Answer:
∣
a−b
b−a
c−a
b+c
c+a
a+b
a
b
c
C
1
→C
3
−C
2
Determinant ⇒b(c
−b
)−c(ac−ab)+a(ab−ac)
=bc
−c
a+abc+a
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Answers & Comments
Answer:
∣
∣
∣
∣
∣
∣
∣
∣
a−b
b−a
c−a
b+c
c+a
a+b
a
b
c
∣
∣
∣
∣
∣
∣
∣
∣
C
1
→C
3
−C
1
∣
∣
∣
∣
∣
∣
∣
∣
b
a
a
b+c
c+a
a+b
a
b
c
∣
∣
∣
∣
∣
∣
∣
∣
C
2
→C
2
−C
1
∣
∣
∣
∣
∣
∣
∣
∣
b
a
a
c
c
b
a
b
c
∣
∣
∣
∣
∣
∣
∣
∣
Determinant ⇒b(c
2
−b
2
)−c(ac−ab)+a(ab−ac)
=bc
2
−b
3
−c
2
a+abc+a
2
b−a
2
c