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CBSE Questions for Class 12 Commerce Maths Determinants Quiz 4 - MCQExams.com
CBSE
Class 12 Commerce Maths
Determinants
Quiz 4
If the points
(
a
,
0
)
,
(
0
,
b
)
and
(
1
,
1
)
are collinear, then
1
a
+
1
b
equal to -
Report Question
0%
1
0%
2
0%
3
0%
4
Explanation
Given points are
(
a
,
0
)
,
(
0
,
b
)
and
(
1
,
1
)
x
1
=
a
,
y
1
=
0
,
x
2
=
0
,
y
2
=
b
and
x
−
3
=
1
,
y
3
=
1
Condition for collinearity
x
1
y
2
+
x
2
y
3
+
x
3
y
1
=
x
2
y
1
+
x
3
y
2
+
x
1
y
3
gives
a
b
+
0
+
0
=
0
+
1.
b
+
a
.1
⇒
a
b
=
a
+
b
⇒
1
=
1
b
+
1
a
⇒
1
a
+
1
b
=
1
If A is a square matrix so that
A
a
d
j
A
=
d
i
a
g
(
k
,
k
,
k
)
then
|
a
d
j
A
|
=
Report Question
0%
k
0%
k
2
0%
k
3
0%
k
4
Explanation
Given ,
A
a
d
j
A
=
d
i
a
g
(
k
,
k
,
k
)
⇒
|
A
|
I
3
=
[
k
0
0
0
k
0
0
0
k
]
⇒
|
A
|
=
k
Now,
|
a
d
j
A
|
=
|
A
|
2
(
∵
|
a
d
j
A
|
=
|
A
|
n
−
1
)
⇒
|
a
d
j
A
|
=
k
2
If A is a square matrix such that
|
4
0
0
0
4
0
0
0
4
|
=
Report Question
0%
4
0%
16
0%
64
0%
256
Explanation
We know that the determinant of a matrix
A
=
[
a
b
c
d
e
f
g
h
i
]
is:
|
A
|
=
a
(
e
i
−
f
h
)
−
b
(
d
i
−
f
g
)
+
c
(
d
h
−
e
g
)
Here, the given matrix is
A
=
[
4
0
0
0
4
0
0
0
4
]
, so, let us find the
determinant
of matrix
A
as shown below:
|
A
|
=
4
[
(
4
×
4
)
−
(
0
×
0
)
]
−
0
[
(
0
×
4
)
−
(
0
×
0
)
]
+
0
[
(
0
×
0
)
−
(
0
×
4
)
]
=
4
(
16
−
0
)
−
0
(
0
−
0
)
+
0
(
0
−
0
)
=
4
×
16
=
64
Hence, the
determinant
of the matrix
[
4
0
0
0
4
0
0
0
4
]
is
64
.
If
A
=
[
1
x
x
2
4
y
]
,
B
=
[
−
3
1
1
0
]
and
a
d
j
(
A
)
+
B
=
[
1
0
0
1
]
,
then the values of
x
and
y
are respectively
Report Question
0%
(
1
,
1
)
0%
(
−
1
,
1
)
0%
(
1
,
0
)
0%
none of these
Explanation
Given,
A
=
[
1
x
x
2
4
y
]
,
B
=
[
−
3
1
1
0
]
a
d
j
(
A
)
=
[
4
y
−
x
2
−
x
1
]
T
=
[
4
y
−
x
−
x
2
1
]
But given,
a
d
j
(
A
)
+
B
=
[
1
0
0
1
]
⇒
[
4
y
−
3
−
x
+
1
−
x
2
+
1
1
]
=
[
1
0
0
1
]
⇒
−
x
+
1
=
0
,
−
x
2
+
1
=
0
,
4
y
−
3
=
1
⇒
x
=
1
,
y
=
1
If
A
is a non-singular matrix of order
3
×
3
, then adj
(
a
d
j
A
)
is equal to
Report Question
0%
|
A
|
A
0%
|
A
|
2
A
0%
|
A
|
−
1
A
0%
none of these
Explanation
Given A is a matrix of order 3.
Now,
a
d
j
(
a
d
j
A
)
=
|
A
|
n
−
2
A
So,
a
d
j
(
a
d
j
A
)
=
|
A
|
3
−
2
A
=
|
A
|
A
A
=
[
1
0
0
2
1
0
3
2
1
]
,
U
1
,
U
2
and
U
3
are columns matrices satisfying
A
U
1
=
[
1
0
0
]
,
A
U
2
=
[
2
3
0
]
,
A
U
3
=
[
2
3
1
]
and
U
is
3
×
3
matrix whose columns are
U
1
,
U
2
,
U
3
then answer the following question
The value of
|
U
|
is
Report Question
0%
3
0%
−
3
0%
3
2
0%
2
Explanation
Given,
A
U
1
=
[
1
0
0
]
⇒
U
1
=
A
−
1
[
1
0
0
]
...(1)
A
U
2
=
[
2
3
0
]
⇒
U
2
=
A
−
1
[
2
3
0
]
...(2)
A
U
3
=
[
2
3
1
]
⇒
U
3
=
A
−
1
[
2
3
1
]
...(3)
Also, given
A
=
[
1
0
0
2
1
0
3
2
1
]
|
A
|
=
1
Now,
a
d
j
A
=
C
T
=
[
1
−
2
1
0
1
−
2
0
0
1
]
T
⇒
a
d
j
A
=
[
1
0
0
−
2
1
0
1
−
2
1
]
⇒
A
−
1
=
a
d
j
A
|
A
|
=
[
1
0
0
−
2
1
0
1
−
2
1
]
Put the value of
A
−
1
in (1), (2) and (3)
U
1
=
[
1
0
0
−
2
1
0
1
−
2
1
]
[
1
0
0
]
⇒
U
1
=
[
1
−
2
1
]
U
2
=
[
1
0
0
−
2
1
0
1
−
2
1
]
[
2
3
0
]
⇒
U
2
=
[
2
−
1
−
4
]
U
3
=
A
−
1
[
2
3
1
]
⇒
U
3
=
[
2
−
1
−
3
]
∴
U
=
[
1
2
2
−
2
−
1
−
1
1
−
4
−
3
]
⇒
|
U
|
=
−
1
−
14
+
18
=
3
Hence, option A.
