First slide

Introduction to Determinants

Question

If $x, y, z a r e n o t e q u a l t o z e r o$ and $|\begin{array}{ccc} 1 + x & 1 & 1 \\ 1 + y & 1 + 2 y & 1 \\ 1 + z & 1 + z & 1 + 3 z \end{array}| = 0$ , then $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} =$

Moderate

A

-1
B

-2
C

-3
D

none of these

Solution

$|\begin{array}{ccc} 1 + x & 1 & 1 \\ 1 + y & 1 + 2 y & 1 \\ 1 + z & 1 + z & 3 + 3 z \end{array}|$
$\begin{matrix} = xyz |\begin{array}{ccc} 1 + \frac{1}{x} & \frac{1}{x} & \frac{1}{x} \\ 1 + \frac{1}{y} & 2 + \frac{1}{y} & \frac{1}{y} \\ 1 + \frac{1}{z} & 1 + \frac{1}{z} & 3 + \frac{1}{z} \end{array}| \\ = xyz (3 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}) |\begin{array}{ccc} 1 & 1 & 1 \\ 1 + \frac{1}{y} & 2 + \frac{1}{y} & \frac{1}{y} \\ 1 + \frac{1}{z} & 1 + \frac{1}{z} & 3 + \frac{1}{z} \end{array}| \\ = xyz (3 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}) |\begin{array}{ccc} 1 & 0 & 0 \\ 1 + \frac{1}{y} & 1 & - 1 \\ 1 + \frac{1}{z} & 0 & 2 \end{array}| \\ = 2 xyz (3 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}) \end{matrix}$
Hence, the given equation gives $x^{- 1} + y^{- 1} + z^{- 1} = - 3$ .

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App