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If A=[αβrΔ]A = \begin{bmatrix}\alpha & \beta\\r & \Delta\end{bmatrix}, such that ∣A∣=25|A| = 25, then the value of ∣15A∣\left|\frac{1}{5}A\right| is.

Solution

✅ Correct Option: 1

For a square matrix of order nn, ∣kA∣=kn∣A∣|kA|=k^n|A|.

Here n=2n=2 and k=15k=\frac15, so ∣15A∣=(15)2(25)=2525=1\left|\frac15A\right|=\left(\frac15\right)^2(25)=\frac{25}{25}=1.

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