1

### JEE Main 2016 (Online) 9th April Morning Slot

The number of distinct real roots of the equation,

$\left| {\matrix{ {\cos x} & {\sin x} & {\sin x} \cr {\sin x} & {\cos x} & {\sin x} \cr {\sin x} & {\sin x} & {\cos x} \cr } } \right| = 0$ in the interval $\left[ { - {\pi \over 4},{\pi \over 4}} \right]$ is :
A
4
B
3
C
2
D
1

## Explanation

Given,

$\left| {\matrix{ {\cos x} & {\sin x} & {\sin x} \cr {\sin x} & {\cos x} & {\sin x} \cr {\sin x} & {\sin x} & {\cos x} \cr } } \right| = 0$

R1  $\to$  R1  $-$  R3

R1  $\to$  R2  $-$  R3

$\left| {\matrix{ {\cos x - \sin x} & 0 & {\sin x - \cos x} \cr 0 & {\cos x - \sin x} & {\sin x - \cos x} \cr {\sin x} & {\sin x} & { \cos x} \cr } } \right| = 0$

C3  $\to$  C3  +  C2

$\left| {\matrix{ {\cos x - \sin x} & 0 & {\sin x - \cos x} \cr 0 & {\cos x - \sin x} & 0 \cr {\sin x} & {\sin x} & {\sin x + \cos x} \cr } } \right| = 0$

Expanding using first column,

(cosx $-$ sinx)(cos $-$ sinx) (sinx + cos x)

+ sinx (cosx $-$ sinx) (sinx $-$ cosx) = 0

$\Rightarrow$   (cosx $-$ sinx)2 (sinx + cosx)

$+$ sinx (cosx $-$ sinx)2 = 0

$\Rightarrow$    (cosx $-$ sinx)2 (sinx + cosx $+$ sinx) = 0

$\Rightarrow$    (2sinx + cosx )(cosx $-$ sinx)2 = 0

$\therefore$   cosx = -2sinx   or  cosx   =  sinx

$\Rightarrow$ tanx = $- {1 \over 2}$ or tanx = 1

$\therefore$ x = $- {\tan ^{ - 1}}\left( {{1 \over 2}} \right)$, ${\pi \over 4}$

$\therefore$   Number of solutions   =  2
2

### JEE Main 2016 (Online) 10th April Morning Slot

Let A be a 3 $\times$ 3 matrix such that A2 $-$ 5A + 7I = 0

Statement - I :

A$-$1 = ${1 \over 7}$ (5I $-$ A).

Statement - II :

The polynomial A3 $-$ 2A2 $-$ 3A + I can be reduced to 5(A $-$ 4I).

Then :
A
Statement-I is true, but Statement-II is false.
B
Statement-I is false, but Statement-II is true.
C
Both the statements are true.
D
Both the statements are false

## Explanation

Given,

A2 $-$ 5A + 7I = 0

$\Rightarrow$   A2 $-$ 5A = $-$ 7I

$\Rightarrow$   AAA$-$1 $-$ 5AA$-$1 = $-$ 7IA$-$1

$\Rightarrow$   AI $-$ 5I = $-$ 7A$-$1

$\Rightarrow$   A $-$ 5I = $-$ 7A$-$1

$\Rightarrow$   A$-$1 = ${1 \over 7}$(5I $-$ A)

Hence, statement 1 is true.

