

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

Explanation

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

\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.

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