**Ex 4.2, 10**

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 4.2, 10 By using properties of determinants, show that: (i) x+42x2x2xx+42x2x2𝑥x+4= (5x + 4) (4 – x)2 Taking L.H.S x+42x2x2xx+42x2x2𝑥x+4 Applying R1 → R1 + R2 + R2 = x+4+2𝑥+2𝑥2x+x+4+2x2x+2x+x+42xx+42x2x2𝑥x+4 = 𝟓𝒙+𝟒𝟓𝐱+𝟒𝟓𝐱+𝟒2xx+42x2x2𝑥x+4 Taking out (5x + 4) common from R1 = (5x + 4) 1112xx+42x2x2𝑥x+4 Applying C1 → C1 – C2 = (5x + 4) 𝟏−𝟏1−11−12x−x−4x+42x2x−2x2𝑥x+4 = (5x + 4) 𝟎11x−4x+42x02𝑥x+4 Applying C2 → C2 – C3 = (5x + 4) 0𝟏−𝟏1x−4x+4−2x2x02𝑥−𝑥−4x+4 = (5x + 4) 0𝟎1x−4−(𝑥−4)2x0(𝑥−4)(𝑥−4)x+4 Taking common (x – 4) from C1 & C2 = (5x + 4) (x – 4) (x – 4) 0011−12x01x+4 Expanding Determinant along R1 = (5x + 4) (x – 4) (x – 4) 0 −12𝑥1𝑥+4−0 12𝑥0𝑥+4+1 1−101 = (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+𝑘yyyy+𝑘yy𝑦y+𝑘= k3 (3y + k) Taking L.H.S y+𝑘yyyy+𝑘yy𝑦y+𝑘 Applying R1 → R1 + R2 + R3 = y+𝑘+𝑦+𝑦y+y+k+yy+y+y+kyy+𝑘yy𝑦y+𝑘 = 𝟑𝐲+𝒌𝟑𝐲+𝐤𝟑𝐲+𝐤yy+𝑘yy𝑦y+𝑘 Taking out (3y + k) common from R1 = (3y + k) 111yy+𝑘yy𝑦y+𝑘 Applying C1 → C1 – C2 = (3y + k) 𝟏−𝟏11𝑦−𝑦−𝑘y+𝑘y𝑦−𝑦𝑦y+𝑘 = (3y + k) 𝟎11−𝑘y+ky0𝑦y+𝑘 Applying C2 → C2 – C3 = (3y + k) 0𝟏−𝟏1−𝑘−0y+k−𝑦y0𝑦−𝑦−𝑘y+𝑘 = (3y + k) 0𝟎1−𝑘𝑘y0−𝑘y+𝑘 Expanding Determinant along R1 = (3y + k) 0 𝑘𝑦−𝑘𝑦+𝑘−0 −𝑘𝑦0𝑦+𝑘+1 𝑘𝑘0−𝑘 = (3y + k) (0 – 0 + 1 (k2 – 0)) = (3y + k) (k2) = k2 (3y + k) = R.H.S Hence Proved

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.