Example 32 - Show that Determinant = 2xyz (x + y + z)^3 - Class 12

Example 32 - Chapter 4 Class 12 Determinants - Part 2
Example 32 - Chapter 4 Class 12 Determinants - Part 3
Example 32 - Chapter 4 Class 12 Determinants - Part 4
Example 32 - Chapter 4 Class 12 Determinants - Part 5 Example 32 - Chapter 4 Class 12 Determinants - Part 6 Example 32 - Chapter 4 Class 12 Determinants - Part 7

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Question 14 Show that ฮ” = |โ– 8((๐‘ฆ+๐‘ง)2&๐‘ฅ๐‘ฆ&๐‘ง๐‘ฅ@๐‘ฅ๐‘ฆ&(๐‘ฅ+๐‘ง)2&๐‘ฆ๐‘ง@๐‘ฅ๐‘ง&๐‘ฆ๐‘ง&(๐‘ฅ+๐‘ฆ)2)| = 2xyz (x + y + z)3 Solving L.H.S ฮ” = |โ– 8((๐‘ฆ+๐‘ง)^2&๐‘ฅ๐‘ฆ&๐‘ง๐‘ฅ@๐‘ฅ๐‘ฆ&(๐‘ฅ+๐‘ง)2&๐‘ฆ๐‘ง@๐‘ฅ๐‘ง&๐‘ฆ๐‘ง&(๐‘ฅ+๐‘ฆ)2)| Divide & Multiply by xyz = ๐‘ฅ๐‘ฆ๐‘ง/๐‘ฅ๐‘ฆ๐‘ง |โ– 8((๐‘ฆ+๐‘ง)2&๐‘ฅ๐‘ฆ&๐‘ง๐‘ฅ@๐‘ฅ๐‘ฆ&(๐‘ฅ+๐‘ง)2&๐‘ฆ๐‘ง@๐‘ฅ๐‘ง&๐‘ฆ๐‘ง&(๐‘ฅ+๐‘ฆ)2)| = 1/๐‘ฅ๐‘ฆ๐‘ง x. y. z |โ– 8((๐‘ฆ+๐‘ง)2&๐‘ฅ๐‘ฆ&๐‘ง๐‘ฅ@๐‘ฅ๐‘ฆ&(๐‘ฅ+๐‘ง)2&๐‘ฆ๐‘ง@๐‘ฅ๐‘ง&๐‘ฆ๐‘ง&(๐‘ฅ+๐‘ฆ)2)| Multiplying R1 by x , R2 by y & R3 by z = 1/๐‘ฅ๐‘ฆ๐‘ง |โ– 8(๐’™(๐‘ฆ+๐‘ง)2&๐’™(๐‘ฅ๐‘ฆ)&๐’™(๐‘ง๐‘ฅ)@๐’š(๐‘ฅ๐‘ฆ)&๐’š(๐‘ฅ+๐‘ง)2&๐’š(๐‘ฆ๐‘ง)@๐’›(๐‘ฅ๐‘ง)&๐‘ฆ๐’›2&๐’›(๐‘ฅ+๐‘ฆ)2)| Taking out x common from C1, y common from C2 & z common from C3 = ๐‘ฅ๐‘ฆ๐‘ง/๐‘ฅ๐‘ฆ๐‘ง |โ– 8((๐‘ฆ+๐‘ง)2&๐‘ฅ2&๐‘ฅ2@๐‘ฆ2&(๐‘ฅ+๐‘ง)2&๐‘ฆ2@๐‘ง2&๐‘ง2&(๐‘ฅ+๐‘ฆ)2)| Applying C2 โ†’ C2 โ€“ C1 = |โ– 