1. Class 12
2. Important Question for exams Class 12

Transcript

Example 16 Show that 1+a﷮1﷮1﷮1﷮1+b﷮1﷮1﷮1﷮1+c﷯﷯ = abc (1+ 1﷮a﷯ + 1﷮b﷯ + 1﷮c﷯) = abc + bc + ca + ab Taking L.H.S 1+a﷮1﷮1﷮1﷮1+b﷮1﷮1﷮1﷮1+c﷯﷯ Taking out a, b, c, common from R1, R2, & R3 respectively = abc 1﷮a﷯+1﷮ 1﷮a﷯﷮ 1﷮a﷯﷮ 1﷮b﷯﷮ 1﷮b﷯+1﷮ 1﷮b﷯﷮ 1﷮c﷯﷮ 1﷮c﷯﷮ 1﷮a﷯+1﷯﷯ Applying R1→ R1 + R2 + R3 = abc 1+ 1﷮a﷯+ 1﷮b﷯+ 1﷮c﷯﷮ 1﷮a﷯+ 1﷮b﷯+1+ 1﷮c﷯﷮ 1﷮a﷯+ 1﷮b﷯+ 1﷮c﷯+1﷮ 1﷮b﷯﷮ 1﷮b﷯+1﷮ 1﷮b﷯﷮ 1﷮c﷯﷮ 1﷮c﷯﷮ 1﷮c﷯+1﷯﷯ = abc 𝟏+ 𝟏﷮𝐚﷯+ 𝟏﷮𝐛﷯+ 𝟏﷮𝐜﷯﷮𝟏+ 𝟏﷮𝐚﷯+ 𝟏﷮𝐛﷯+ 𝟏﷮𝐜﷯﷮𝟏+ 𝟏﷮𝐚﷯+ 𝟏﷮𝐛﷯+ 𝟏﷮𝐜﷯﷮ 1﷮b﷯﷮ 1﷮b﷯+1﷮ 1﷮b﷯﷮ 1﷮c﷯﷮ 1﷮c﷯﷮ 1﷮c﷯+1﷯﷯ Taking 1+ 1﷮𝑎﷯+ 1﷮𝑏﷯+ 1﷮𝑐﷯﷯ common from R1 = abc 𝟏+ 𝟏﷮𝐚﷯+ 𝟏﷮𝐛﷯+ 𝟏﷮𝐜﷯﷯ 1﷮1﷮1﷮ 1﷮b﷯﷮ 1﷮b﷯+1﷮ 1﷮b﷯﷮ 1﷮c﷯﷮ 1﷮c﷯﷮ 1﷮c﷯+1﷯﷯ = abc 𝟏+ 𝟏﷮𝐚﷯+ 𝟏﷮𝐛﷯+ 𝟏﷮𝐜﷯﷯ 1﷮1﷮1﷮ 1﷮b﷯﷮ 1﷮b﷯+1﷮ 1﷮b﷯﷮ 1﷮c﷯﷮ 1﷮c﷯﷮ 1﷮c﷯+1﷯﷯ Applying C3 → C3 – C1 = abc 1+ 1﷮a﷯+ 1﷮b﷯+ 1﷮c﷯﷯ 1﷮0﷮𝟏−𝟏﷮ 1﷮b﷯﷮1﷮ 1﷮b﷯− 1﷮b﷯﷮ 1﷮c﷯﷮0﷮ 1﷮c﷯+1− 1﷮𝑐﷯﷯﷯ = abc 1+ 1﷮a﷯+ 1﷮b﷯+ 1﷮c﷯﷯ 1﷮0﷮𝟎﷮ 1﷮b﷯﷮1﷮0﷮ 1﷮c﷯﷮0﷮1﷯﷯ Expanding determinant along R1 = abc 1+ 1﷮a﷯+ 1﷮b﷯+ 1﷮c﷯﷯ 1 1﷮0﷮0﷮1﷯﷯−0 1﷮𝑏﷯﷮0﷮ 1﷮𝑐﷯﷮1﷯﷯+0 1﷮𝑏﷯﷮1﷮ 1﷮𝑐﷯﷮0﷯﷯﷯ = abc 1+ 1﷮a﷯+ 1﷮b﷯+ 1﷮c﷯﷯ (1(1 – 0) – 0 + 0) = abc 1+ 1﷮a﷯+ 1﷮b﷯+ 1﷮c﷯﷯ (1(1)) = abc 1+ 1﷮a﷯+ 1﷮b﷯+ 1﷮c﷯﷯ = abc 𝑎𝑏𝑐+𝑏𝑐+𝑎𝑐+𝑎𝑏﷮𝑎𝑏𝑐﷯﷯ = 𝑎𝑏𝑐+𝑏𝑐+𝑎𝑐+𝑎𝑏 = R.H.S Hence Proved

Class 12
Important Question for exams Class 12