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Ex 4.2, 10 - Show: (i) |x+4 2x 2x 2x x+4 2x 2x 2x x+4| - Ex 4.2

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  1. Chapter 4 Class 12 Determinants
  2. Serial order wise
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Ex 4.2, 10 By using properties of determinants, show that: (i) x+4﷮2x﷮2x﷮2x﷮x+4﷮2x﷮2x﷮2𝑥﷮x+4﷯﷯= (5x + 4) (4 – x)2 Taking L.H.S x+4﷮2x﷮2x﷮2x﷮x+4﷮2x﷮2x﷮2𝑥﷮x+4﷯﷯ Applying R1 → R1 + R2 + R2 = x+4+2𝑥+2𝑥﷮2x+x+4+2x﷮2x+2x+x+4﷮2x﷮x+4﷮2x﷮2x﷮2𝑥﷮x+4﷯﷯ = 𝟓𝒙+𝟒﷮𝟓𝐱+𝟒﷮𝟓𝐱+𝟒﷮2x﷮x+4﷮2x﷮2x﷮2𝑥﷮x+4﷯﷯ Taking out (5x + 4) common from R1 = (5x + 4) 1﷮1﷮1﷮2x﷮x+4﷮2x﷮2x﷮2𝑥﷮x+4﷯﷯ Applying C1 → C1 – C2 = (5x + 4) 𝟏−𝟏﷮1−1﷮1−1﷮2x−x−4﷮x+4﷮2x﷮2x−2x﷮2𝑥﷮x+4﷯﷯ = (5x + 4) 𝟎﷮1﷮1﷮x−4﷮x+4﷮2x﷮0﷮2𝑥﷮x+4﷯﷯ Applying C2 → C2 – C3 = (5x + 4) 0﷮𝟏−𝟏﷮1﷮x−4﷮x+4−2x﷮2x﷮0﷮2𝑥−𝑥−4﷮x+4﷯﷯ = (5x + 4) 0﷮𝟎﷮1﷮x−4﷮−(𝑥−4)﷮2x﷮0(𝑥−4)﷮(𝑥−4)﷮x+4﷯﷯ Taking common (x – 4) from C1 & C2 = (5x + 4) (x – 4) (x – 4) 0﷮0﷮1﷮1﷮−1﷮2x﷮0﷮1﷮x+4﷯﷯ Expanding Determinant along R1 = (5x + 4) (x – 4) (x – 4) 0 −1﷮2𝑥﷮1﷮𝑥+4﷯﷯−0 1﷮2𝑥﷮0﷮𝑥+4﷯﷯+1 1﷮−1﷮0﷮1﷯﷯﷯ = (5x – 4) (x – 4)2 (0 – 0 + (1 – 0)) = (5x – 4) (x – 4)2 (1) = (5x – 4) (x – 4)2 = R.H.S Hence Proved Ex 4.2 , 10 By using properties of determinants, show that: (ii) y+𝑘﷮y﷮y﷮y﷮y+𝑘﷮y﷮y﷮𝑦﷮y+𝑘﷯﷯= k3 (3y + k) Taking L.H.S y+𝑘﷮y﷮y﷮y﷮y+𝑘﷮y﷮y﷮𝑦﷮y+𝑘﷯﷯ Applying R1 → R1 + R2 + R3 = y+𝑘+𝑦+𝑦﷮y+y+k+y﷮y+y+y+k﷮y﷮y+𝑘﷮y﷮y﷮𝑦﷮y+𝑘﷯﷯ = 𝟑𝐲+𝒌﷮𝟑𝐲+𝐤﷮𝟑𝐲+𝐤﷮y﷮y+𝑘﷮y﷮y﷮𝑦﷮y+𝑘﷯﷯ Taking out (3y + k) common from R1 = (3y + k) 1﷮1﷮1﷮y﷮y+𝑘﷮y﷮y﷮𝑦﷮y+𝑘﷯﷯ Applying C1 → C1 – C2 = (3y + k) 𝟏−𝟏﷮1﷮1﷮𝑦−𝑦−𝑘﷮y+𝑘﷮y﷮𝑦−𝑦﷮𝑦﷮y+𝑘﷯﷯ = (3y + k) 𝟎﷮1﷮1﷮−𝑘﷮y+k﷮y﷮0﷮𝑦﷮y+𝑘﷯﷯ Applying C2 → C2 – C3 = (3y + k) 0﷮𝟏−𝟏﷮1﷮−𝑘−0﷮y+k−𝑦﷮y﷮0﷮𝑦−𝑦−𝑘﷮y+𝑘﷯﷯ = (3y + k) 0﷮𝟎﷮1﷮−𝑘﷮𝑘﷮y﷮0﷮−𝑘﷮y+𝑘﷯﷯ Expanding Determinant along R1 = (3y + k) 0 𝑘﷮𝑦﷮−𝑘﷮𝑦+𝑘﷯﷯−0 −𝑘﷮𝑦﷮0﷮𝑦+𝑘﷯﷯+1 𝑘﷮𝑘﷮0﷮−𝑘﷯﷯﷯ = (3y + k) (0 – 0 + 1 (k2 – 0)) = (3y + k) (k2) = k2 (3y + k) = R.H.S Hence Proved

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