1

### JEE Main 2018 (Online) 16th April Morning Slot

The number of values of k for which the system of linear equations,
(k + 2)x + 10y = k
kx + (k +3)y = k -1
has no solution, is :
A
1
B
2
C
3
D
infinitely many

## Explanation

System of linear equation have no solution,

$\therefore\,\,\,$ determinant of coefficient = 0

$\left| {\matrix{ {k + 2} & {10} \cr k & {k + 3} \cr } } \right| = 0$

$\Rightarrow$ $\,\,\,\,$ (k + 2) (k + 3) $-$ 10 K = 0

$\Rightarrow$ $\,\,\,\,$ k2 $-$ 5k + 6 = 0

$\therefore\,\,\,\,$ k = 2, 3

When, k = 2 then equations become,

4x + 10y = 2

and 2x + 5y = 1

It has in finite number of solutions.

When k = 3, equations becomes

5x + 10y = 3

3x + 6y = 2

Those equation has no solutions.

$\therefore\,\,\,\,$ When k = 3, then system of equations have no solutions.
2

### JEE Main 2018 (Online) 16th April Morning Slot

Let A = $\left[ {\matrix{ 1 & 0 & 0 \cr 1 & 1 & 0 \cr 1 & 1 & 1 \cr } } \right]$ and B = A20. Then the sum of the elements of the first column of B is :
A
210
B
211
C
231
D
251

=

## Explanation

A = $\left[ {\matrix{ 1 & 0 & 0 \cr 1 & 1 & 0 \cr 1 & 1 & 1 \cr } } \right]$

A2 = A.A = $\left[ {\matrix{ 1 & 0 & 0 \cr 1 & 1 & 0 \cr 1 & 1 & 1 \cr } } \right] \times \left[ {\matrix{ 1 & 0 & 0 \cr 1 & 1 & 0 \cr 1 & 1 & 1 \cr } } \right]$

=   $\left[ {\matrix{ 1 & 0 & 0 \cr 2 & 1 & 0 \cr 3 & 2 & 1 \cr } } \right]$

A3 = A2.A =  $\left[ {\matrix{ 1 & 0 & 0 \cr 2 & 1 & 0 \cr 3 & 2 & 1 \cr } } \right] \times \left[ {\matrix{ 1 & 0 & 0 \cr 1 & 1 & 0 \cr 1 & 1 & 1 \cr } } \right]$

=   $\left[ {\matrix{ 1 & 0 & 0 \cr 3 & 1 & 0 \cr 6 & 3 & 1 \cr } } \right]$

Similarly

A4 =   $\left[ {\matrix{ 1 & 0 & 0 \cr 4 & 1 & 0 \cr {10} & 4 & 1 \cr } } \right]$

From this we can say,

An =   $\left[ {\matrix{ 1 & 0 & 0 \cr n & 1 & 0 \cr {{{n\left( {n + 1} \right)} \over 2}} & n & 1 \cr } } \right]$

$\therefore\,\,\,$ A20 =   $\left[ {\matrix{ 1 & 0 & 0 \cr {20} & 1 & 0 \cr {210} & {20} & 1 \cr } } \right]$

$\therefore\,\,\,$ Sum of the first column

= 1 + 20 + 210

= 231
3

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

If $A = \left[ {\matrix{ {\cos \theta } & { - \sin \theta } \cr {\sin \theta } & {\cos \theta } \cr } } \right]$, then the matrix A–50 when $\theta$ = $\pi \over 12$, is equal to :
A
$\left[ {\matrix{ { {{\sqrt 3 } \over 2}} & { - {1 \over 2}} \cr {{{ 1} \over 2}} & {{{\sqrt 3 } \over 2}} \cr } } \right]$
B
$\left[ {\matrix{ {{1 \over 2}} & -{{{\sqrt 3 } \over 2}} \cr {{{\sqrt 3 } \over 2}} & {{{ - 1} \over 2}} \cr } } \right]$
C
$\left[ {\matrix{ {{{\sqrt 3 } \over 2}} & {{1 \over 2}} \cr -{{1 \over 2}} & {{{\sqrt 3 } \over 2}} \cr } } \right]$
D
$\left[ {\matrix{ {{1 \over 2}} & {{{\sqrt 3 } \over 2}} \cr {-{{\sqrt 3 } \over 2}} & {{{ 1} \over 2}} \cr } } \right]$

## Explanation

(A$-$50) = (A$-$1)50

We know,

A$-$1 = ${{adjA} \over {\left| A \right|}}$

$\left| A \right|$ = cos2$\theta$ + sin2$\theta$ = 1

cofactor of A = $\left[ {\matrix{ {\cos \theta } & { - \sin \theta } \cr {\sin \theta } & {\cos \theta } \cr } } \right]$

Adjoint of A = Transpose of cofactor matrix

$\therefore$  Adj A = $\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$

$\therefore$  A$-$1 = $\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$

$\therefore$  (A$-$1)2 = $\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$

= $\left[ {\matrix{ {\cos 2\theta } & {\sin 2\theta } \cr { - \sin 2\theta } & {\cos 2\theta } \cr } } \right]$

Similarly,

(A$-$1)3 = $\left[ {\matrix{ {\cos 3\theta } & {\sin 2\theta } \cr { - \sin 3\theta } & {\cos 3\theta } \cr } } \right]$

:

:

:

(A$-$1)50 = $\left[ {\matrix{ {\cos 50\theta } & {\sin 50\theta } \cr { - \sin 50\theta } & {\cos 50\theta } \cr } } \right]$

when $\theta$ = ${\pi \over {12}}$ then

${A^{ - 50}}$ = $\left[ {\matrix{ {\cos {{25\pi } \over 6}} & {\sin {{25\pi } \over 6}} \cr { - \sin {{25\pi } \over 6}} & {\cos {{25\pi } \over 6}} \cr } } \right]$

= $\left[ {\matrix{ {{{\sqrt 3 } \over 2}} & {{1 \over 2}} \cr { - {1 \over 2}} & {{{\sqrt 3 } \over 2}} \cr } } \right]$

Note:

$\cos {{25\pi } \over 6} = \cos \left( {4\pi + {\pi \over 6}} \right) = \cos {\pi \over 6} = {{\sqrt 3 } \over 2}$
4

### 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