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Byju's Answer
Standard XII
Mathematics
Properties of Determinants
Using the pro...
Question
Using the property of determinants , prove that,
∣
∣ ∣
∣
b
+
c
p
+
r
y
+
z
c
+
a
r
+
p
z
+
x
a
+
b
p
+
q
x
+
y
∣
∣ ∣
∣
=
2
∣
∣ ∣
∣
a
p
x
b
q
y
c
r
z
∣
∣ ∣
∣
Open in App
Solution
△
=
∣
∣ ∣
∣
b
+
c
p
+
r
y
+
z
c
+
a
r
+
p
z
+
x
a
+
b
p
+
q
x
+
y
∣
∣ ∣
∣
=
∣
∣ ∣
∣
b
p
+
r
y
+
z
c
r
+
p
z
+
x
a
p
+
q
x
+
y
∣
∣ ∣
∣
+
∣
∣ ∣
∣
c
p
+
r
y
+
z
a
r
+
p
z
+
x
b
p
+
q
x
+
y
∣
∣ ∣
∣
Again Splitting
=
∣
∣ ∣
∣
b
p
y
+
z
c
r
z
+
x
a
p
x
+
y
∣
∣ ∣
∣
+
∣
∣ ∣
∣
c
r
y
+
z
a
p
z
+
x
b
q
x
+
y
∣
∣ ∣
∣
+
∣
∣ ∣
∣
c
r
y
+
z
a
p
z
+
x
b
q
x
+
y
∣
∣ ∣
∣
On splitting all above Determinants together, we get
=
∣
∣ ∣
∣
b
c
a
q
r
p
y
z
x
∣
∣ ∣
∣
+
∣
∣ ∣
∣
b
c
b
q
r
q
y
z
y
∣
∣ ∣
∣
+
∣
∣ ∣
∣
b
a
a
q
p
p
y
x
x
∣
∣ ∣
∣
+
∣
∣ ∣
∣
b
a
b
q
p
q
y
x
y
∣
∣ ∣
∣
+
∣
∣ ∣
∣
c
a
a
r
p
p
z
x
x
∣
∣ ∣
∣
+
∣
∣ ∣
∣
c
a
b
r
p
q
z
x
y
∣
∣ ∣
∣
△
=
∣
∣ ∣
∣
b
c
a
q
r
p
y
z
x
∣
∣ ∣
∣
+
∣
∣ ∣
∣
c
a
b
r
p
q
z
x
y
∣
∣ ∣
∣
(
1
)
C
1
↔
C
3
&
C
1
↔
2
(
2
)
△
=
∣
∣ ∣
∣
a
c
b
p
r
q
x
z
y
∣
∣ ∣
∣
−
∣
∣ ∣
∣
a
c
b
p
r
q
x
z
y
∣
∣ ∣
∣
C
2
↔
C
3
△
=
2
∣
∣ ∣
∣
a
b
c
p
q
r
x
y
z
∣
∣ ∣
∣
By taking transpose, we get
△
=
2
∣
∣ ∣
∣
a
p
x
b
q
y
c
r
z
∣
∣ ∣
∣
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Similar questions
Q.
Using the property of determinants and without expanding, prove that
∣
∣ ∣
∣
b
+
c
q
+
r
y
+
z
c
+
a
r
+
p
z
+
x
a
+
b
p
+
q
x
+
y
∣
∣ ∣
∣
=
2
∣
∣ ∣
∣
a
p
x
b
q
y
c
r
z
∣
∣ ∣
∣
Q.
Using properties of determinants, prove that
⎡
⎢
⎣
b
+
c
q
+
r
y
+
z
c
+
a
r
+
p
z
+
x
a
+
b
p
+
q
x
+
y
⎤
⎥
⎦
=
2
⎡
⎢
⎣
a
p
x
b
q
y
c
r
z
⎤
⎥
⎦
Q.
Using the property of determinants and without expanding, prove that: