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Evaluation of a Determinant

If =1 x x...

Question

If $△ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & x & x^{2} 1 & y & y^{2} 1 & z & z^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$ and $△_{1} = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 y z & z x & x y x & y & z \end{matrix} ∣ ∣ ∣ ∣,$ then the value of $△_{1} + △ =$

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Solution

Given : $△_{1} = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 y z & z x & x y x & y & z \end{matrix} ∣ ∣ ∣ ∣$
Taking transpose of above determinant, we get
$△_{1} = ∣ ∣ ∣ ∣ \begin{matrix} 1 & y z & x 1 & z x & y 1 & x y & z \end{matrix} ∣ ∣ ∣ ∣$
Now multiplying $x$ in $R_{1}, y$ in $R_{2}$ and $z$ in $R_{3},$ we get
$△_{1} = \frac{1}{x y z} ∣ ∣ ∣ ∣ ∣ \begin{matrix} x & x y z & x^{2} y & x y z & y^{2} z & x y z & z^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$
$\Rightarrow △_{1} = \frac{x y z}{x y z} ∣ ∣ ∣ ∣ ∣ \begin{matrix} x & 1 & x^{2} y & 1 & y^{2} z & 1 & z^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$
Now interchanging $C_{1}$ and $C_{2},$ we get
$△_{1} = (- 1) ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & x & x^{2} 1 & y & y^{2} 1 & z & z^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = - △$
$∴ △_{1} + △ = 0$

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Evaluation of a Determinant

Standard XII Mathematics

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