1

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

The system of linear equations
x + y + z = 2
2x + 3y + 2z = 5
2x + 3y + (a2 – 1) z = a + 1 then
A
has infinitely many solutions for a = 4
B
has a unique solution for |a| = $\sqrt3$
C
is inconsistent when |a| = $\sqrt3$
D
is inconsistent when a = 4

## Explanation

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

D = 3$a$2 $-$ 3 $-$ 6 $-$ 2$a$2 + 2 + 4 + 2$a$2 $-$ 2 $-$ 4

D = ($a$2 $-$ 3)

When D $\ne$ 0 then system of equiation has unique solution.

$\therefore$  3($a$2 $-$ 3) $\ne$ 0

$\Rightarrow$  $\left| a \right|$ $\ne \sqrt 3$

When $3({a^2} - 3) = 0$

$\Rightarrow$  $\left| a \right| = \sqrt 3$ then D = 0

If D = 0 then two cases possible

(1)  System of equation has infinite many solution.

(2) System of equation has no solution and inconsistent.

Here D1 = $\left| {\matrix{ 2 & 1 & 1 \cr 5 & 3 & 2 \cr {a + 1} & 3 & {{a^2} - 1} \cr } } \right| = {a^2} - a + 1$

D2 = $\left| {\matrix{ 1 & 2 & 1 \cr 2 & 5 & 2 \cr 2 & {a + 1} & {{a^2} - 1} \cr } } \right| = {a^2} - 3$

D3 = $\left| {\matrix{ 1 & 1 & 2 \cr 2 & 3 & 5 \cr 2 & 3 & {a + 1} \cr } } \right| = a - 4$

System of equation will have infinite solution if D1 = D2 = D3 = 0.

And system of equation will have no solution if at last one of D1, D2, D2 is non zero.

At $\left| a \right| = \sqrt 3$ we get D = 0

But D1 = 3 $\pm$ $\sqrt 3 + 1$ $\ne$ 0

and D3 = $\pm$ $\sqrt 3 - 4$ $\ne$ 0

So, system of equations has no solution at $\left| a \right| = \sqrt 3$ then system is in consistent
2

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

If the system of linear equations
x $-$ 4y + 7z = g
3y $-$ 5z = h
$-$2x + 5y $-$ 9z = k
is consistent, then :
A
g + 2h + k = 0
B
g + h + 2k = 0
C
2g + h + k = 0
D
g + h + k = 0

## Explanation

x $-$ $4y + 7z = g$
$3y$ $-$ $5z = h$
$-$$2x + 5y$ $-$ $9z = k$

$D = \left| {\matrix{ 1 & { - 4} & 7 \cr 0 & 3 & { - 5} \cr { - 2} & 5 & { - 9} \cr } } \right|$

$D = 1\left( { - 27 + 25} \right) - 2\left( {20 - 21} \right)$

$D = - 2 + 2 = 0$

If system is consistent then ${D_1} = {D_2} = {D_3} = 0$

$\left| {\matrix{ 1 & { - 4} & g \cr 0 & 3 & h \cr { - 2} & 5 & k \cr } } \right| = 0$

$1\left( {3k - 5h} \right) - 2\left( { - 4h - 3g} \right) = 0$

$3k - 5h + 8h + 6g = 0$

$6g + 3h + 3k = 0$

$2g + h + k = 0$
3

### 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
4

### 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 ${D_y} = {D_z} = 0$

$\alpha = 5,\beta = 13$