If
ω
is an imaginary cube root of unity,then the value of
|
a
b
ω
2
a
ω
b
ω
c
b
ω
2
c
ω
2
a
ω
c
|
,is ?
Report Question
0%
a
3
+
b
3
+
c
3
0%
a
2
b
−
b
2
c
0%
0
0%
a
3
b
+
b
3
+
3
a
b
c
Explanation
△
=
|
a
b
ω
2
a
ω
b
ω
c
b
ω
2
c
ω
2
a
ω
c
|
1
+
ω
+
ω
2
=
0
;
ω
3
=
1
;
ω
4
=
ω
⇒
a
(
c
2
−
a
b
ω
3
)
−
b
ω
2
(
b
c
ω
−
b
c
ω
4
)
+
a
ω
(
a
b
ω
2
−
c
2
ω
2
)
⇒
a
(
c
2
−
a
b
ω
3
)
−
b
ω
2
(
b
c
ω
−
b
c
ω
)
+
a
(
a
b
−
c
2
)
=
a
(
c
2
−
a
b
)
−
a
(
c
2
−
a
b
)
=
0
∴
△
=
0
Option C
If
A
=
|
a
b
c
x
y
z
p
q
r
|
and
B
=
|
q
−
b
y
−
p
a
−
x
r
−
c
z
|
, then
Report Question
0%
A
=
2
B
0%
A
=
B
0%
A
=
−
B
0%
none of these
Explanation
B
=
|
q
−
b
y
−
p
a
−
x
r
−
c
z
|
=
−
|
q
b
y
−
p
−
a
−
x
r
c
z
|
=
|
q
b
y
p
a
x
r
c
z
|
=
−
|
p
a
x
q
b
y
r
c
z
|
=
−
|
a
x
p
b
y
q
c
z
r
|
=
−
|
a
b
c
x
y
z
p
q
r
|
=
−
A
(
∵
|
A
|
=
|
A
T
|
)
Hence, option C.
The value of the determinant
|
1
ω
3
ω
5
ω
3
1
ω
4
ω
5
ω
4
1
|
, where
ω
is an imaginary cube root of unity,is
Report Question
0%
(
1
−
ω
)
2
0%
3
0%
-3
0%
none of these
Explanation
|
1
ω
3
ω
5
ω
3
1
ω
4
ω
5
ω
4
1
|
=
|
1
1
ω
2
1
1
ω
ω
2
ω
1
|
⇒
ω
2
−
2
ω
+
1
(
∵
ω
3
=
1
a
n
d
1
+
ω
+
ω
2
=
0
)
⇒
(
ω
−
1
)
2
Let
ω
=
−
1
2
+
i
√
3
2
,then the value of the determinant
|
1
1
1
1
−
1
−
ω
2
ω
2
1
ω
2
ω
4
|
,
is
Report Question
0%
3
ω
0%
3
ω
(
ω
−
1
)
0%
3
ω
2
0%
3
(
−
2
ω
−
1
)
Explanation
|
1
1
1
1
−
1
−
ω
2
ω
2
1
ω
2
ω
4
|
ω
=
cube root of unity
⇒
ω
3
=
1
1
+
ω
+
ω
2
=
0
⇒
−
1
−
ω
2
=
ω
ω
4
=
(
ω
3
)
ω
=
ω
△
=
|
1
1
1
1
ω
ω
2
1
ω
2
ω
|
=
1
(
ω
2
−
ω
4
)
−
1
(
ω
−
ω
2
)
+
1
(
ω
2
−
ω
)
=
ω
2
−
ω
−
+
ω
2
+
ω
2
−
ω
=
3
(
ω
2
−
ω
)
=
3
ω
(
ω
−
1
)
=
3
(
−
2
ω
−
1
)
Option B and D
Let
A
=
[
1
0
0
2
1
0
3
2
1
]
and
U
1
,
U
2
,
U
3
be column
matrices satisfying
A
U
1
=
[
1
0
0
]
,
A
U
2
=
[
2
3
0
]
,
A
U
3
=
[
2
3
1
]
. If U is
3
×
3
matrix whose columns are
U
1
,
U
2
,
U
3
, then
|
U
|
=
Report Question
0%
3
0%
−
3
0%
3
2
0%
2
Explanation
A
=
[
1
0
0
2
1
0
3
2
1
]
Let
U
1
=
[
a
1
b
1
c
1
]
,
U
2
=
[
a
2
b
2
c
2
]
and
U
3
=
[
a
3
b
3
c
3
]
Given
A
U
1
=
[
1
0
0
]
⇒
a
1
=
1
;
2
a
1
+
b
1
=
0
;
3
a
1
+
2
b
1
+
c
1
=
0
simplifying gives
a
1
=
1
;
b
1
=
−
2
;
c
1
=
1
A
U
2
=
[
2
3
0
]
⇒
a
2
=
2
;
2
a
2
+
b
2
=
3
;
3
a
2
+
2
b
2
+
c
2
=
0
simplifying gives
a
2
=
2
;
b
2
=
−
1
;
c
2
=
−
4
A
U
3
=
[
2
3
1
]
⇒
a
3
=
2
;
2
a
3
+
b
3
=
3
;
3
a
3
+
2
b
3
+
c
3
=
1
simplifying gives
a
3
=
2
;
b
3
=
−
1
;
c
3
=
−
3
∴
U
=
[
1
2
2
−
2
−
1
−
1
1
−
4
−
3
]
|
U
|
=
|
1
2
2
−
2
−
1
−
1
1
−
4
−
3
|
=
3
Hence, option A.
