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Byju's Answer
Standard XII
Mathematics
Properties of Determinants
Prove the fol...
Question
Prove the following identities:
∣
∣ ∣
∣
b
+
c
c
+
a
a
+
b
q
+
r
r
+
p
p
+
q
y
+
z
z
+
x
x
+
y
∣
∣ ∣
∣
=
2
∣
∣ ∣
∣
a
b
c
p
q
r
x
y
z
∣
∣ ∣
∣
.
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Solution
Δ
=
∣
∣ ∣
∣
b
+
c
c
+
a
a
+
b
q
+
r
r
+
p
p
+
q
y
+
z
z
+
x
x
+
y
∣
∣ ∣
∣
⇒
Δ
=
∣
∣ ∣
∣
b
c
a
q
r
p
y
z
x
∣
∣ ∣
∣
+
∣
∣ ∣
∣
c
a
b
r
p
q
z
x
y
∣
∣ ∣
∣
⇒
Δ
=
Δ
1
+
Δ
2
Δ
1
=
∣
∣ ∣
∣
b
c
a
q
r
p
y
z
x
∣
∣ ∣
∣
Interchanging
C
1
and
C
3
⇒
Δ
1
=
−
∣
∣ ∣
∣
a
c
b
p
r
q
x
z
y
∣
∣ ∣
∣
Interchanging
C
2
and
C
3
⇒
Δ
1
=
∣
∣ ∣
∣
a
b
c
p
q
r
x
y
z
∣
∣ ∣
∣
similarly in
Δ
2
interchange
C
2
and
C
1
and then
C
2
and
C
3
⇒
Δ
2
=
∣
∣ ∣
∣
a
b
c
p
q
r
x
y
z
∣
∣ ∣
∣
Δ
=
Δ
1
+
Δ
2
⇒
Δ
=
2
∣
∣ ∣
∣
a
b
c
p
q
r
x
y
z
∣
∣ ∣
∣
[
h
e
n
c
e
p
r
o
v
e
d
]
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Similar questions
Q.
Using properties of determinants, prove that
∣
∣ ∣
∣
b
+
c
c
+
a
a
+
b
q
+
r
r
+
p
p
+
q
y
+
z
z
+
x
x
+
y
∣
∣ ∣
∣
=
2
∣
∣ ∣
∣
a
b
c
p
q
r
x
y
z
∣
∣ ∣
∣
.
Q.
The
∣
∣ ∣
∣
b
+
c
c
+
a
a
+
b
q
+
r
r
+
p
p
+
q
y
+
z
z
+
x
x
+
y
∣
∣ ∣
∣
can be simplified to
Q.
Prove the following identities:
2
a
a
+
b
+
2
b
b
+
c
+
2
c
c
+
a
+
(
b
−
c
)
(
c
−
a
)
(
a
−
b
)
(
b
+
c
)
(
c
+
a
)
(
a
+
b
)
=
3
.
Q.
Prove the following identity:
(
b
+
c
)
3
+
(
c
+
a
)
3
+
(
a
+
b
)
3
−
3
(
b
+
c
)
(
c
+
a
)
(
a
+
b
)
=
2
(
a
3
+
b
3
+
c
3
−
3
a
b
c
)
.
Q.
Show that:
∣
∣ ∣ ∣
∣
(
a
−
x
)
2
(
a
−
y
)
2
(
a
−
z
)
2
(
b
−
x
)
2
(
b
−
y
)
2
(
b
−
z
)
2
(
c
−
x
)
2
(
c
−
y
)
2
(
c
−
z
)
2
∣
∣ ∣ ∣
∣
=
2
(
b
−
c
)
(
c
−
a
)
(
a
−
b
)
(
y
−
z
)
(
z
−
x
)
(
x
−
y
)
.
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