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IfA is an invertible matrix of order $\mathrm{n}$ and $\mathrm{k}$ is any positive real number, then the value of $[\operatorname{det}(\mathrm{kA})]^{-1}$ det $\mathrm{A}$ is
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The correct answer is:
$\mathrm{k}^{-\mathrm{n}}$
If A is a matrix that is invertible then det- $(\mathrm{kA})$ will be $\mathrm{k}^{\mathrm{n}} \cdot \operatorname{det}(\mathrm{A})$, where $\mathrm{n}$ is the order.
$\therefore [\operatorname{det}(\mathrm{KA})]^{-1} \operatorname{det}(\mathrm{A})$
$=\left[(\mathrm{K})^{\mathrm{n}} \times \operatorname{det}(\mathrm{A})\right]^{-1} \cdot \operatorname{det}(\mathrm{A})$
$=\mathrm{K}^{-\mathrm{n}} \times \frac{1}{\operatorname{det}(\mathrm{A})} \times \operatorname{det}(\mathrm{A})$
$=\mathrm{K}^{-\mathrm{n}}$
$\therefore [\operatorname{det}(\mathrm{KA})]^{-1} \operatorname{det}(\mathrm{A})$
$=\left[(\mathrm{K})^{\mathrm{n}} \times \operatorname{det}(\mathrm{A})\right]^{-1} \cdot \operatorname{det}(\mathrm{A})$
$=\mathrm{K}^{-\mathrm{n}} \times \frac{1}{\operatorname{det}(\mathrm{A})} \times \operatorname{det}(\mathrm{A})$
$=\mathrm{K}^{-\mathrm{n}}$
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