8((๐‘ฆ+๐‘ง)2&๐‘ฅ2โˆ’(๐‘ฆ+๐‘ง)2&๐‘ฅ2@๐‘ฆ2&(๐‘ฅ+๐‘ง)2โˆ’๐‘ฆ2&๐‘ฆ2@๐‘ง2&๐‘ง2โˆ’๐‘ง2&(๐‘ฅ+๐‘ฆ)2)| = |โ– 8((๐‘ฆ+๐‘ง)2&(๐‘ฅโˆ’(๐‘ฆ+๐‘ง))(๐‘ฅ+(๐‘ฆ+๐‘ง))&๐‘ฅ2@๐‘ฆ2&((๐‘ฅ+๐‘ง)โˆ’๐‘ฆ)(๐‘ฅ+๐‘ง+๐‘ฆ)&๐‘ฆ2@๐‘ง2&0&(๐‘ฅ+๐‘ฆ)2)| = |โ– 8((๐‘ฆ+๐‘ง)2&(๐‘ฅโˆ’๐‘ฆโˆ’๐‘ง)(๐’™+๐’š+๐’›)&๐‘ฅ2@๐‘ฆ2&(๐‘ฅ+๐‘งโˆ’๐‘ฆ)(๐’™+๐’š+๐’›)&๐‘ฆ2@๐‘ง2&0&(๐‘ฅ+๐‘ฆ)2)| Taking out (๐’™+๐’š+๐’›) common from C2 = (๐‘ฅ+๐‘ฆ+๐‘ง)|โ– 8((y+z)2&๐‘ฅโˆ’๐‘ฆโˆ’๐‘ง&x2@y2&๐‘ฅ+๐‘งโˆ’๐‘ฆ&y2@๐‘ง2&0&(x+y)2)| Applying C3 โ†’ C3 โ€“ C1 = (๐‘ฅ+๐‘ฆ+๐‘ง)|โ– 8((y+z)2&๐‘ฅโˆ’๐‘ฆโˆ’๐‘ง&๐‘ฅ2 โˆ’(๐‘ฆ+๐‘ง)2@y2&๐‘ฅ+๐‘งโˆ’๐‘ฆ&๐‘ฆ2โˆ’๐‘ฆ2@๐‘ง2&0&(๐‘ฅ+๐‘ฆ)2โˆ’๐‘ง2)| = (๐‘ฅ+๐‘ฆ+๐‘ง)|โ– 8((y+z)2&๐‘ฅโˆ’๐‘ฆโˆ’๐‘ง&(๐’™+๐’š+๐’›)(๐‘ฅโˆ’(๐‘ฆ+๐‘ง))@y2&๐‘ฅ+๐‘งโˆ’๐‘ฆ&0@๐‘ง2&0&(๐’™+๐’š+๐’›)((๐‘ฅ+๐‘ฆ)โˆ’๐‘ง))| Taking out (๐’™+๐’š+๐’›) Common from C3 = (๐‘ฅ+๐‘ฆ+๐‘ง)(๐‘ฅ+๐‘ฆ+๐‘ง)|โ– 8((y+z)2&๐‘ฅโˆ’๐‘ฆโˆ’๐‘ง&๐‘ฅโˆ’๐‘ฆโˆ’๐‘ง@y2&๐‘ฅ+๐‘งโˆ’๐‘ฆ&0@๐‘ง2&0&๐‘ฅ+๐‘ฆโˆ’๐‘ง)| = (๐‘ฅ+๐‘ฆ+๐‘ง)2 |โ– 8((y+z)2&๐‘ฅโˆ’๐‘ฆโˆ’๐‘ง&๐‘ฅโˆ’๐‘ฆโˆ’๐‘ง@y2&๐‘ฅ+๐‘งโˆ’๐‘ฆ&0@๐‘ง2&0&๐‘ฅ+๐‘ฆโˆ’๐‘ง)| Applying R1โ†’ R1 โ€“ R2 โ€“ R3 = (๐‘ฅ+๐‘ฆ+๐‘ง)2|โ– 8((y+z)2โˆ’๐‘ฆ2โˆ’๐‘ง2&(๐‘ฅโˆ’๐‘ฆโˆ’๐‘ง)โˆ’(๐‘ฅ+๐‘งโˆ’๐‘ฆ)โˆ’0&(๐‘ฅโˆ’๐‘ฆโˆ’๐‘ง)โˆ’0โˆ’(๐‘ฅ+๐‘ฆโˆ’๐‘ง)@y2&๐‘ฅ+๐‘งโˆ’๐‘ฆ&0@๐‘ง2&0&๐‘ฅ+๐‘ฆโˆ’๐‘ง)| = (๐‘ฅ+๐‘ฆ+๐‘ง)2|โ– 8(๐‘ฆ2+๐‘ง2+2๐‘ฆ๐‘งโˆ’๐‘ฆ2โˆ’๐‘ง2&๐‘ฅโˆ’๐‘ฅโˆ’๐‘ฆ+๐‘ฆโˆ’๐‘งโˆ’๐‘ง&๐‘ฅโˆ’๐‘ฅโˆ’๐‘ฆโˆ’๐‘ฆโˆ’๐‘ง+๐‘ง@y2&๐‘ฅ+๐‘งโˆ’๐‘ฆ&0@๐‘ง2&0&๐‘ฅ+๐‘ฆโˆ’๐‘ง)| = (๐‘ฅ+๐‘ฆ+๐‘ง)2|โ– 