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Using the properties of determinants in prove that:
\left|\begin{array}{lll} y^{2} z^{2} & y z & y+z \\ z^{2} x^{2} & z x & z+x \\ x^{2} y^{2} & x y & x+y \end{array}\right|=0

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Taking LHS, \left|\begin{array}{lll} y^{2} z^{2} & y z & y+z \\ z^{2} x^{2} & z x & z+x \\ x^{2} y^{2} & x y & x+y \end{array}\right|

\\\begin{aligned} &\text { Multiply and divide } \mathrm{R}_{1} \mathrm{R}_{2}, \mathrm{R}_{3} \text { respectively by } \mathrm{x}, \mathrm{y}, \mathrm{z}\\ &=\frac{1}{x y z}\left|\begin{array}{lll} x y^{2} z^{2} & x y z & x(y+z) \\ y z^{2} x^{2} & y z x & y(z+x) \\ z x^{2} y^{2} & z x y & z(x+y) \end{array}\right|\\ &=\frac{1}{x y z}\left|\begin{array}{lll} x y^{2} z^{2} & x y z & x y+x z \\ x^{2} y z^{2} & x y z & y z+x y \\ x^{2} y^{2} z & x y z & x z+y z \end{array}\right|\\ &\text { Now, take xyz common from the first and second Column }\\ &=\frac{1}{\mathrm{xyz}} \times \mathrm{xyz} \times \mathrm{xyz}\left|\begin{array}{lll} \mathrm{yz} & 1 & \mathrm{xy}+\mathrm{xz} \\ \mathrm{xz} & 1 & \mathrm{yz}+\mathrm{xy} \\ \mathrm{xy} & 1 & \mathrm{xz}+\mathrm{yz} \end{array}\right|\\ &=\operatorname{xyz}\left|\begin{array}{lll} y z & 1 & x y+x z \\ x z & 1 & y z+x y \\ x y & 1 & x z+y z \end{array}\right| \end{aligned}

\\\begin{aligned} &\text { Apply, } \mathrm{C}_{3} \rightarrow \mathrm{C}_{3}+\mathrm{C}_{2}\\ &=\operatorname{xyz}\left|\begin{array}{lll} y z & 1 & x y+x z+y z \\ x z & 1 & y z+x y+x z \\ x y & 1 & x z+y z+x y \end{array}\right|\\ &\text { Take }(\mathrm{xy}+\mathrm{yz}+\mathrm{xz}) \text { common from } \mathrm{C}_{3}\\ &=(\text { xyz })(\mathrm{xy}+\mathrm{yz}+\mathrm{xz})\left|\begin{array}{lll} \mathrm{yz} & 1 & 1 \\ \mathrm{xz} & 1 & 1 \\ \mathrm{xy} & 1 & 1 \end{array}\right| \end{aligned}

Whenever any the values in any two rows or columns of a determinant are identical, the resultant value of that determinant is 0

Hence, \left|\begin{array}{lll} y^{2} z^{2} & y z & y+z \\ z^{2} x^{2} & z x & z+x \\ x^{2} y^{2} & x y & x+y \end{array}\right|=0

∴ LHS = RHS

 

 

 

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