1

JEE Main 2016 (Online) 10th April Morning Slot

Let A be a 3 $\times$ 3 matrix such that A2 $-$ 5A + 7I = 0

Statement - I :

A$-$1 = ${1 \over 7}$ (5I $-$ A).

Statement - II :

The polynomial A3 $-$ 2A2 $-$ 3A + I can be reduced to 5(A $-$ 4I).

Then :
A
Statement-I is true, but Statement-II is false.
B
Statement-I is false, but Statement-II is true.
C
Both the statements are true.
D
Both the statements are false

Explanation

Given,

A2 $-$ 5A + 7I = 0

$\Rightarrow$   A2 $-$ 5A = $-$ 7I

$\Rightarrow$   AAA$-$1 $-$ 5AA$-$1 = $-$ 7IA$-$1

$\Rightarrow$   AI $-$ 5I = $-$ 7A$-$1

$\Rightarrow$   A $-$ 5I = $-$ 7A$-$1

$\Rightarrow$   A$-$1 = ${1 \over 7}$(5I $-$ A)

Hence, statement 1 is true.

Now A3 $-$ 2A2 $-$ 3A + I

=   A(A2) $-$ 2A2 $-$ 3A + I

=   A(5A $-$ 7I) $-$ 2A2 $-$ 3A + I

=   5A2 $-$ 7A $-$ 2A2 $-$ 3A + I

=   3A2 $-$ 10A + I

=   3(5A $-$ 7I) $-$ 10A + I

=   15A $-$ 21A $-$ 10A + I

=   5A $-$ 20I

=   5(A $-$ 4I)

So, statement 2 is also correct.
2

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
3

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) = $\left[ {\matrix{ {51} & {63} \cr {84} & {72} \cr } } \right]$
4

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.