Question 33 (OR 1st Question) - CBSE Class 12 Sample Paper for 2020 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards
Last updated at May 29, 2023 by Teachoo

Using the properties of determinants, prove that
| (y + z)
^{
2
}
x
^{
2
}
x
^{
2
}
y
^{
2
}
(z + x)
^{
2
}
y
^{
2
}
z
^{
2
}
z
^{
2
}
(x + y)
^{
2
}
| = 2xyz (x + y + z)
^{
3
}

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Transcript
Question 33 (OR 1st Question) Using the properties of determinants, prove that [ (y + z)2 x2 x2 y2 (z + x)2 y2 z2 z2 (x + y)2 ] 2xyz (x + y + z)3
[ (y + z)2 x2 x2 y2 (z + x)2 y2 z2 z2 (x + y)2 ]
Applying C2 -> C2 - C1
= | (y + z)2 x2 - (y + z)2 x2 y2 (z + x)2 - y2 y2 z2 z2 - z2 (x + y)2
= | y + z |2 (x - (y + z)) (x + (y + z)) x2 y2 ((z + x) - y ((z + x ) + y y2 z2 0 (x + y)2
= | (y + z)2 (x - y - z) (x + y + z) x2 y2 (x + z - y) (x + y + z) y2 z2 0 (x + y)2
Applying C3-> C3 - C1
| (y + z)2 (x - y - z) (x + y + z) x2 - (y + z)2 y2 (x + z - y) (x + y + z) y2 - y2 z2 0 (x + y)2 - z2
= (y + z)2 (x - y - z) (x + y + z) (x - (y + z) (x + (y + z) y2 (x + z - y) (x + y + z) 0 z2 0 ((x + y) - z) ((x + y) + z)
= | (y + z)2 (x - y - z) (x + y + z) (x - y - z) (x + y + z) y2 (x + z - y) (x + y + z) 0 z2 0 (x + y - z) (x + y + z)
Taking (x + y + z) common from both c2 and c3
= (x + y + z)2 | (y + z)2 (x - y - z) (x - y - z) y2 (x + z - y) 0 z2 0 (x + y - z)
Applying R1 -> R1 - R2
= (x + y + z)2 | (y + z)2 - y2 (x - y - z) - (x + z - y) (x - y - z) - 0 y2 (x + z - y) 0 z2 0 (x + y - z)
= (x + y + z)2 | (y + z) - y (y + z) + y) -2z (x - y - z) y2 (x + z - y) 0 z2 0 (x + y - z)
= (x + y + z)2 | z (2y + z) - 2z (x - y - z) y2 (x + z - y) 0 z2 0 (x + y - z)
= (x + y + z)2 | 2yz2 - z2 (x - y - z) y2 (x + z - y) 0 z2 0 (x + y - z)
Applying R1 -> R1 - R3
= (x + y + z)2 | 2yz +z2 - Z2 -2z - 0 (x - y - z) - (x + y - z) y2 (x + z - y) 0 z2 0 (x + y - z)
= (x + y + z)2 |2yz -2z -2y y2 (x + z - y) 0 z2 0 (x + y - z)
Applying C2 -> C2 + 1/y + C1
= (x + y + z) | 2yz -2z + 1/y + 2yz -2y y2 (x + z - y) + 1/y x y2 0 z2 0 + 1/y + z2 (x + y - z)
= (x + y + z) | 2yz -2z + 2z -2y y2 (x + z - y) + y 0 z2 z2/y (x + y - z) |
Applying C3 -> C3 + 1/z + C1
= (x + y + z )2 | 2yz 0 -2y + 1/z + 2yz y2 (x + z) 0 + 1/z x y2 z2 z2/y (x + y - z) + 1/z x z2
= (x + y + z)2 | 2yz 0 -2y + 2y y2 (x + z) y2/z z2 z2/y (x + y - z) + z
= (x + y + z)2 |2yz 0 0 y2 (x + z) y2/z z2 z2/y (x + y) |
Expanding Determinant along R1
= (𝑥 + 𝑦 + 𝑧)^2 × [𝟐𝒚𝒛 ((𝒙 + 𝒛)(𝒙 + 𝒚) − 𝒛^𝟐/𝒚 × 𝒚^𝟐/𝒛) − 𝟎 − 𝟎]
= (𝑥 + 𝑦 + 𝑧)^2 × 2𝑦𝑧 ((𝑥 + 𝑧)(𝑥 + 𝑦) − 𝑧𝑦)
= (𝑥 + 𝑦 + 𝑧)^2 × 2𝑦𝑧 (𝑥(𝑥 + 𝑦) + 𝑧(𝑥 + 𝑦) − 𝑧𝑦)
= (𝑥 + 𝑦 + 𝑧)^2 × 2𝑦𝑧 (𝑥^2 + 𝑥𝑦 + 𝑧𝑥 + 𝑧𝑦 − 𝑧𝑦)
= (𝑥 + 𝑦 + 𝑧)^2 × 2𝑦𝑧 (𝑥^2 + 𝑥𝑦 + 𝑧𝑥)
= (𝑥 + 𝑦 + 𝑧)^2 × 2𝑥𝑦𝑧 × (𝑥 + 𝑦 + 𝑧)
= 2𝑥𝑦𝑧(𝑥 + 𝑦 + 𝑧)^3
Hence proved

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