The value of
|
11
12
13
12
13
14
13
14
15
|
,is
Report Question
0%
1
0%
0
0%
-1
0%
67
Explanation
|
11
12
13
12
13
14
13
14
15
|
R
2
→
R
2
−
R
1
,
R
3
→
R
3
−
R
1
=
|
11
12
13
1
1
1
2
2
2
|
=
2
|
11
12
13
1
1
1
1
1
1
|
=
0
If the lines
L
1
:
λ
2
x
−
y
−
1
=
0
L
2
:
x
−
λ
2
y
+
1
=
0
L
3
:
x
+
y
−
λ
2
=
0
pass through the same point the value(s) of
λ
equals
Report Question
0%
1
0%
√
2
0%
2
0%
0
Explanation
The given equations passes through same point.So, they are concurrent lines
⇒
|
λ
2
−
1
−
1
1
−
λ
2
1
1
1
−
λ
2
|
=
0
⇒
λ
6
−
3
λ
2
−
2
=
0
By doing synthetic division we get,
(
λ
4
−
2
λ
2
+
1
)
(
λ
2
−
2
)
=
0
(
λ
2
−
1
)
2
(
λ
2
−
2
)
=
0
λ
=
±
√
2
o
r
λ
=
±
1
But here
λ
=
±
1
d
o
e
s
n
o
t
s
a
t
i
s
f
i
e
s
,
h
e
n
c
e
λ
=
√
2
Option B satisfies above equation
When the determinant
|
cos
2
x
sin
2
x
cos
4
x
sin
2
x
cos
2
x
cos
2
x
cos
4
x
cos
2
x
cos
2
x
|
is expanded in powers of
sin
x
, then the constant term in that expression is
Report Question
0%
1
0%
0
0%
-1
0%
2
Explanation
|
cos
2
x
sin
2
x
cos
4
x
sin
2
x
cos
2
x
cos
2
x
cos
4
x
cos
2
x
cos
2
x
|
=
a
0
+
a
1
sin
x
+
a
2
sin
2
x
+
.
.
.
.
.
Put
x
=
0
⇒
|
1
0
1
0
1
1
1
1
1
|
=
a
0
⇒
a
0
=
−
1
If
p
+
q
+
r
=
0
=
a
+
b
+
c
, then the value of the determinant
|
p
a
q
b
r
c
q
c
r
a
p
b
r
b
p
c
q
a
|
is
Report Question
0%
0
0%
p
q
+
q
b
+
r
a
0%
1
0%
none of these
Explanation
Given ,
p
+
q
+
r
=
0
=
a
+
b
+
c
|
p
a
q
b
r
c
q
c
r
a
p
b
r
b
p
c
q
a
|
=
p
a
(
q
r
a
2
−
p
2
b
c
)
−
q
b
(
q
2
a
c
−
p
r
b
2
)
+
r
c
(
p
q
c
2
−
r
2
a
b
)
=
p
q
r
a
3
−
p
3
a
b
c
−
q
3
a
b
c
+
p
q
r
b
3
+
p
q
r
c
3
−
r
3
a
b
c
=
p
q
r
(
a
3
+
b
3
+
c
3
)
−
a
b
c
(
p
3
+
q
3
+
r
3
)
=
p
q
r
[
(
a
+
b
+
c
)
3
+
3
a
b
c
(
a
+
b
+
c
)
]
−
a
b
c
[
(
p
+
q
+
r
)
3
+
3
p
q
r
(
p
+
q
+
r
)
]
=
0
If
a
≠
b
≠
c
,
are value of x which satisfies the equation
|
0
x
−
a
x
−
b
x
+
a
0
x
−
c
x
+
b
x
+
c
0
|
=
0
is given by
Report Question
0%
x
=
0
0%
x
=
c
0%
x
=
b
0%
x
=
a
Explanation
|
0
x
−
a
x
−
b
x
+
a
0
x
−
c
x
+
b
x
+
c
0
|
=
0
⇒
(
x
−
a
)
(
x
+
b
)
(
x
−
c
)
+
(
x
−
b
)
(
x
+
a
)
(
x
+
c
)
=
0
We can now check by options
f
o
r
o
p
t
i
o
n
A
,
put
x
=
0
in above equation we get,
⇒
(
0
−
a
)
(
0
+
b
)
(
0
−
c
)
+
(
0
−
b
)
(
0
+
a
)
(
0
+
c
)
=
0
⇒
a
b
c
−
a
b
c
=
0
So,
x
=
0
satisfies the equation.
