4
You visited us
4
times! Enjoying our articles?
Unlock Full Access!
Byju's Answer
Standard XII
Mathematics
Cramer's Rule
Solve the fol...
Question
Solve the following determinant equations:
(i)
x
+
a
b
c
a
x
+
b
c
a
b
x
+
c
=
0
(ii)
x
+
a
x
x
x
x
+
a
x
x
x
x
+
a
=
0
,
a
≠
0
(iii)
3
x
-
8
3
3
3
3
x
-
8
3
3
3
3
x
-
8
=
0
(iv)
1
x
x
2
1
a
a
2
1
b
b
2
=
0
,
a
≠
b
(v)
x
+
1
3
5
2
x
+
2
5
2
3
x
+
4
=
0
(vi)
1
x
x
3
1
b
b
3
1
c
c
3
=
0
,
b
≠
c
(vii)
15
-
2
x
11
-
3
x
7
-
x
11
17
14
10
16
13
=
0
(viii)
1
1
x
p
+
1
p
+
1
p
+
x
3
x
+
1
x
+
2
=
0
Open in App
Solution
(i)
Let
∆
=
x
+
a
b
c
a
x
+
b
c
a
b
x
+
c
=
x
+
a
+
b
+
c
b
c
x
+
a
+
b
+
c
x
+
b
c
x
+
a
+
b
+
c
b
x
+
c
Applying
C
1
→
C
1
+
C
2
+
C
3
=
x
+
a
+
b
+
c
1
b
c
1
x
+
b
c
1
b
x
+
c
=
x
+
a
+
b
+
c
1
b
c
0
x
0
1
b
x
+
c
Applying
R
2
→
R
2
-
R
1
=
x
+
a
+
b
+
c
1
b
c
0
x
0
0
0
x
Applying
R
3
→
R
3
-
R
1
∆
=
x
+
a
+
b
+
c
x
2
-
0
=
0
Given
⇒
x
2
=
0
or
x
+
a
+
b
+
c
=
0
⇒
x
=
0
or
x
=
-
a
+
b
+
c
(ii)
Let
∆
=
x
+
a
x
x
x
x
+
a
x
x
x
x
+
a
=
3
x
+
a
x
x
3
x
+
a
x
+
a
x
3
x
+
a
x
x
+
a
Applying
C
1
→
C
1
+
C
2
+
C
3
=
3
x
+
a
1
x
x
1
x
+
a
x
1
x
x
+
a
=
3
x
+
a
1
x
x
0
a
0
1
x
x
+
a
Applying
R
2
→
R
2
-
R
1
=
3
x
+
a
1
x
x
0
a
0
0
0
a
Applying
R
3
→
R
3
-
R
1
∆
=
3
x
+
a
a
2
-
0
=
0
x
=
-
a
3
(iii)
Let
∆
=
3
x
-
8
3
3
3
3
x
-
8
3
3
3
3
x
-
8
=
3
x
-
2
3
3
3
x
-
2
3
x
-
8
3
3
x
-
2
3
3
x
-
8
Applying
C
1
=
C
1
+
C
2
+
C
3
=
3
x
-
2
1
3
3
1
3
x
-
8
3
1
3
3
x
-
8
=
3
x
-
2
1
3
3
0
3
x
-
11
0
1
3
3
x
-
8
Applying
R
2
→
R
2
-
R
1
=
3
x
-
2
1
3
3
0
3
x
-
11
0
0
0
3
x
-
11
Applying
R
3
→
R
3
-
R
1
∆
=
3
x
-
2
3
x
-
11
2
=
0
x
=
2
3
,
11
3
,
11
3
(iv)
Let
∆
=
1
x
x
2
1
a
a
2
1
b
b
2
=
1
x
x
2
0
x
-
a
x
2
-
a
2
1
b
b
2
Applying
R
2
→
R
1
-
R
2
=
1
x
x
2
0
x
-
a
x
2
-
a
2
0
x
-
b
x
2
-
b
2
Applying
R
3
→
R
1
-
R
3
=
x
-
a
x
-
b
1
x
x
2
0
1
x
+
a
0
1
x
+
b
∆
=
x
-
a
x
-
b
x
+
b
-
x
-
a
=
0
x
=
a
,
b
(v)
Let
∆
=
x
+
1
3
5
2
x
+
2
5
2
3
x
+
4
=
x
+
9
3
5
x
+
9
x
+
2
5
x
+
9
3
x
+
4
Applying
C
1
=
C
1
+
C
2
+
C
3
=
x
+
9
1
3
5
1
x
+
2
5
1
3
x
+
4
=
x
+
9
1
3
5
0
x
-
1
0
1
3
x
+
4
Applying
R
2
→
R
2
-
R
1
=
x
+
9
1
3
5
0
x
-
1
0
0
0
x
-
1
Applying
R
3
→
R
3
-
R
1
∆
=
x
+
9
x
-
1
2
=
0
x
=
-
9
,
1
,
1
(vi)
Let
∆
=
1
x
x
3
1
b
b
3
1
c
c
3
=
1
x
x
3
0
b
-
x
b
3
-
x
3
1
c
c
3
Applying
R
2
→
R
2
-
R
1
=
1
x
x
3
0
b
-
x
b
3
-
x
3
0
c
-
x
c
3
-
x
3
Applying
R
3
→
R
