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Question Number 10793 by j.masanja06@gmail.com last updated on 25/Feb/17

$$\mathrm{let}\mathrm{A}=\left|\begin{array}{ccc}4& 4\mathrm{k}& \mathrm{k}\\ 0& \mathrm{k}& 4\mathrm{k}\\ 0& 0& 4\end{array}\right|\mathrm{if}\det \left({\mathrm{A}}^2\right)=16$$$$\mathrm{then}\mid \mathrm{k}\mid \mathrm{is}?$$

Answered by sandy_suhendra last updated on 25/Feb/17

$$\det \mathrm{A}=16\mathrm{k}+0+0-(0+0+0)=16\mathrm{k}$$$$\det \left({\mathrm{A}}^2\right)={\left(\det \mathrm{A}\right)}^2={\left(16\mathrm{k}\right)}^2={256\mathrm{k}}^2$$$${256\mathrm{k}}^2=16$$$${\mathrm{k}}^2=\frac{1}{16}\Rightarrow \mathrm{k}=\pm \frac{1}{4}$$$$\mathrm{so}\mid \mathrm{k}\mid =\frac{1}{4}$$

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