1
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 15th April Evening Slot

If the system of linear equations
x + ay + z = 3
x + 2y + 2z = 6
x + 5y + 3z = b
has no solution, then :
A
a = $$-$$ 1,    b = 9
B
a = $$-$$ 1,    b $$ \ne $$ 9
C
a $$ \ne $$ $$-$$ 1,    b = 9
D
a = 1,    b $$ \ne $$ 9

Explanation

As the given system of equations has no solution then

$$\Delta $$ = 0 and at least one of $$\Delta $$1, $$\Delta $$2 and $$\Delta $$2 should not be zero.

$$ \therefore $$ $$\Delta $$ = $$\left| {\matrix{ 1 & a & 1 \cr 1 & 2 & 2 \cr 1 & 5 & 3 \cr } } \right| = 0$$

$$ \Rightarrow $$ - $$a$$ - 1 = 0

$$ \Rightarrow $$ a = - 1

$$\Delta $$2 = $$\left| {\matrix{ 1 & 3 & 1 \cr 1 & 6 & 2 \cr 1 & b & 3 \cr } } \right| \ne 0$$

$$ \Rightarrow $$ b $$ \ne $$ 0
2
MCQ (Single Correct Answer)

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.
3
MCQ (Single Correct Answer)

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

Answer

=

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
4
MCQ (Single Correct Answer)

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:

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

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