Now A3 $-$ 2A2 $-$ 3A + I

=   A(A2) $-$ 2A2 $-$ 3A + I

=   A(5A $-$ 7I) $-$ 2A2 $-$ 3A + I

=   5A2 $-$ 7A $-$ 2A2 $-$ 3A + I

=   3A2 $-$ 10A + I

=   3(5A $-$ 7I) $-$ 10A + I

=   15A $-$ 21A $-$ 10A + I

=   5A $-$ 20I

=   5(A $-$ 4I)

So, statement 2 is also correct.
3

### JEE Main 2016 (Online) 10th April Morning Slot

If    A = $\left[ {\matrix{ { - 4} & { - 1} \cr 3 & 1 \cr } } \right]$,

then the determinant of the matrix (A2016 − 2A2015 − A2014) is :
A
2014
B
$-$ 175
C
2016
D
$-$ 25

## Explanation

Given,

$A = \left[ {\matrix{ { - 4} & { - 1} \cr 3 & 1 \cr } } \right]$

${A^2} = \left[ {\matrix{ { - 4} & { - 1} \cr 3 & 1 \cr } } \right]\left[ {\matrix{ { - 4} & { - 1} \cr 3 & 1 \cr } } \right]$

$= \left[ {\matrix{ {13} & 3 \cr { - 9} & { - 2} \cr } } \right]$

A2 $-$ 2A $-$ I

$= \left[ {\matrix{ {13} & 3 \cr { - 9} & { - 2} \cr } } \right] - \left[ {\matrix{ { - 8} & { - 2} \cr 6 & 2 \cr } } \right] - \left[ {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right]$

$= \left[ {\matrix{ {20} & 5 \cr { - 15} & { - 5} \cr } } \right]$

$\left| A \right| = \left| {\matrix{ { - 4} & { - 1} \cr 3 & 1 \cr } } \right|$ $=$ $-$ 4 + 3 $=$ $-$ 1

Now,

$\left| {{A^{2016}} - 2{A^{2015}} - {A^{2014}}} \right|$

$=$ ${\left| A \right|^{2014}}\left| {{A^2} - 2A - {\rm I}} \right|$

$= {\left( { - 1} \right)^{2014}}\left| {\matrix{ {20} & 5 \cr { - 15} & { - 5} \cr } } \right|$

$=$ 1 $\times$ ($-$ 100 + 75)

$=$ $-$ 25
4

### JEE Main 2017 (Offline)

If $A = \left[ {\matrix{ 2 & { - 3} \cr { - 4} & 1 \cr } } \right]$,

then adj(3A2 + 12A) is equal to
A
$\left[ {\matrix{ {51} & {63} \cr {84} & {72} \cr } } \right]$
B
$\left[ {\matrix{ {51} & {84} \cr {63} & {72} \cr } } \right]$
C
$\left[ {\matrix{ {72} & {-63} \cr {-84} & {51} \cr } } \right]$
D
$\left[ {\matrix{ {72} & {-84} \cr {-63} & {51} \cr } } \right]$

## Explanation

We have, $A = \left[ {\matrix{ 2 & { - 3} \cr { - 4} & 1 \cr } } \right]$

$\therefore$ A2 = A.A = $\left[ {\matrix{ 2 & { - 3} \cr { - 4} & 1 \cr } } \right]\left[ {\matrix{ 2 & { - 3} \cr { - 4} & 1 \cr } } \right]$

= $\left[ {\matrix{ {4 + 12} & { - 6 - 3} \cr { - 8 - 4} & {12 + 1} \cr } } \right]$

= $\left[ {\matrix{ {16} & { - 9} \cr { - 12} & {13} \cr } } \right]$

Now, 3A2 + 12A

= $3\left[ {\matrix{ {16} & { - 9} \cr { - 12} & {13} \cr } } \right] + 12\left[ {\matrix{ 2 & { - 3} \cr { - 4} & 1 \cr } } \right]$

= $\left[ {\matrix{ {48} & { - 27} \cr { - 36} & {39} \cr } } \right] + \left[ {\matrix{ {24} & { - 36} \cr { - 48} & {12} \cr } } \right]$

= $\left[ {\matrix{ {72} & { - 63} \cr { - 84} & {51} \cr } } \right]$

$\therefore$ adj(3A2 + 12A) = $\left[ {\matrix{ {51} & {63} \cr {84} & {72} \cr } } \right]$