The value of
|
−
1
2
1
3
+
2
√
2
2
+
2
√
2
1
3
−
2
√
2
2
−
2
√
2
1
|
is equal to
Report Question
0%
zero
0%
−
16
√
2
0%
−
8
√
2
0%
one of these
Explanation
|
−
1
2
1
3
+
2
√
2
2
+
2
√
2
1
3
−
2
√
2
2
−
2
√
2
1
|
−
1
(
2
+
2
√
2
−
2
+
2
√
2
)
−
2
(
3
+
2
√
2
−
3
+
2
√
2
)
+
1
[
(
3
+
2
√
2
)
(
2
−
2
√
2
)
−
(
2
+
2
√
2
)
(
3
−
2
√
2
)
]
=
−
1
(
4
√
2
)
−
2
(
4
√
2
)
+
1
(
−
4
√
2
)
=
−
16
√
2
Number of values of
a
for which the lines
2
x
+
y
−
1
=
0
,
a
x
+
3
y
−
3
=
0
,
3
x
+
2
y
−
2
=
0
are concurrent is
Report Question
0%
0
0%
1
0%
2
0%
∞
Explanation
Here coefficient matrix,
Δ
=
|
2
1
−
1
a
3
−
3
3
2
−
2
|
Using
C
2
→
C
2
+
C
3
Δ
=
|
2
0
−
1
a
0
3
3
0
−
2
|
=
0
Clearly
Δ
=
0
, Hence given lines are concurrent for all values of
a
Let
|
x
2
x
x
2
x
6
x
x
6
|
=
A
x
4
+
B
x
3
+
C
x
2
+
D
x
+
E
. Then the value of
5
A
+
4
B
+
3
C
+
2
D
+
E
is equal to
Report Question
0%
zero
0%
-16
0%
16
0%
-11
Explanation
|
x
2
x
x
2
x
6
x
x
6
|
=
A
x
4
+
B
x
3
+
C
x
2
+
D
x
+
E
Consider
L
H
S
=
|
x
2
x
x
2
x
6
x
x
6
|
=
x
(
6
x
−
6
x
)
−
2
(
6
x
2
−
6
x
)
+
x
(
x
3
−
x
2
)
=
x
4
−
x
3
−
12
x
2
+
12
x
Comparing with RHS, we get
A
=
1
,
B
=
−
1
,
C
=
−
12
,
D
=
12
,
E
=
0
So,
5
A
+
4
B
+
3
C
+
2
D
+
E
=
−
11
In triangle
A
B
C
, if
|
1
1
1
cot
A
2
cot
B
2
cot
C
2
tan
B
2
+
tan
C
2
tan
C
2
+
tan
A
2
tan
A
2
+
tan
B
2
|
=
0
, then the triangle must be
Report Question
0%
equilateral
0%
isosceles
0%
obtuse angled
0%
none of these
Explanation
|
1
1
1
cot
A
2
cot
B
2
cot
C
2
tan
B
2
+
tan
C
2
tan
C
2
+
tan
A
2
tan
A
2
+
tan
B
2
|
=
0
⇒
tan
A
2
cot
B
2
−
tan
A
2
cot
C
2
+
tan
B
2
cot
C
2
−
tan
B
2
cot
A
2
+
tan
C
2
cot
A
2
−
tan
C
2
cot
B
2
=
0
tan
A
2
(
cot
B
2
−
cot
C
2
)
+
tan
B
2
(
cot
C
2
−
cot
A
2
)
+
tan
C
2
(
cot
A
2
−
cot
B
2
)
=
0
(
cot
B
2
−
cot
C
2
)
=
0
,
(
cot
C
2
−
cot
A
2
)
=
0
,
(
cot
A
2
−
cot
B
2
)
=
0
⇒
A
=
B
=
C
The value of the determinant
|
1
1
1
m
C
1
m
+
1
C
1
m
+
2
C
1
m
C
2
m
+
1
C
2
m
+
2
C
2
|
is equal to
Report Question
0%
1
0%
−
1
0%
0
0%
none of these
Explanation
|
1
1
1
m
C
1
m
+
1
C
1
m
+
2
C
1
m
C
2
m
+
1
C
2
m
+
2
C
2
|
=
|
1
1
1
m
m
+
1
m
+
2
m
(
m
−
1
)
2
m
(
m
+
1
)
2
(
m
+
2
)
(
m
+
1
)
2
|
C
1
→
C
1
−
C
2
,
C
2
→
C
2
−
C
3
|
0
0
1
−
1
−
1
m
+
2
−
m
−
(
m
+
1
)
(
m
+
2
)
(
m
+
1
)
2
|
1
(
−
1
(
−
(
m
+
1
)
)
−
(
−
1
)
(
−
m
)
)
=
1
If
a
,
b
,
c
are different, then the value of
x
satisfying
|
0
x
2
−
a
x
3
−
b
x
2
+
a
0
x
2
+
c
x
4
+
b
x
−
c
0
|
=
0
is
Report Question
0%
a
0%
c
0%
b
0%
0
Δ
=
|
a
a
2
0
1
2
a
+
b
(
a
+
b
)
0
1
2
a
+
3
b
|
is divisible by
Report Question
0%
a
+
b
0%
a
+
2
b
0%
2
a
+
3
b
0%
a
2
Explanation
Expanding by
R
3
,
=
−
