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If x = a/(b-c), y = b/(c-a), z = c/(a-b) then what is the value of the following?(a). 0(b). 1(c). abc(d). ab + bc + ca |
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Answer» We have x = \(\frac{a}{b-c},\) y = \(\frac{b}{c-a},\) z = \(\frac{c}{z-b}\) Now, \(\begin{vmatrix}1&-x&x\\1&1&-y\\1&z&1\end{vmatrix}\) = \(\begin{vmatrix}1&-x&x\\0&1+x&-(x+y)\\0&x+z&1-x\end{vmatrix}\) (By applying R2 - 1 R2 - R1 and R3 - 1, R3 - R2) = (1 + x) (1 - x) + (x + y)(x + z) (By expanding determinant along C1) = 1 - x2 + x2 + xy + xz + yz = 1 + \(\frac{a}{b-c}\times\frac{b}{c-a}\) + \(\frac{a}{b-c}\times\frac{c}{a-b}\) + \(\frac{b}{c-a}\times\frac{c}{a-b}\) = 1 + \(\frac{ab(a-b)+ac(c-a)+bc(b-c)}{(a-b)(b-c)(c-a)}\) = 1 + \(\frac{a^2b-ab^2+ac^2-a^2c+b^2c-b^2c}{(a-b)(b-c)(c-a)}\) = \(\frac{(a-b)(b-c)(c-a)+a^2b-ab^2+ac^2-a^2c+b^2c-bc^2}{(a-b)(b-c)(c-a)}\) = \(\frac{(-a^2b-ac^2+a^2c-b^2c+ab^2+bc^2)+(a^2b-ab^2+ac^2-a^2c+b^2c-bc^2)}{(a-b)(b-c)(c-a)}\) = \(\frac{0}{(a-c)(b-c)(c-a)}\) = 0 |
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