3
-
R
1
=
1
x
x
3
0
x
-
b
x
3
-
b
3
0
x
-
c
x
3
-
c
3
=
x
-
b
x
-
c
1
x
x
2
0
1
x
2
+
x
b
+
b
2
0
1
x
2
+
x
c
+
c
2
∆
=
x
-
b
x
-
c
x
c
-
b
-
b
2
+
c
2
=
0
x
=
b
,
c
,
-
b
+
c
(vii)
Let
Δ
=
15
-
2
x
11
-
3
x
7
-
x
11
17
14
10
16
13
=
0
⇒
15
-
2
x
-
14
+
2
x
11
-
3
x
7
-
x
11
-
28
17
14
10
-
26
16
13
=
0
Applying
C
1
→
C
1
-
2
C
3
⇒
1
11
-
3
x
7
-
x
-
17
17
14
-
16
16
13
=
0
⇒
12
-
3
x
4
-
2
x
7
-
x
0
3
14
0
3
13
=
0
Applying
C
1
→
C
1
+
C
2
and
C
2
→
C
2
-
C
3
⇒
12
-
3
x
3
×
13
-
3
×
14
=
0
⇒
12
-
3
x
-
3
=
0
⇒
12
-
3
x
=
0
⇒
3
x
=
12
⇒
x
=
4
(viii)
Let
∆
=
1
1
x
p
+
1
p
+
1
p
+
x
3
x
+
1
x
+
2
=
1
1
x
p
p
p
3
x
+
1
x
+
2
Applying
R
2
→
R
2
-
R
1
=
p
1
1
x
1
1
1
3
x
+
1
x
+
2
=
p
1
1
x
1
1
1
2
x
2
Applying
R
3
→
R
3
-
R
1
=
p
0
1
x
0
1
1
2
-
x
x
2
Applying
C
1
→
C
1
-
C
2
=
p
2
-
x
×
1
x
1
1
Expanding
along
C
1
=
p
2
-
x
1
-
x
=
0
x
=
1
,
2
Suggest Corrections
0
Similar questions
Q.
Solve the following determinant equations:
(i)
x
+
a
b
c
a
x
+
b
c
a
b
x
+
c
=
0
(ii)
x
+
a
x
x
x
x
+
a
x
x
x
x
+
a
=
0
,
a
≠
0
(iii)
3
x
-
8
3
3
3
3
x
-
8
3
3
3
3
x
-
8
=
0
(iv)
1
x
x
2
1
a
a
2
1
b
b
2
=
0
,
a
≠
b
(v)
x
+
1
3
5
2
x
+
2
5
2
3
x
+
4
=
0
(vi)
1
x
x
3
1
b
b
3
1
c
c
3
=
0
,
b
≠
c
(vii)
15
-
2
x
11
-
3
x
7
-
x
11
17
14
10
16
13
=
0
(viii)
1
1
x
p
+
1
p
+
1
p
+
x
3
x
+
1
x
+
2
=
0
(ix)
3
-
2
sin
3
θ
-
7
8
cos
2
θ
-
11
14
2
=
0
Q.
Solve the following determinant equations:
(i)
x
+
a
b
c
a
x
+
b
c
a
b
x
+
c
=
0
(ii)
x
+
a
x
x
x
x
+
a
x
x
x
x
+
a
=
0
,
a
≠
0
(iii)
3
x
-
8
3
3
3
3
x
-
8
3
3
3
3
x
-
8
=
0
(iv)
1
x
x
2
1
a
a
2
1
b
b
2
=
0
,
a
≠
b
(v)
x
+
1
3
5
2
x
+
2
5
2
3
x
+
4
=
0
(vi)
1
x
x
3
1
b
b
3
1
c
c
3
=
0
,
b
≠
c
(vii)
15
-
2
x
11
-
3
x
7
-
x
11
17
14
10
16
13
=
0
(viii)
1
1
x
p
+
1
p
+
1
p
+
x
3
x
+
1
x
+
2
=
0
(ix)
3
-
2
sin
3
θ
-
7
8
cos
2
θ
-
11
14
2
=
0
(x)
4
-
x
4
+
x
4
+
x
4
+
x
4
-
x
4
+
x
4
+
x
4
+
x
4
+
x
=
0
Q.
If
p
≠
0
solution set of the equation
Δ
=
∣
∣ ∣
∣
1
1
x
p
+
1
p
+
1
p
+
x
3
x
+
1
x
+
2
∣
∣ ∣
∣
=
0
is
Q.
Determine the nature of the roots of the following quadratic equations:
(i) 2x
2
− 3x + 5 = 0
(ii) 2x
2
− 6x + 3 = 0
(iii)
3
5
x
2
-
2
3
x
+
1
=
0
(iv)
3
x
2
-
4
3
x
+
4
=
0
(v)
3
x
2
-
2
6
x
+
2
=
0
(vi)
x
-
2
a
x
-
2
b
=
4
a
b
(vii)
9
a
2
b
2
x
2
-
24
a
b
c
d
x
+
16
c
2
d
2
=
0
,
a
≠
0
,
b
≠
0
(viii)
2
a
2
+
b
2
x
2
+
2
a
+
b
x
+
1
=
0
(ix)
b
+
c
x
2
-
a
+
b
+
c
x
+
a
=
0