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Byju's Answer
Standard XII
Mathematics
In Radius
If Δ 1=111 ab...
Question
If
∆
1
=
1
1
1
a
b
c
a
2
b
2
c
2
,
∆
2
=
1
b
c
a
1
c
a
b
1
a
b
c
,
then
(a)
∆
1
+
∆
2
=
0
(b)
∆
1
+
2
∆
2
=
0
(c)
∆
1
=
∆
2
(d) none of these
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Solution
(a)
∆
1
+
∆
2
=
0
Δ
2
=
1
b
c
a
1
c
a
b
1
a
b
c
=
1
a
b
c
a
a
b
c
a
2
b
b
c
a
b
2
c
c
a
b
c
2
[
R
1
,
R
2
,
R
3
are
multiplied
by
a
,
b
and
c
respectively
,
therefore
we
divide
by
abc
]
=
a
b
c
a
b
c
a
1
a
2
b
1
b
2
c
1
c
2
Taking
a
b
c
common
from
C
2
=
-
1
a
a
2
1
b
b
2
1
c
c
2
C
1
↔
C
2
We
know
that
the
value
of
a
determinant
remains
unchanged
if
its
rows
and
columns
are
interchanged
.
So
,
∆
2
=
-
1
1
1
a
b
c
a
2
b
2
c
2
=
-
Δ
1
⇒
Δ
1
+
Δ
2
=
0
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0
Similar questions
Q.
If
Δ
1
,
∣
∣ ∣
∣
1
1
1
a
b
c
a
2
b
2
c
2
∣
∣ ∣
∣
,
Δ
2
=
∣
∣ ∣
∣
1
b
c
a
1
c
a
b
1
a
b
c
∣
∣ ∣
∣
then
Q.
If
Δ
1
=
∣
∣ ∣
∣
x
sin
θ
cos
θ
−
sin
θ
−
x
1
cos
θ
1
x
∣
∣ ∣
∣
and
Δ
2
=
∣
∣ ∣
∣
x
sin
2
θ
cos
2
θ
−
sin
2
θ
−
x
1
cos
2
θ
1
x
∣
∣ ∣
∣
,
x
≠
0
; then for all
θ
∈
(
0
,
π
2
)
:
Q.
If
Δ
1
=
∣
∣ ∣
∣
a
b
c
x
y
z
p
q
r
∣
∣ ∣
∣
and
Δ
2
=
∣
∣ ∣
∣
y
b
q
x
a
p
z
c
r
∣
∣ ∣
∣
then
Δ
1
is equal to
Q.
Let
Δ
1
=
∣
∣ ∣ ∣
∣
1
cos
α
cos
β
cos
α
1
cos
γ
cos
β
cos
γ
1
∣
∣ ∣ ∣
∣
and
Δ
2
=
∣
∣ ∣ ∣
∣
0
cos
α
cos
β
cos
α
0
cos
γ
cos
β
cos
γ
0
∣
∣ ∣ ∣
∣
.
If
Δ
1
=
Δ
2
, find
sin
2
α
+
sin
2
β
+
sin
2
γ
Q.
If
Δ
1
=
∣
∣ ∣
∣
7
x
2
−
5
x
+
1
3
4
x
7
∣
∣ ∣
∣
,
Δ
2
=
∣
∣ ∣
∣
x
2
7
x
+
1
3
−
5
x
7
4
∣
∣ ∣
∣
then
Δ
1
−
Δ
2
=
0
for
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