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Textbooks
NCERT Mathematics Part-I Std 12
Determinants
All Exercises
Exercise 4.1
Exercise 4.2
Exercise 4.3
Exercise 4.4
Exercise 4.5
Exercise 4.6
Miscellaneous Exercises on Chapter 4
Chapter 4 : Determinants
Q.
If
a
,
b
,
c
are in A.P, then the determinant
∣
∣ ∣
∣
x
+
2
x
+
3
x
+
2
a
x
+
3
x
+
4
x
+
2
b
x
+
4
x
+
5
x
+
2
c
∣
∣ ∣
∣
is
0
1
x
2
x
View Solution
Q.
Evaluate the determinants
(i)
∣
∣ ∣
∣
3
−
1
−
2
0
0
−
1
3
−
5
−
0
∣
∣ ∣
∣
(ii)
∣
∣ ∣
∣
3
−
4
5
1
1
−
2
2
3
1
∣
∣ ∣
∣
(iii)
∣
∣ ∣
∣
0
1
2
−
1
0
−
3
−
2
3
0
∣
∣ ∣
∣
(iv)
∣
∣ ∣
∣
2
−
1
−
2
0
2
−
1
3
−
5
0
∣
∣ ∣
∣
View Solution
Q.
Let
A
=
⎡
⎢
⎣
1
sin
θ
1
−
sin
θ
1
sin
θ
−
1
−
sin
θ
1
⎤
⎥
⎦
, where
0
≤
θ
≤
2
π
. Then
|
A
|
lies between
View Solution
Q.
Prove that:
∣
∣ ∣ ∣
∣
sin
α
cos
α
cos
(
α
+
δ
)
sin
β
cos
β
cos
(
β
+
δ
)
sin
γ
cos
γ
cos
(
γ
+
δ
)
∣
∣ ∣ ∣
∣
=
0
View Solution
Q.
If
∣
∣
∣
x
2
18
x
∣
∣
∣
=
∣
∣
∣
6
2
3
x
6
∣
∣
∣
, then
x
is equal to
−
6
±
6
6
0
View Solution
Q.
If true Enter
′
1
′
else
′
0
′
.
∣
∣ ∣
∣
x
y
x
+
y
y
x
+
y
x
x
+
y
x
y
∣
∣ ∣
∣
=
−
2
(
x
+
y
)
(
x
2
+
y
2
−
x
y
)
View Solution
Q.
If
x
,
y
,
z
are non zero real numbers, then the inverse of matrix
A
=
⎡
⎢
⎣
x
0
0
0
y
0
0
0
z
⎤
⎥
⎦
is
⎡
⎢
⎣
x
−
1
0
0
0
y
−
1
0
0
0
z
−
1
⎤
⎥
⎦
x
y
z
⎡
⎢
⎣
x
−
1
0
0
0
y
−
1
0
0
0
z
−
1
⎤
⎥
⎦
1
x
y
z
⎡
⎢
⎣
x
0
0
0
y
0
0
0
z
⎤
⎥
⎦
1
x
y
z
⎡
⎢
⎣
1
0
0
0
1
0
0
0
1
⎤
⎥
⎦
View Solution
Q.
Prove that
∣
∣ ∣
∣
1
1
+
p
1
+
p
+
q
2
3
+
2
p
4
+
3
p
+
2
q
3
6
+
3
p
10
+
6
p
+
3
q
∣
∣ ∣
∣
=
1
View Solution
Q.
Let
A
=
⎡
⎢
⎣
1
sin
θ
1
−
sin
θ
1
sin
θ
−
1
−
sin
θ
1
⎤
⎥
⎦
, where
0
≤
θ
≤
2
π
. Then
D
e
t
(
A
)
=
0
D
e
t
(
A
)
∈
(
2
,
∞
)
D
e
t
(
A
)
∈
(
2
,
4
)
D
e
t
(
A
)
∈
[
2
,
4
]
View Solution
Q.
If
A
=
⎡
⎢
⎣
1
1
−
2
2
1
−
3
5
4
−
9
⎤
⎥
⎦
, find |A|.
View Solution
Q.
Evaluate
∣
∣ ∣
∣
1
x
y
1
x
+
y
y
x
+
y
x
y
∣
∣ ∣
∣
View Solution
Q.
Using properties of determinant, prove
that:
∣
∣ ∣
∣
3
a
−
a
+
b
−
a
+
c
−
b
+
a
3
b
−
b
+
c
−
c
+
a
−
c
+
b
3
c
∣
∣ ∣
∣
=
3
(
a
+
b
+
c
)
(
a
b
+
b
c
+
c
a
)
View Solution
Q.
Find the value of determinant.
(i)
∣
∣
∣
cos
θ
−
sin
θ
sin
θ
cos
θ
∣
∣
∣
(ii)
∣
∣
∣
x
2
−
x
+
1
x
−
1
x
+
1
x
+
1
∣
∣
∣
View Solution
Q.
If A=
⎡
⎢
⎣
1
0
1
0
1
2
0
0
4
⎤
⎥
⎦
then show that |3A|=27|A|
View Solution
Q.
Find the value of determinant
∣
∣
∣
2
4
−
5
−
1
∣
∣
∣
.
View Solution
Q.
If A=
∣
∣
∣
1
2
4
2
∣
∣
∣
, then show that
|
2
A
|
=
4
|
A
|
View Solution
Q.
Find the values of
x
, if
(i)
∣
∣
∣
2
4
5
1
∣
∣
∣
=
∣
∣
∣
2
x
4
6
x
∣
∣
∣
(ii)
∣
∣
∣
2
3
4
5
∣
∣
∣
=
∣
∣
∣
x
3
2
x
5
∣
∣
∣
View Solution
Q.
Using properties of determinant, prove
that:
∣
∣ ∣ ∣
∣
x
x
2
1
+
p
x
3
y
y
2
1
+
p
y
3
z
z
2
1
+
p
z
3
∣
∣ ∣ ∣
∣
=
(
1
+
p
x
y
z
)
(
x
−
y
)
(
y
−
z
)
(
z
−
x
)
View Solution
Q.
Using the property of determinants and without expanding, prove that
∣
∣ ∣
∣
x
a
x
+
a
y
b
y
+
b
z
c
z
+
c
∣
∣ ∣
∣
=
0
View Solution
Q.
Using the property of determinants and without expanding, find
∣
∣ ∣
∣
a
−
b
b
−
c
c
−
a
b
−
c
c
−
a
a
−
b
c
−
a
a
−
b
b
−
c
∣
∣ ∣
∣
=
0
View Solution
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