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### 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 $$a=-2$$

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

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. 3 ### JEE Main 2014 (Offline) 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: A $${B^{ - 1}}$$ B $$\left( {{B^{ - 1}}} \right)'$$ C $$I+B$$ D $$I$$ ## Explanation $$BB' = B\left( {{A^{ - 1}}A'} \right)' = B\left( {A'} \right)'\left( {{A^{ - 1}}} \right)' = BA\left( {{A^{ - 1}}} \right)'$$ $$= \left( {{A^{ - 1}}A'} \right)\left( {A\left( {{A^{ - 1}}} \right)'} \right)$$ $$= {A^{ - 1}}A.A'.\left( {{A^{ - 1}}} \right)'\,\,\,\,\,\,$$ $$\left\{ {} \right.$$ as $$\,\,\,\,\,\,$$ $$AA' = A'A$$ $$\left. \, \right\}$$ $$= I\left( {{A^{ - 1}}A} \right)'$$ $$= I.I = {I^2} = I$$ 4 ### JEE Main 2014 (Offline) 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 :
A
$$1$$
B
$$-1$$
C
$$\alpha \beta$$
D
$${1 \over {\alpha \beta }}$$

## Explanation

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, $$k=1$$

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