Question
If $\left|\begin{array}{ccc}x^{n} & x^{n+2} & x^{n+3} \\ y^{n} & y^{n+2} & y^{n+3} \\ z^{n} & z^{n+2} & z^{n+3}\end{array}\right|$$=(x-y)(y-z)(z-x),\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$ then $n=$(A) $-2$(B) $-1$(C) 0(D) 1
Step 1
The determinant is a 3x3 matrix with elements $x^n$, $x^{n+2}$, $x^{n+3}$ in the first row, $y^n$, $y^{n+2}$, $y^{n+3}$ in the second row, and $z^n$, $z^{n+2}$, $z^{n+3}$ in the third row. Show more…
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Engineering Mathematics
Linear Algebra
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