1

### JEE Main 2017 (Online) 8th April Morning Slot

Let A be any 3 $\times$ 3 invertible matrix. Then which one of the following is not always true ?
A
adj (A) = $\left| \right.$A$\left| \right.$.A$-$1
B
adj (adj(A)) = $\left| \right.$A$\left| \right.$.A
C
adj (adj(A)) = $\left| \right.$A$\left| \right.$2.(adj(A))$-$1
D
adj (adj(A)) = $\left| \, \right.$A $\left| \, \right.$.(adj(A))$-$1

## Explanation

We know, the formula

A-1 = ${{adj\left( A \right)} \over {\left| A \right|}}$

$\therefore$ adj (A) = $\left| \right.$A$\left| \right.$.A$-$1

So, Option (A) is true.

We know, the formula

adj (adj (A)) = ${\left| A \right|^{n - 2}}.A$

Now if we put n = 3 as given that A is a 3 $\times$ 3 matrix, we get

adj (adj (A)) = ${\left| A \right|^{3 - 2}}.A$ = $\left| A \right|.A$

So, Option (B) is also true.

We know, the formula

adj (adj (A)) = ${\left| A \right|^{n - 1}}{\left( {adj\left( A \right)} \right)^{ - 1}}$

Now if we put n = 3 as given that A is a 3 $\times$ 3 matrix, we get

adj (adj (A)) = ${\left| A \right|^{3 - 1}}{\left( {adj\left( A \right)} \right)^{ - 1}}$ = ${\left| A \right|^{2}}{\left( {adj\left( A \right)} \right)^{ - 1}}$

So, Option (C) is also true.

Now in this formula

adj (adj (A)) = ${\left| A \right|^{n - 1}}{\left( {adj\left( A \right)} \right)^{ - 1}}$

if we put n = 2, we get

adj (adj (A)) = ${\left| A \right|^{2 - 1}}{\left( {adj\left( A \right)} \right)^{ - 1}}$ = ${\left| A \right|}{\left( {adj\left( A \right)} \right)^{ - 1}}$

But as A is a 3 $\times$ 3 matrix so we can not take n = 2, so we can say for a 3 $\times$ 3 matrix option (D) is not true.

So, Option (D) is false.
2

### JEE Main 2017 (Online) 8th April Morning Slot

The number of real values of $\lambda$ for which the system of linear equations

2x + 4y $-$ $\lambda$z = 0

4x + $\lambda$y + 2z = 0

$\lambda$x + 2y + 2z = 0

has infinitely many solutions, is :
A
0
B
1
C
2
D
3
3

### JEE Main 2017 (Online) 8th April Morning Slot

If

$S = \left\{ {x \in \left[ {0,2\pi } \right]:\left| {\matrix{ 0 & {\cos x} & { - \sin x} \cr {\sin x} & 0 & {\cos x} \cr {\cos x} & {\sin x} & 0 \cr } } \right| = 0} \right\},$

then $\sum\limits_{x \in S} {\tan \left( {{\pi \over 3} + x} \right)}$ is equal to :
A
$4 + 2\sqrt 3$
B
$- 2 + \sqrt 3$
C
$- 2 - \sqrt 3$
D
$-\,\,4 - 2\sqrt 3$

## Explanation

Given,

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

$\Rightarrow $$\,\,\, 0 (0 - cosx sinx) - cosx (0 - cos2x) - sinx(sin2x) = 0 \Rightarrow$$\,\,\,$ cos3x $-$ sin3 x = 0

$\Rightarrow $$\,\,\, tan3x = 1 \Rightarrow$$\,\,\,$ tanx = 1

$\therefore$ $\,\,\,$ $\sum\limits_{x\, \in \,\,S} {\,\tan \left( {{\pi \over 3} + x} \right)}$

= ${{\tan {\pi \over 3} + \tan x} \over {1 - \tan {\pi \over 3}\tan x}}$

= ${{\sqrt 3 + 1} \over {1 - \sqrt 3 }}$

= ${{\left( {\sqrt 3 + 1} \right)} \over {\left( {1 - \sqrt 3 } \right)}} \times {{1 + \sqrt 3 } \over {1 + \sqrt 3 }}$

= ${{1 + 3 + 2\sqrt 3 } \over { - 2}}$

= $-$ 2 $-$ $\sqrt 3$
4

### JEE Main 2017 (Online) 9th April Morning Slot

For two 3 × 3 matrices A and B, let A + B = 2BT and 3A + 2B = I3, where BT is the transpose of B and I3 is 3 × 3 identity matrix. Then :
A
5A + 10B = 2I3
B
10A + 5B = 3I3
C
B + 2A = I3
D
3A + 6B = 2I3

## Explanation

Given, A + B = 2BT .......(1)

$\Rightarrow$ (A + B)T = (2BT)T

$\Rightarrow$ AT + BT = 2B

$\Rightarrow$ B = ${{{A^T} + {B^T}} \over 2}$

Now put this in equation (1)

So, A + ${{{A^T} + {B^T}} \over 2}$ = 2BT

$\Rightarrow$2A + AT = 3BT

$\Rightarrow$ A = ${{3{B^T} - {A^T}} \over 2}$

Also, 3A + 2B = I3 .......(2)

$\Rightarrow$ $3\left( {{{3{B^T} - {A^T}} \over 2}} \right) + 2\left( {{{{A^T} + {B^T}} \over 2}} \right)$ = I3

$\Rightarrow$ 11BT - AT = 2I3

$\Rightarrow$ (11BT - AT)T = (2I3)T

$\Rightarrow$ 11B - A = 2I3 ........(3)

Multiply (3) by 3 and then adding (2) and (3) we get,

35B = 7I3

$\Rightarrow$ B = ${{{I_3}} \over 5}$

From (3), 11${{{I_3}} \over 5}$ - A = 2I3

$\Rightarrow$ A = ${{{I_3}} \over 5}$

$\therefore$ 5A = 5B = I3

$\Rightarrow$ 10A + 5B = 3I3