Question 14 - Solving by simplifying det. - Chapter 4 Class 12 Determinants
Last updated at April 16, 2024 by Teachoo
Solving by simplifying det.
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Question 14 Important Deleted for CBSE Board 2025 Exams You are here
Solving by simplifying det.
Last updated at April 16, 2024 by Teachoo
Question 14 Show that ฮ = |โ 8((๐ฆ+๐ง)2&๐ฅ๐ฆ&๐ง๐ฅ@๐ฅ๐ฆ&(๐ฅ+๐ง)2&๐ฆ๐ง@๐ฅ๐ง&๐ฆ๐ง&(๐ฅ+๐ฆ)2)| = 2xyz (x + y + z)3 Solving L.H.S ฮ = |โ 8((๐ฆ+๐ง)^2&๐ฅ๐ฆ&๐ง๐ฅ@๐ฅ๐ฆ&(๐ฅ+๐ง)2&๐ฆ๐ง@๐ฅ๐ง&๐ฆ๐ง&(๐ฅ+๐ฆ)2)| Divide & Multiply by xyz = ๐ฅ๐ฆ๐ง/๐ฅ๐ฆ๐ง |โ 8((๐ฆ+๐ง)2&๐ฅ๐ฆ&๐ง๐ฅ@๐ฅ๐ฆ&(๐ฅ+๐ง)2&๐ฆ๐ง@๐ฅ๐ง&๐ฆ๐ง&(๐ฅ+๐ฆ)2)| = 1/๐ฅ๐ฆ๐ง x. y. z |โ 8((๐ฆ+๐ง)2&๐ฅ๐ฆ&๐ง๐ฅ@๐ฅ๐ฆ&(๐ฅ+๐ง)2&๐ฆ๐ง@๐ฅ๐ง&๐ฆ๐ง&(๐ฅ+๐ฆ)2)| Multiplying R1 by x , R2 by y & R3 by z = 1/๐ฅ๐ฆ๐ง |โ 8(๐(๐ฆ+๐ง)2&๐(๐ฅ๐ฆ)&๐(๐ง๐ฅ)@๐(๐ฅ๐ฆ)&๐(๐ฅ+๐ง)2&๐(๐ฆ๐ง)@๐(๐ฅ๐ง)&๐ฆ๐2&๐(๐ฅ+๐ฆ)2)| Taking out x common from C1, y common from C2 & z common from C3 = ๐ฅ๐ฆ๐ง/๐ฅ๐ฆ๐ง |โ 8((๐ฆ+๐ง)2&๐ฅ2&๐ฅ2@๐ฆ2&(๐ฅ+๐ง)2&๐ฆ2@๐ง2&๐ง2&(๐ฅ+๐ฆ)2)| Applying C2 โ C2 โ C1 = |โ 8((๐ฆ+๐ง)2&๐ฅ2โ(๐ฆ+๐ง)2&๐ฅ2@๐ฆ2&(๐ฅ+๐ง)2โ๐ฆ2&๐ฆ2@๐ง2&๐ง2โ๐ง2&(๐ฅ+๐ฆ)2)| = |โ 8((๐ฆ+๐ง)2&(๐ฅโ(๐ฆ+๐ง))(๐ฅ+(๐ฆ+๐ง))&๐ฅ2@๐ฆ2&((๐ฅ+๐ง)โ๐ฆ)(๐ฅ+๐ง+๐ฆ)&๐ฆ2@๐ง2&0&(๐ฅ+๐ฆ)2)| = |โ 8((๐ฆ+๐ง)2&(๐ฅโ๐ฆโ๐ง)(๐+๐+๐)&๐ฅ2@๐ฆ2&(๐ฅ+๐งโ๐ฆ)(๐+๐+๐)&๐ฆ2@๐ง2&0&(๐ฅ+๐ฆ)2)| Taking out (๐+๐+๐) common from C2 = (๐ฅ+๐ฆ+๐ง)|โ 8((y+z)2&๐ฅโ๐ฆโ๐ง&x2@y2&๐ฅ+๐งโ๐ฆ&y2@๐ง2&0&(x+y)2)| Applying C3 โ C3 โ C1 = (๐ฅ+๐ฆ+๐ง)|โ 8((y+z)2&๐ฅโ๐ฆโ๐ง&๐ฅ2 โ(๐ฆ+๐ง)2@y2&๐ฅ+๐งโ๐ฆ&๐ฆ2โ๐ฆ2@๐ง2&0&(๐ฅ+๐ฆ)2โ๐ง2)| = (๐ฅ+๐ฆ+๐ง)|โ 