Using properties solve the determinants:
1+x 2 3
1 2+x 3
1 2 3+x
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Answered by
0
Explanation :
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12th
Maths
Determinants
Properties of Determinants
Using properties of determi...
MATHS
Using properties of determinants prove the following :
∣
∣
∣
∣
∣
∣
∣
∣
3x
x−y
x−z
−x+y
3y
y−z
−x+z
z−y
3z
∣
∣
∣
∣
∣
∣
∣
∣
=3(x+y+z)(xy+yz+xz).
October 15, 2019avatar
Sanghati Sarfraz
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VIDEO EXPLANATION
ANSWER
First take LHS: let Δ=
∣
∣
∣
∣
∣
∣
∣
∣
3x
x−y
x−z
−x+y
3y
y−z
−x+z
z−y
3z
∣
∣
∣
∣
∣
∣
∣
∣
[Applying C
1
→C
1
+C
2
+C
3
]
=
∣
∣
∣
∣
∣
∣
∣
∣
x+y+z
x+y+z
x+y+z
−x+y
3y
y−z
−x+z
z−y
3z
∣
∣
∣
∣
∣
∣
∣
∣
Takine x + y + z common from C
1
=(x+y+z)
∣
∣
∣
∣
∣
∣
∣
∣
1
1
1
−x+y
3y
y−z
−x+z
z−y
3z
∣
∣
∣
∣
∣
∣
∣
∣
Applying R
1
→R
1
−R
3
,R
2
→R
2
−R
3
=(x+y+z)
∣
∣
∣
∣
∣
∣
∣
∣
0
0
1
−x+y
3y
y−z
−x+z
z−y
3z
∣
∣
∣
∣
∣
∣
∣
∣
Expanding along C
1
=(x+y+z)(1×
∣
∣
∣
∣
∣
∣
−x+z
2y+z
−x−2z
−2z−y
∣
∣
∣
∣
∣
∣
)
=(x+y+z)[(−x
z
)(−2z−y)−(2y+z)(−x−2z)]
=(x+y+z)(3xy+3yz+3zx)
=3(x+y+z)(xy+yz+zx)
LHS = RHS
Hence proved.
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Answered by
3
Answer:
LHS=RHS IS answer of question
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