1

### JEE Main 2019 (Online) 9th January Evening Slot

If   $A = \left[ {\matrix{ {{e^t}} & {{e^{ - t}}\cos t} & {{e^{ - t}}\sin t} \cr {{e^t}} & { - {e^{ - t}}\cos t - {e^{ - t}}\sin t} & { - {e^{ - t}}\sin t + {e^{ - t}}co{\mathop{\rm s}\nolimits} t} \cr {{e^t}} & {2{e^{ - t}}\sin t} & { - 2{e^{ - t}}\cos t} \cr } } \right]$

then A is :
A
invertible for all t$\in$R.
B
invertible only if t $=$ $\pi$
C
not invertible for any t$\in$R
D
invertible only if t $=$ ${\pi \over 2}$.

## Explanation

$A = \left[ {\matrix{ {{e^t}} & {{e^{ - t}}\cos t} & {{e^{ - t}}\sin t} \cr {{e^t}} & { - {e^{ - t}}\cos t - {e^{ - t}}\sin t} & { - {e^{ - t}}\sin t + {e^{ - t}}co{\mathop{\rm s}\nolimits} t} \cr {{e^t}} & {2{e^{ - t}}\sin t} & { - 2{e^{ - t}}\cos t} \cr } } \right]$

$\left| A \right| = {e^t}.\,{e^{ - t}}.{e^{ - t}}\left| {\matrix{ 1 & {\cos t} & {\sin t} \cr 1 & { - \cos t - \sin t} & { - \sin t + \cos t} \cr 1 & {2\sin t} & { - 2\cos t} \cr } } \right|$

Apply operations R2 < R2 $-$R1, R3 < R3 $-$ R1, R1 < R1

$\left| A \right| = {e^{ - t}}\left| {\matrix{ 1 & {\cos t} & {\sin t} \cr 0 & { - \sin t - 2\cos t} & { - 2\sin t + \cos t} \cr 0 & {2\sin t - \cos t} & { - 2\cos t - \sin t} \cr } } \right|$

Open the determinant by R1

$\left| A \right| = 5{e^{ - t}}$

Invertible for all t $\in$ R
2

### JEE Main 2019 (Online) 10th January Morning Slot

If the system of equations

x + y + z = 5

x + 2y + 3z = 9

x + 3y + az = $\beta$

has infinitely many solutions, then $\beta$ $-$ $\alpha$ equals -
A
8
B
21
C
18
D
5

## Explanation

$D = \left| {\matrix{ 1 & 1 & 1 \cr 1 & 2 & 3 \cr 1 & 3 & \alpha \cr } } \right| = \left| {\matrix{ 1 & 1 & 1 \cr 0 & 1 & 2 \cr 0 & 2 & {\alpha - 1} \cr } } \right|$

$= \left( {\alpha - 1} \right) - 4 = \left( {\alpha - 5} \right)$

for infinite solutions $D = 0 \Rightarrow \alpha = 5$

${D_x} = 0 \Rightarrow \left| {\matrix{ 5 & 1 & 1 \cr 9 & 2 & 3 \cr \beta & 3 & 5 \cr } } \right| = 0$

$\Rightarrow \left| {\matrix{ 0 & 0 & 1 \cr { - 1} & { - 1} & 3 \cr {\beta - 15} & { - 2} & 5 \cr } } \right| = 0$

$\Rightarrow 2 + \beta - 15 = 0 \Rightarrow \beta - 13 = 0$

on $\beta = 13$ we get ${D_y} = {D_z} = 0$

$\alpha = 5,\beta = 13$
3

### JEE Main 2019 (Online) 10th January Morning Slot

Let  d $\in$ R, and

$A = \left[ {\matrix{ { - 2} & {4 + d} & {\left( {\sin \theta } \right) - 2} \cr 1 & {\left( {\sin \theta } \right) + 2} & d \cr 5 & {\left( {2\sin \theta } \right) - d} & {\left( { - \sin \theta } \right) + 2 + 2d} \cr } } \right],$

$\theta \in \left[ {0,2\pi } \right]$ If the minimum value of det(A) is 8, then a value of d is -
A
$-$ 7
B
$2\left( {\sqrt 2 + 2} \right)$
C
$-$ 5
D
$2\left( {\sqrt 2 + 1} \right)$

## Explanation

$\det A = \left| {\matrix{ { - 2} & {4 + d} & {\sin \theta - 2} \cr 1 & {\sin \theta + 2} & d \cr 5 & {2\sin \theta - d} & { - \sin \theta + 2 + 2d} \cr } } \right|$

(R1 $\to$ R1 + R3 $-$ 2R2)

$= \left| {\matrix{ 1 & 0 & 0 \cr 1 & {\sin \theta + 2} & d \cr 5 & {2\sin \theta - d} & {2 + 2d - \sin \theta } \cr } } \right|$

= (2 + sin $\theta$) ( 2 + 2d $-$ sin$\theta$) $-$ d(2sin$\theta$ $-$ d)

= 4 + 4d $-$ 2sin$\theta$ + 2sin$\theta$ + 2dsin$\theta$ $-$ sin2$\theta$ $-$ 2dsin$\theta$ + d2

d2 + 4d + 4 $-$ sin2$\theta$

= (d + 2)2 $-$ sin2$\theta$

For a given d, minimum value of

det(A) = (d + 2)2 $-$ 1 = 8

$\Rightarrow$  d = 1 or $-$ 5
4

### JEE Main 2019 (Online) 10th January Evening Slot

Let A = $\left[ {\matrix{ 2 & b & 1 \cr b & {{b^2} + 1} & b \cr 1 & b & 2 \cr } } \right]$ where b > 0.

Then the minimum value of ${{\det \left( A \right)} \over b}$ is -
A
$\sqrt 3$
B
$-$ $2\sqrt 3$
C
$- \sqrt 3$
D
$2\sqrt 3$

## Explanation

A = $\left[ {\matrix{ 2 & b & 1 \cr b & {{b^2} + 1} & b \cr 1 & b & 2 \cr } } \right]$ (b > 0)

$\left| A \right|$ = 2(2b2 + 2 $-$ b2) $-$ b(2b $-$ b) + 1(b2 $-$ b2 $-$ 1)

$\left| A \right|$ = 2(b2 + 2) $-$ b2 $-$ 1

$\left| A \right|$ = b2 + 3

${{\left| A \right|} \over b} = b + {3 \over b} \Rightarrow {{b + {3 \over b}} \over 2} \ge \sqrt 3$

$b + {3 \over b} \ge 2\sqrt 3$