If - x y3- 2 0 - 1 8- 2 0 - then - x y- 2 0 -1 is equal to

The correct option is D [ 0 1/2; 1/2 1/4 ] Given, [ x amp; y^3; 2 amp; 0 ]=[ 1 amp; 2; 2 amp; 0 ] ⇒ x=1 and ...

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Standard X

Mathematics

Square Matrix

If [ x y3;...

Question

If $[\begin{matrix} x & y^{3} 2 & 0 \end{matrix}] = [\begin{matrix} 1 & 8 2 & 0 \end{matrix}]$ , then ${[\begin{matrix} x & y 2 & 0 \end{matrix}]}^{- 1}$ is equal to

[\begin{matrix} 0 & - 2 - 2 & 1 \end{matrix}]

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[\begin{matrix} 1 & 0 0 & 1 \end{matrix}]

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[\begin{matrix} 0 & - 8 - 2 & 1 \end{matrix}]

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[\begin{matrix} 0 & 1 / 2 1 / 2 & - 1 / 4 \end{matrix}]

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Solution

The correct option is D $[\begin{matrix} 0 & 1 / 2 1 / 2 & - 1 / 4 \end{matrix}]$
Given, $[\begin{matrix} x & y^{3} 2 & 0 \end{matrix}] = [\begin{matrix} 1 & 2 2 & 0 \end{matrix}]$
$\Rightarrow x = 1$ and $y = 2$
$∴ {[\begin{matrix} 1 & 2 2 & 0 \end{matrix}]}^{- 1} = \frac{1}{- 4} [\begin{matrix} 0 & - 2 - 2 & 1 \end{matrix}] = ⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 0 & \frac{1}{2} \frac{1}{2} & \frac{1}{- 4} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦$

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If [xy320]=[1820], then [xy20]−1 is equal to

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If $[\begin{matrix} x & y^{3} 2 & 0 \end{matrix}] = [\begin{matrix} 1 & 8 2 & 0 \end{matrix}]$ , then ${[\begin{matrix} x & y 2 & 0 \end{matrix}]}^{- 1}$ is equal to