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1

### JEE Main 2014 (Offline)

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 $$\,\,\,\,\,\,$$ $$AA' = A'A$$ $$\left. \, \right\}$$

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

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

### JEE Main 2014 (Offline)

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, $$k=1$$
3

### JEE Main 2013 (Offline)

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 :
A
$$4$$
B
$$11$$
C
$$5$$
D
$$0$$

## Explanation

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

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

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

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

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

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

$$\Rightarrow \alpha = 11$$
4

### AIEEE 2012

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

## Explanation

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

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

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

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

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