If z = ax + by +cz +d = 0 then Independent variables is
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For given equation equation to posses non-trivial solution we must have,
∣
∣
∣
∣
∣
∣
∣
∣
a
1
1
1
b
1
1
1
c
∣
∣
∣
∣
∣
∣
∣
∣
=0
Now use the property of R
1
→R
1
−R
2
and R
2
→R
2
−R
3
∣
∣
∣
∣
∣
∣
∣
∣
a−1
0
1
1−b
b−1
1
0
1−c
c
∣
∣
∣
∣
∣
∣
∣
∣
=0
Now expand along first row
⇒(a−1)[(b−1)c−(1−c)]−(1−b)[0−(1−c)]=0
⇒(a−1)(b−1)c−(a−1)(1−c)+(1−b)(1−c)=0
⇒(1−a)(1−b)c+(1−a)(1−c)+(1−b)(1−c)=0
Divide both sides (1−a)(1−b)(1−c)
⇒
1−c
c
+
1−b
1
+
1−a
1
=0
⇒−
1−c
1−c−1
+
1−b
1
+
1−a
1
=0
⇒−1+
1−c
1
+
1−b
1
+
1−a
1
=0
∴
1−a
1
+
1−b
1
+
1−c
1
=1
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