1
MCQ (Single Correct Answer)

JEE Main 2016 (Online) 10th April Morning Slot

If    A = $$\left[ {\matrix{ { - 4} & { - 1} \cr 3 & 1 \cr } } \right]$$,

then the determinant of the matrix (A2016 − 2A2015 − A2014) is :
A
2014
B
$$-$$ 175
C
2016
D
$$-$$ 25

Explanation

Given,

$$A = \left[ {\matrix{ { - 4} & { - 1} \cr 3 & 1 \cr } } \right]$$

$${A^2} = \left[ {\matrix{ { - 4} & { - 1} \cr 3 & 1 \cr } } \right]\left[ {\matrix{ { - 4} & { - 1} \cr 3 & 1 \cr } } \right]$$

$$ = \left[ {\matrix{ {13} & 3 \cr { - 9} & { - 2} \cr } } \right]$$

A2 $$-$$ 2A $$-$$ I

$$ = \left[ {\matrix{ {13} & 3 \cr { - 9} & { - 2} \cr } } \right] - \left[ {\matrix{ { - 8} & { - 2} \cr 6 & 2 \cr } } \right] - \left[ {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right]$$

$$ = \left[ {\matrix{ {20} & 5 \cr { - 15} & { - 5} \cr } } \right]$$

$$\left| A \right| = \left| {\matrix{ { - 4} & { - 1} \cr 3 & 1 \cr } } \right|$$ $$=$$ $$-$$ 4 + 3 $$=$$ $$-$$ 1

Now,

$$\left| {{A^{2016}} - 2{A^{2015}} - {A^{2014}}} \right|$$

$$=$$ $${\left| A \right|^{2014}}\left| {{A^2} - 2A - {\rm I}} \right|$$

$$ = {\left( { - 1} \right)^{2014}}\left| {\matrix{ {20} & 5 \cr { - 15} & { - 5} \cr } } \right|$$

$$=$$ 1 $$ \times $$ ($$-$$ 100 + 75)

$$=$$ $$-$$ 25
2
MCQ (Single Correct Answer)

JEE Main 2017 (Offline)

If $$A = \left[ {\matrix{ 2 & { - 3} \cr { - 4} & 1 \cr } } \right]$$,

then adj(3A2 + 12A) is equal to
A
$$\left[ {\matrix{ {51} & {63} \cr {84} & {72} \cr } } \right]$$
B
$$\left[ {\matrix{ {51} & {84} \cr {63} & {72} \cr } } \right]$$
C
$$\left[ {\matrix{ {72} & {-63} \cr {-84} & {51} \cr } } \right]$$
D
$$\left[ {\matrix{ {72} & {-84} \cr {-63} & {51} \cr } } \right]$$

Explanation

We have, $$A = \left[ {\matrix{ 2 & { - 3} \cr { - 4} & 1 \cr } } \right]$$

$$ \therefore $$ A2 = A.A = $$\left[ {\matrix{ 2 & { - 3} \cr { - 4} & 1 \cr } } \right]\left[ {\matrix{ 2 & { - 3} \cr { - 4} & 1 \cr } } \right]$$

= $$\left[ {\matrix{ {4 + 12} & { - 6 - 3} \cr { - 8 - 4} & {12 + 1} \cr } } \right]$$

= $$\left[ {\matrix{ {16} & { - 9} \cr { - 12} & {13} \cr } } \right]$$

Now, 3A2 + 12A

= $$3\left[ {\matrix{ {16} & { - 9} \cr { - 12} & {13} \cr } } \right] + 12\left[ {\matrix{ 2 & { - 3} \cr { - 4} & 1 \cr } } \right]$$

= $$\left[ {\matrix{ {48} & { - 27} \cr { - 36} & {39} \cr } } \right] + \left[ {\matrix{ {24} & { - 36} \cr { - 48} & {12} \cr } } \right]$$

= $$\left[ {\matrix{ {72} & { - 63} \cr { - 84} & {51} \cr } } \right]$$

$$ \therefore $$ adj(3A2 + 12A) = $$\left[ {\matrix{ {51} & {63} \cr {84} & {72} \cr } } \right]$$
3
MCQ (Single Correct Answer)

JEE Main 2017 (Offline)

If S is the set of distinct values of 'b' for which the following system of linear equations

x + y + z = 1
x + ay + z = 1
ax + by + z = 0

has no solution, then S is :
A
an empty set
B
an infinite set
C
a finite set containing two or more elements
D
a singleton

Explanation

$$\left| {\matrix{ 1 & 1 & 1 \cr 1 & a & 1 \cr a & b & 1 \cr } } \right| = 0$$

$$ \Rightarrow $$ 1 [a – b] – 1 [1 – a] + 1 [b – a2] = 0

$$ \Rightarrow $$ (a - 1)2 = 0

$$ \Rightarrow $$ a = 1

For a = 1, the equations become

x + y + z = 1

x + y + z = 1

x + by + z = 0

These equations give no solution for b = 1

$$ \Rightarrow $$ S is singleton set.
4
MCQ (Single Correct Answer)

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.

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