1
(
a
(
a
+
b
)
)
+
(
2
a
+
3
b
)
(
2
a
2
+
a
b
−
a
2
)
=
−
1
(
a
(
a
+
b
)
)
+
(
2
a
+
3
b
)
a
(
a
+
b
)
=
a
(
a
+
b
)
(
2
a
+
3
b
−
1
)
If
Δ
=
|
sin
θ
cos
ϕ
sin
θ
sin
ϕ
cos
θ
cos
θ
cos
ϕ
cos
θ
sin
ϕ
−
sin
θ
−
sin
θ
sin
ϕ
sin
θ
cos
ϕ
0
|
, then
Report Question
0%
Δ
is independent of
θ
0%
Δ
is independent of
ϕ
0%
Δ
is a constant
0%
d
Δ
d
θ
|
θ
=
π
/
2
=
0
Explanation
Δ
=
|
s
i
n
θ
c
o
s
ϕ
s
i
n
θ
s
i
n
ϕ
c
o
s
θ
c
o
s
θ
c
o
s
ϕ
c
o
s
θ
s
i
n
ϕ
−
s
i
n
θ
−
s
i
n
θ
s
i
n
ϕ
s
i
n
θ
c
o
s
ϕ
0
|
Expanding by
C
3
,
Δ
=
c
o
s
θ
(
c
o
s
θ
s
i
n
θ
(
c
o
s
2
ϕ
+
s
i
n
2
ϕ
)
)
+
s
i
n
θ
(
s
i
n
2
θ
(
c
o
s
2
ϕ
+
s
i
n
2
ϕ
)
)
=
c
o
s
θ
(
c
o
s
θ
s
i
n
θ
)
+
s
i
n
θ
(
s
i
n
2
θ
)
=
s
i
n
θ
So,
Δ
is independent of
ϕ
d
Δ
d
θ
=
c
o
s
θ
|
π
2
=
0
If
α
is a characteristic root of a nonsingular matrix, then the corresponding characteristic root of adj A is
Report Question
0%
|
A
|
α
0%
|
A
α
|
0%
|
a
d
j
A
|
α
0%
|
a
d
j
A
α
|
Explanation
Since
α
is a characteristic root of a nonsingular matrix,
therefore
α
≠
0
. Also
α
is a characteristic root of
A implies that there exists a nonzero vector X such that
A
X
=
α
X
(
a
d
j
A
)
(
A
X
)
=
(
a
d
j
A
)
(
α
X
)
[
(
a
d
j
A
)
A
]
X
=
α
(
a
d
j
A
)
X
|
A
|
I
X
=
α
(
a
d
j
A
)
X
[
∵
(
a
d
j
A
)
A
=
|
A
|
I
]
|
A
|
X
=
α
(
a
d
j
A
)
X
|
A
|
X
=
α
(
a
d
j
A
)
X
|
A
|
α
X
=
(
a
d
j
A
)
X
(
a
d
j
A
)
X
=
|
A
|
α
X
Since X is a nonzero vector,
|
A
/
α
|
is a characteristic root of the matrix adj A.
If
A
is a square matrix of order
m
×
n
, then
a
d
j
(
a
d
j
A
)
is equal to
Report Question
0%
|
A
|
n
A
0%
|
A
|
n
−
1
A
0%
|
A
|
n
−
2
A
0%
|
A
|
n
−
3
A
Explanation
(
a
d
j
A
)
−
1
=
a
d
j
(
a
d
j
A
)
|
a
d
j
A
|
⇒
(
|
A
|
A
−
)
−
1
=
a
d
j
(
a
d
j
A
)
|
a
d
j
A
|
⇒
A
|
A
|
=
a
d
j
(
a
d
j
A
)
|
a
d
j
A
|
⇒
a
d
j
(
a
d
j
A
)
=
|
a
d
j
A
|
A
|
A
|
∴
a
d
j
(
a
d
j
A
)
=
|
A
|
n
−
2
A
Hence, option C.
If
A
=
[
−
1
−
2
−
2
2
1
−
2
2
−
2
1
]
, Then
a
d
j
(
A
)
equals
Report Question
0%
A
0%
A
T
0%
3
A
0%
3
A
T
Explanation
Given
A
=
[
−
1
−
2
−
2
2
1
−
2
2
−
2
1
]
A
T
=
[
−
1
2
2
−
2
1
−
2
−
2
−
2
1
]
Now,
a
d
j
A
=
C
T
=
[
−
3
−
6
−
6
6
3
−
6
6
−
6
3
]
T
⇒
a
d
j
A
=
[
−
3
6
6
−
6
3
−
6
−
6
−
6
3
]
⇒
a
d
j
A
=
3
[
−
1
2
2
−
2
1
−
2
−
2
−
2
1
]
⇒
a
d
j
A
=
3
A
T
Find the determinants of minors and cofactors of the determinant
|
2
3
4
7
2
−
5
8
−
1
3
|
Report Question
0%
|
1
61
−
23
13
−
26
−
26
−
23
−
38
−
17
|
and
|
1
61
−
23
−
13
−
26
26
−
23
38
−
17
|
0%
|
1
61
−
23
13
−
26
−
26
−
23
−
38
−
17
|
and
|
1
−
61
23
−
13
−
26
26
−
23
38
−
17
|
0%
|
1
61
−
23
13
−
26
−
26
−
23
−
38
−
17
|
and
|
1
61
23
−
13
−
26
26
−
23
38
−
17
|
0%
None of these.