8((y+z)2&๐ฅโ๐ฆโ๐ง&(๐+๐+๐)(๐ฅโ(๐ฆ+๐ง))@y2&๐ฅ+๐งโ๐ฆ&0@๐ง2&0&(๐+๐+๐)((๐ฅ+๐ฆ)โ๐ง))| Taking out (๐+๐+๐) Common from C3 = (๐ฅ+๐ฆ+๐ง)(๐ฅ+๐ฆ+๐ง)|โ 8((y+z)2&๐ฅโ๐ฆโ๐ง&๐ฅโ๐ฆโ๐ง@y2&๐ฅ+๐งโ๐ฆ&0@๐ง2&0&๐ฅ+๐ฆโ๐ง)| = (๐ฅ+๐ฆ+๐ง)2 |โ 8((y+z)2&๐ฅโ๐ฆโ๐ง&๐ฅโ๐ฆโ๐ง@y2&๐ฅ+๐งโ๐ฆ&0@๐ง2&0&๐ฅ+๐ฆโ๐ง)| Applying R1โ R1 โ R2 โ R3 = (๐ฅ+๐ฆ+๐ง)2|โ 8((y+z)2โ๐ฆ2โ๐ง2&(๐ฅโ๐ฆโ๐ง)โ(๐ฅ+๐งโ๐ฆ)โ0&(๐ฅโ๐ฆโ๐ง)โ0โ(๐ฅ+๐ฆโ๐ง)@y2&๐ฅ+๐งโ๐ฆ&0@๐ง2&0&๐ฅ+๐ฆโ๐ง)| = (๐ฅ+๐ฆ+๐ง)2|โ 8(๐ฆ2+๐ง2+2๐ฆ๐งโ๐ฆ2โ๐ง2&๐ฅโ๐ฅโ๐ฆ+๐ฆโ๐งโ๐ง&๐ฅโ๐ฅโ๐ฆโ๐ฆโ๐ง+๐ง@y2&๐ฅ+๐งโ๐ฆ&0@๐ง2&0&๐ฅ+๐ฆโ๐ง)| = (๐ฅ+๐ฆ+๐ง)2|โ 8(2๐ฆ๐ง&โ2๐ง&โ2๐ฆ@y2&๐ฅ+๐งโ๐ฆ&0@๐ง2&0&๐ฅ+๐ฆโ๐ง)| Applying C2โ C2 + ๐/๐ C1 = (๐ฅ+๐ง+๐ฆ)2|โ 8(2๐ฆ๐ง&โ2๐ง+๐/๐(๐๐๐)&2๐ฆ@y2&xโ๐ฆ+๐ง+๐/๐ (๐๐)&0@๐ง2&0+๐/๐(๐๐)&๐ฅ+๐ฆโ๐ง)| = (๐ฅ+๐ง+๐ฆ)2|โ 8(2๐ฆ๐ง&0&2๐ฆ@y2&x+๐ง&0@๐ง2&๐ง^2/๐ฆ&๐ฅ+๐ฆโ๐ง)| Applying C3โ C3 + ๐/๐ C1 = (๐ฅ+๐ฆ+๐ง)2|โ 8(2๐ฆ๐ง&0&โ2๐ฆ+๐/๐(๐๐๐)@y2&๐ฅ+๐ง&0+๐/๐ (๐๐)@๐ง2&๐ง^2/๐ฆ&(๐ฅ+๐ฆโ๐ง)+๐/๐ (๐๐))| = (๐ฅ+๐ฆ+๐ง)2|โ 8(2๐ฆ๐ง&0&0@y2&๐ฅ+๐ง&๐ฆ^2/๐ง @๐ง2&๐ง^2/๐ฆ&๐ฅ+๐ฆ)| Expanding Determinant along R1 = (๐ฅ+๐ฆ+๐ง)2(2๐ฆ๐ง|โ 8(๐ฅ+๐ง&๐ฆ^2/๐ง@๐ง^2/๐ฆ&๐ฅ+๐ฆ)|โ0|โ 8(๐ฆ2&๐ฆ^2/๐ง@๐ง^2&๐ฅ+๐ฆ)|+0|โ 8(๐ฆ2&๐ฅ+๐ฆ@๐ง^2&๐ง^2/๐ฆ)|) = (๐ฅ+๐ฆ+๐ง)2(2๐ฆ๐ง|โ 8(๐ฅ+๐ง&๐ฆ^2/๐ง@๐ง^2/๐ฆ&๐ฅ+๐ฆ)|โ0+0) = (๐ฅ+๐ฆ+๐ง)2 ("2yz " ("(x + z) (x + y) โ " ๐ง2/๐ฆ " " (๐ฆ2/๐ง))" โ 0 + 0" ) = (๐ฅ+๐ฆ+๐ง)2 (2yz ((x + z) (x + y) โ zy ) = (๐ฅ+๐ฆ+๐ง)2 (2yz) ((x + z) (x + y) โ zy ) = (๐ฅ+๐ฆ+๐ง)2 (2yz) (x2 + xy + zx + zy โ zy) = (๐ฅ+๐ฆ+๐ง)2 (2yz) (x2 + xy + xz) = (๐ฅ+๐ฆ+๐ง)2 (2yz) . x (x + y + z) = (๐ฅ+๐ฆ+๐ง)3 (2xyz) = (2xyz) (๐ฅ+๐ฆ+๐ง)^3 = R.H.S Hence Proved