8(2๐‘ฆ๐‘ง&โˆ’2๐‘ง&โˆ’2๐‘ฆ@y2&๐‘ฅ+๐‘งโˆ’๐‘ฆ&0@๐‘ง2&0&๐‘ฅ+๐‘ฆโˆ’๐‘ง)| Applying C2โ†’ C2 + ๐Ÿ/๐’š C1 = (๐‘ฅ+๐‘ง+๐‘ฆ)2|โ– 8(2๐‘ฆ๐‘ง&โˆ’2๐‘ง+๐Ÿ/๐’š(๐Ÿ๐’š๐’›)&2๐‘ฆ@y2&xโˆ’๐‘ฆ+๐‘ง+๐Ÿ/๐’š (๐’š๐Ÿ)&0@๐‘ง2&0+๐Ÿ/๐’š(๐’›๐Ÿ)&๐‘ฅ+๐‘ฆโˆ’๐‘ง)| = (๐‘ฅ+๐‘ง+๐‘ฆ)2|โ– 8(2๐‘ฆ๐‘ง&0&2๐‘ฆ@y2&x+๐‘ง&0@๐‘ง2&๐‘ง^2/๐‘ฆ&๐‘ฅ+๐‘ฆโˆ’๐‘ง)| Applying C3โ†’ C3 + ๐Ÿ/๐’› C1 = (๐‘ฅ+๐‘ฆ+๐‘ง)2|โ– 8(2๐‘ฆ๐‘ง&0&โˆ’2๐‘ฆ+๐Ÿ/๐’›(๐Ÿ๐’š๐’›)@y2&๐‘ฅ+๐‘ง&0+๐Ÿ/๐’› (๐’š๐Ÿ)@๐‘ง2&๐‘ง^2/๐‘ฆ&(๐‘ฅ+๐‘ฆโˆ’๐‘ง)+๐Ÿ/๐’› (๐’›๐Ÿ))| = (๐‘ฅ+๐‘ฆ+๐‘ง)2|โ– 8(2๐‘ฆ๐‘ง&0&0@y2&๐‘ฅ+๐‘ง&๐‘ฆ^2/๐‘ง @๐‘ง2&๐‘ง^2/๐‘ฆ&๐‘ฅ+๐‘ฆ)| Expanding Determinant along R1 = (๐‘ฅ+๐‘ฆ+๐‘ง)2(2๐‘ฆ๐‘ง|โ– 8(๐‘ฅ+๐‘ง&๐‘ฆ^2/๐‘ง@๐‘ง^2/๐‘ฆ&๐‘ฅ+๐‘ฆ)|โˆ’0|โ– 8(๐‘ฆ2&๐‘ฆ^2/๐‘ง@๐‘ง^2&๐‘ฅ+๐‘ฆ)|+0|โ– 8(๐‘ฆ2&๐‘ฅ+๐‘ฆ@๐‘ง^2&๐‘ง^2/๐‘ฆ)|) = (๐‘ฅ+๐‘ฆ+๐‘ง)2(2๐‘ฆ๐‘ง|โ– 8(๐‘ฅ+๐‘ง&๐‘ฆ^2/๐‘ง@๐‘ง^2/๐‘ฆ&๐‘ฅ+๐‘ฆ)|โˆ’0+0) = (๐‘ฅ+๐‘ฆ+๐‘ง)2 ("2yz " ("(x + z) (x + y) โ€“ " ๐‘ง2/๐‘ฆ " " (๐‘ฆ2/๐‘ง))" โ€“ 0 + 0" ) = (๐‘ฅ+๐‘ฆ+๐‘ง)2 (2yz ((x + z) (x + y) โ€“ zy ) = (๐‘ฅ+๐‘ฆ+๐‘ง)2 (2yz) ((x + z) (x + y) โ€“ zy ) = (๐‘ฅ+๐‘ฆ+๐‘ง)2 (2yz) (x2 + xy + zx + zy โ€“ zy) = (๐‘ฅ+๐‘ฆ+๐‘ง)2 (2yz) (x2 + xy + xz) = (๐‘ฅ+๐‘ฆ+๐‘ง)2 (2yz) . x (x + y + z) = (๐‘ฅ+๐‘ฆ+๐‘ง)3 (2xyz) = (2xyz) (๐‘ฅ+๐‘ฆ+๐‘ง)^3 = R.H.S Hence Proved

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.