Explanation
Here
M
11
=
|
2
−
5
−
1
3
|
(Delete 1st row and first column)
=
6
−
5
M
11
=
1
∴
C
11
=
1
(
∵
(
−
1
)
1
+
1
=
1
)
M
2
=
|
7
−
5
8
3
|
(Delete 1st row and 2nd column)
=
21
−
(
−
40
)
M
2
=
61
∴
C
12
=
−
61
,
(
∵
(
−
1
)
1
+
2
=
−
1
)
M
13
=
|
7
2
8
−
1
|
(Delete 1st row and 3rd column)
=
−
7
−
16
M
13
=
−
23
∴
C
13
=
−
23
,
(
∵
(
−
1
)
1
+
3
=
1
)
M
21
=
|
3
4
−
1
3
|
(Delete 2nd row and 1st column)
=
9
−
(
−
4
)
M
21
=
13
∴
C
21
=
−
13
,
(
∵
(
−
1
)
2
+
1
=
−
1
)
M
22
=
|
2
4
8
3
|
(Delete 2nd row and 2nd column)
=
6
−
32
M
22
=
−
26
∴
C
22
=
−
26
,
(
∵
(
−
1
)
2
+
2
=
1
)
M
23
=
|
2
3
8
−
1
|
(Delete 2nd row and 3rd column)
=
−
2
−
24
M
23
=
−
26
∴
C
23
=
26
,
(
∵
(
−
1
)
2
+
3
=
−
1
)
M
31
=
|
3
4
2
−
5
|
(Delete 3rd row and 1st column)
=
−
15
−
8
M
31
=
−
23
∴
C
31
=
−
23
,
(
∵
(
−
1
)
3
+
1
=
1
)
M
32
=
|
2
4
7
−
5
|
(Delete 3rd row and 2nd column)
=
−
10
−
28
M
32
=
−
38
∴
C
32
=
38
,
(
∵
(
−
1
)
3
+
2
=
−
1
)
M
33
=
|
2
3
7
2
|
(Delete 3rd row and 3rd column)
=
4
−
21
M
33
=
−
17
∴
C
33
=
−
17
,
(
∵
(
−
1
)
3
+
3
=
1
)
Hence Determinants of Minors and Cofactors are
|
1
61
−
23
13
−
26
−
26
−
23
−
38
−
17
|
and
|
1
−
61
−
23
−
13
−
26
26
−
23
38
−
17
|
The adjoint of the matrix
A
=
[
1
1
1
2
1
−
3
−
1
2
3
]
is
Report Question
0%
1
11
[
9
−
1
−
4
−
3
4
5
5
−
3
−
1
]
0%
[
9
1
−
4
3
4
−
5
5
3
−
1
]
0%
[
9
−
3
5
−
1
4
−
3
−
4
5
−
1
]
0%
[
9
−
1
−
4
−
3
4
5
5
−
3
−
1
]
Explanation
Let
C
i
j
be a cofactor of
a
i
j
in A. Then, the cofactors of elements of A are given by
C
11
=
[
1
−
3
2
3
]
=
9
C
12
=
−
[
2
−
3
−
1
3
]
=
−
3
C
13
=
[
2
1
−
1
2
]
=
5
C
21
=
−
[
1
1
2
3
]
=
−
1
C
22
=
[
1
1
−
1
3
]
=
4
C
23
=
−
[
1
1
−
1
2
]
=
−
3
C
31
=
[
1
1
1
−
3
]
=
−
4
C
32
=
−
[
1
1
2
−
3
]
=
5
C
33
=
[
1
1
2
1
]
=
−
1
∴
a
d
j
A
=
[
9
−
3
5
−
1
4
−
3
−
4
5
−
1
]
′
=
[
9
−
1
−
4
−
3
4
5
5
−
3
−
1
]
If
A
2
=
I
, then the value of
d
e
t
(
A
−
I
)
is (where
A
has order
3
)
Report Question
0%
1
0%
−
1
0%
0
0%
cannot say anything
Explanation
A
2
=
I
⇒
A
2
−
I
=
O
⇒
(
A
−
I
)
(
A
+
I
)
=
O
⇒
|
A
−
I
|
|
A
+
I
|
=
0
∴
Either
|
A
−
I
|
=
0
or
|
A
+
I
|
=
0
Hence, cannot say anything about
|
A
−
I
|
.
Hence, option D.
If
A
=
|
1
2
2
1
|
and
f
(
x
)
=
1
+
x
1
−
x
, then
f
(
|
A
|
)
is
Report Question
0%
−
1
2
0%
1
2
0%
−
1
3
0%
none of these
Explanation
Here
|
A
|
=
1
×
1
−
2
×
2
=
−
3
∴
f
(
|
A
|
)
=
1
+
(
−
3
)
1
+
3
=
−
1
2
If
A
is a square matrix of order
n
×
n
and
k
is a scalar, then
a
d
j
(
k
A
)
is equal to _____________.
Report Question
0%
k
n
−
1
a
d
j
A
0%
k
n
a
d
j
A
0%
k
n
+
1
a
d
j
A
0%
k
a
d
j
A
Explanation
(
k
A
)
−
1
=
a
d
j
(
k
A
)
|
k
A
|
⇒
1
k
A
−
1
=
a
d
j
(
k
A
)
|
k
A
|
⇒
a
d
j
A
k
|
A
|
=
a
d
j
(
k
A
)
k
n
|
A
|
∴
a
d
j
(
k
A
)
=
k
n
−
1
a
d
j
A
Hence, option A.
Find the adjoint of the matrix
A
=
[
1
2
3
1
3
5
1
5
12
]
.
If
Adjoint A:
[
a
−
9
1
b
9
−
2
2
c
1
]
, find the value of
a
b
c
.
