Byju's Answer
Standard XII
Mathematics
Properties of Determinants
If a-b-c 2a ...
Question
If
∣
∣ ∣
∣
a
−
b
−
c
2
a
2
a
2
b
b
−
c
−
a
2
b
2
c
2
c
c
−
a
−
b
∣
∣ ∣
∣
=
(
a
+
b
+
c
)
(
x
+
a
+
b
+
c
)
2
,
x
≠
0
and
a
+
b
+
c
≠
0
, then
x
is equal to :
A
a
b
c
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B
2
(
a
+
b
+
c
)
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C
−
(
a
+
b
+
c
)
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D
−
2
(
a
+
b
+
c
)
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Solution
The correct option is
D
−
2
(
a
+
b
+
c
)
Δ
=
∣
∣ ∣
∣
a
−
b
−
c
2
a
2
a
2
b
b
−
c
−
a
2
b
2
c
2
c
c
−
a
−
b
∣
∣ ∣
∣
R
1
→
R
1
+
R
2
+
R
3
Δ
=
∣
∣ ∣
∣
a
+
b
+
c
a
+
b
+
c
a
+
b
+
c
2
b
b
−
c
−
a
2
b
2
c
2
c
c
−
a
−
b
∣
∣ ∣
∣
=
(
a
+
b
+
c
)
∣
∣ ∣
∣
1
1
1
2
b
b
−
c
−
a
2
b
2
c
2
c
c
−
a
−
b
∣
∣ ∣
∣
C
1
→
C
1
−
C
2
,
C
2
→
C
2
−
C
3
Δ
=
(
a
+
b
+
c
)
∣
∣ ∣
∣
0
0
1
b
+
c
+
a
−
b
−
c
−
a
2
b
0
c
+
a
+
b
c
−
a
−
b
∣
∣ ∣
∣
=
(
a
+
b
+
c
)
(
a
+
b
+
c
)
2
Hence,
x
=
−
2
(
a
+
b
+
c
)
Suggest Corrections
19
Similar questions
Q.
If
∣
∣ ∣
∣
a
−
b
−
c
2
a
2
a
2
b
b
−
c
−
a
2
b
2
c
2
c
c
−
a
−
b
∣
∣ ∣
∣
=
(
a
+
b
+
c
)
(
x
+
a
+
b
+
c
)
2
,
x
≠
0
and
a
+
b
+
c
≠
0
, then
x
is equal to :
Q.
If
∣
∣ ∣
∣
a
−
b
−
c
2
a
2
a
2
b
b
−
c
−
a
2
b
2
c
2
c
c
−
a
−
b
∣
∣ ∣
∣
=
(
a
+
b
+
c
)
(
x
+
a
+
b
+
c
)
2
,
x
≠
0
and
a
+
b
+
c
≠
0
, then is equal to:-
Q.
∣
∣ ∣
∣
a
−
b
−
c
2
a
2
a
2
b
b
−
c
−
a
2
b
2
c
2
c
c
−
a
−
b
∣
∣ ∣
∣
is equal to
Q.
If
∣
∣ ∣
∣
a
−
b
−
c
2
a
2
a
2
b
b
−
c
−
a
2
b
2
c
2
c
c
−
a
−
b
∣
∣ ∣
∣
=
(
a
+
b
+
c
)
(
x
+
a
+
b
+
c
)
2
,
x
≠
0
and
a
+
b
+
c
≠
0
, then
x
is equal to :
Q.
Using the properties of determinants, show that:
(i)
∣
∣ ∣
∣
a
−
b
−
c
2
a
2
a
2
b
b
−
c
−
a
2
b
2
c
2
c
c
−
a
−
b
∣
∣ ∣
∣
=
(
a
+
b
+
c
)
2
(ii)
∣
∣ ∣
∣
x
+
y
+
2
z
x
y
z
y
+
z
+
2
x
y
z
x
z
+
x
+
2
y
∣
∣ ∣
∣
=
2
(
x
+
y
+
z
)
3
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