Report Question
0%
231
0%
213
0%
321
0%
312
If matrix A is given by
A
=
[
6
11
2
4
]
, then the determinant of
A
2005
−
6
A
2004
is
Report Question
0%
2
2006
0%
(
−
11
)
2
2005
0%
−
2
2005
0%
(
−
9
)
2
2004
Explanation
Given,
A
=
[
6
11
2
4
]
|
A
|
=
2
Now,
|
A
2005
−
6
A
2004
|
=
|
A
|
2004
|
A
−
6
I
|
=
2
2004
|
0
11
2
−
2
|
=
2
2004
(
−
22
)
=
2
2005
(
−
11
)
|
0
p
−
q
p
−
r
q
−
p
0
q
−
r
r
−
p
r
−
q
0
|
is equal to
Report Question
0%
p
+
q
+
r
0%
0
0%
p
−
q
−
r
0%
−
p
+
q
+
r
Explanation
|
0
p
−
q
p
−
r
q
−
p
0
q
−
r
r
−
p
r
−
q
0
|
=
0
(
0
−
(
r
−
q
)
(
q
−
r
)
)
−
(
p
−
q
)
(
0
−
(
r
−
p
)
(
q
−
r
)
)
+
(
p
−
r
)
(
(
q
−
p
)
(
r
−
q
)
−
0
)
=
(
p
−
q
)
(
r
−
p
)
(
q
−
r
)
+
(
p
−
r
)
(
q
−
p
)
(
r
−
q
)
=
0
If
|
6
i
−
3
i
1
4
3
i
−
1
20
3
i
|
=
x
+
i
y
then
Report Question
0%
x
=
3
,
y
=
1
0%
x
=
1
,
y
=
3
0%
x
=
0
,
y
=
3
0%
x
=
0
,
y
=
0
Explanation
Δ
=
|
6
i
−
3
i
1
4
3
i
−
1
20
3
i
|
=
6
i
(
−
3
+
3
)
+
3
i
(
4
i
+
20
)
+
1
(
12
−
60
i
)
=
0
−
12
+
60
i
+
12
−
60
i
=
0
⇒
x
=
0
,
y
=
0
Hence, the option 'D' is correct.
If
A
=
[
3
−
3
4
2
−
3
4
0
−
1
1
]
then find
A
d
j
(
A
d
j
A
)
.
Report Question
0%
[
3
−
3
4
2
−
3
4
0
−
1
1
]
0%
[
3
3
4
2
−
3
−
4
0
−
1
1
]
0%
[
3
3
4
2
−
3
4
0
1
1
]
0%
[
3
−
3
4
2
−
3
−
4
0
1
1
]
Explanation
We know that
A
d
j
(
A
d
j
A
)
=
|
A
|
n
−
2
A
Since
A
=
[
3
−
3
4
2
−
3
4
0
−
1
1
]
Here
n
=
3
(3 order matrix)
∴
|
A
|
=
|
3
−
3
4
2
−
3
4
0
−
1
1
|
=
3
(
1
)
+
3
(
2
−
0
)
+
4
(
−
2
−
0
)
=
1
≠
0
,
∴
A
is non-singular.
∴
A
d
j
(
A
d
j
A
)
=
|
A
|
3
−
2
.
A
=
|
A
|
.
A
=
1.
A
=
A
=
[
3
−
3
4
2
−
3
4
0
−
1
1
]
The matrix
[
1
0
1
2
1
0
3
1
1
]
is
Report Question
0%
non-singular
0%
singular
0%
skew-symmetric
0%
symmetric
Explanation
Given
A
=
[
1
0
1
2
1
0
3
1
1
]
|
A
|
=
|
1
0
1
2
1
0
3
1
1
|
⇒
|
A
|
=
1
−
1
=
0
Hence,
A
is singular.
The value of the
m
t
h
order determinant of a matrix
A
is
15
then the value of determinant formed by the cofactors of
A
will be
Report Question
0%
(
15
)
m
0%
15
2
m
0%
(
15
)
m
−
1
0%
(
15
)
2
m
−
1
Explanation
If A be the matrix of order m and B is matrix of cofactors of a then determinant of B will be equal to
(
m
−
1
)
t
h
power of the determinant of A.
|
B
|
=
|
A
|
m
−
1
∴
|
B
|
=
(
15
)
m
−
1
The points
(
a
,
b
+
c
)
,
(
b
,
c
+
a
)
,
(
c
,
a
+
b
)
are
Report Question
0%
vertices of an equilateral triangle
0%
collinear
0%
concyclic
0%
none of these
Explanation
To check for collineauty, we can check the slope of the two points. Suppose the points be A, B and C then
(
s
l
o
p
e
)
A
B
=
(
c
+
a
)
−
(
b
+
c
)
b
−
a
=
−
1
(
s
l
o
p
e
)
B
C
=
(
a
+
b
)
−
(
c
+
a
)
c
−
b
=
−
1
m
A
B
=
m
B
C
=
points are collinear
In a third order determinant
a
i
j
denotes the element in the ith row and the jth column If
a
i
j
=
{
0
,
i
=
j
1
,
i
>
j
−
1
,
i
<
j
then the value of the determinant
Report Question
0%
0
0%
1
0%
-1
0%
none of these
Explanation
a
i
j
=
{
0
,
i
=
j
1
,
i
>
j
−
1
,
i
<
j
Let
A
=
[
0
−
1
−
1
1
0
−
1
1
1
0
]
Now,
|
A
|
=
|
0
−
1
−
1
1
0
−
1
1
1
0
|
=
1
−
1
=
0
⇒
|
A
|
=
0
The value of the determinant
|
√
6
2
i
3
+
√
6
√
12
√
3
+
√
8
i
3
√
2
+
√
6
i
√
18
√
2
+
√
12
i
√
27
+
2
i
|
is
Report Question
0%
complex
0%
real
0%
irrational
0%
rational
Explanation
|
√
6
2
i
3
+
√
6
√
12
√
3
+
√
8
i
3
√
2
+
√
6
i
√
18
√
2
+
√
12
i
√
27
+
2
i
|
R
2
→
R
2
−
√
2
R
1
R
3
→
R
3
−
√
3
R
1
⇒
△
=
√
6
|
1
2
i
3
+
√
6
0
√
3
√
6
i
−
2
√
3
0
√
2
2
i
−
3
√
2
|
⇒
√
6
|
√
3
√
6
i
−
2
√
3
√
2
2
i
−
3
√
2
|
⇒
△
is real and complex.
A and B
If
|
a
+
x
a
x
a
−
x
a
x
a
−
x
a
−
x
|
=
0
then
x
is
Report Question
0%
0
0%
a
0%
3
0%
2
a
If
Δ
=
|
cos
θ
/
2
1
1
1
cos
θ
/
2
−
cos
θ
/
2
−
cos
θ
/
2
1
−
1
|
, If the minimun of
Δ
is
m
1
and maximum of
Δ
is
m
2
, then
[
m
1
,
m
2
]
are related
Report Question
0%
[-4, -2]
0%
[2, 4]
0%
[-4, 0]
0%
[0, 2]
Explanation
Δ
=
|
cos
θ
/
2
1
1
1
cos
θ
/
2
−
cos
θ
/
2
−
cos
θ
/
2
1
−
1
|
=
−
1
(
−
1
−
cos
2
θ
/
2
)
+
1
(
1
+
cos
2
θ
/
2
)
=
2
+
2
cos
2
θ
/
2
=
3
+
cos
θ
Now,
∵
−
1
≤
cos
θ
≤
1
⇒
2
≤
3
+
cos
θ
≤
4
Hence,
m
1
=
2
,
m
2
=
4
If
A
=
[
2
14
17
0
sin
2
x
cos
2
x
0
cos
2
x
sin
2
x
]
then
|
A
|
equals
Report Question
0%
cos
2
x
0%
−
2
0%
−
2
cos
4
x
0%
sin
4
x
Explanation
A
=
[
2
14
17
0
sin
2
x
cos
2
x
0
cos
2
x
sin
2
x
]
|
A
|
=
2
(
sin
2
2
x
−
cos
2
2
x
)
=
−
2
cos
4
x
If
x
is a non-real cube root of
−
2
, then the value of
|
1
2
x
1
x
2
1
3
x
2
2
2
x
1
|
equals to
Report Question
0%
−
7
0%
−
13
0%
0
0%
−
12
Explanation
Given,
x
=
3
√
−
2
⇒
x
3
=
−
2
|
1
2
x
1
x
2
1
3
x
2
2
2
x
1
|
=
(
1
−
6
x
3
)
−
2
x
(
−
5
x
2
)
+
1
(
2
x
3
−
2
)
=
13
−
20
−
6
=
−
13
If
|
x
2
+
3
x
x
+
1
x
−
2
x
−
1
1
−
2
x
x
+
4
x
+
3
x
−
4
3
x
|
=
A
x
4
+
B
x
3
+
C
x
2
+
D
x
+
ϱ
Then value of
ϱ
equals to,
Report Question
0%
-10
0%
10
0%
0
0%
None of these
Explanation
|
x
2
+
3
x
x
+
1
x
−
2
x
−
1
1
−
2
x
x
+
4
x
+
3
x
−
4
3
x
|
=
A
x
4
+
B
x
3
+
C
x
2
+
D
x
+
ϱ
Substitute
x
=
0
⇒
|
0
1
−
2
−
1
1
4
3
−
4
0
|
=
ϱ
=
12
−
2
=
10
If
A
=
(
1
2
1
−
1
0
3
2
−
1
1
)
then characteristic equation is given by
Report Question
0%
−
λ
3
+
2
λ
2
−
4
λ
+
18
=
0
0%
λ
3
+
2
λ
2
+
4
λ
+
18
=
0
0%
2
λ
3
−
λ
2
+
6
λ
−
2
=
0
0%
None of these
Explanation
The characteristic equation of A is given by
|
A
−
λ
I
|
=
0
|
1
−
λ
2
1
−
1
−
λ
3
2
−
1
1
−
λ
|
=
0
(
1
−
λ
)
[
−
λ
(
1
−
λ
)
+
3
]
−
2
(
−
1
(
1
−
λ
)
−
6
)
)
+
1
(
1
+
2
λ
)
=
0
⇒
−
λ
3
+
2
λ
2
−
4
λ
+
18
=
0
If the determinant
|
a
b
a
t
−
b
b
c
b
t
−
c
2
1
0
|
=
0
, if
a
,
b
,
c
are in
Report Question
0%
A
.
P
.
0%
G
.
P
.
0%
H
.
P
.
0%
k
=
1
/
2
Explanation
|
a
b
a
t
−
b
b
c
b
t
−
c
2
1
0
|
=
0
Expanding it along third row,
⇒
2
[
b
(
b
t
−
c
)
−
c
(
a
t
−
b
)
]
−
1
[
a
(
b
t
−
c
)
−
b
(
a
t
−
b
)
]
=
0
⇒
2
t
(
b
2
−
a
c
)
−
(
b
2
−
a
c
)
=
0
⇒
(
2
t
−
1
)
(
b
2
−
a
c
)
=
0
⇒
t
=
1
2
or
b
2
=
a
c
If
b
2
=
a
c
then
a
,
b
,
c
∈
G.P.
If
A
=
[
−
1
−
3
−
3
3
1
−
3
3
−
3
1
]
then adj (A) is
Report Question
0%
=
4
[
−
2
3
3
−
3
2
−
3
−
3
3
2
]
0%
=
4
[
−
2
3
3
3
2
−
3
−
3
−
3
2
]
0%
=
4
[
−
2
−
3
3
−
3
2
−
3
−
3
−
3
2
]
0%
=
4
[
−
2
3
3
−
3
2
−
3
−
3
−
3
2
]
Explanation
Given
A
=
[
−
1
−
3
−
3
3
1
−
3
3
−
3
1
]
C
=
[
−
8
−
12
−
12
12
8
−
12
12
−
12
8
]
a
d
j
A
=
C
T
=
[
−
8
12
12
−
12
8
−
12
−
12
−
12
8
]
=
4
[
−
2
3
3
−
3
2
−
3
−
3
−
3
2
]
0:0:3
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Answered
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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