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

let  A = determinant ((4,(4k),k),(0,k,(4k)),(0,0,4)) if det(A^2 )=16  then ∣k∣ is?

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

Answered by sandy_suhendra last updated on 25/Feb/17

det A=16k+0+0−(0+0+0)=16k  det(A^2 )=(det A)^2 =(16k)^2 =256k^2   256k^2 =16  k^2 =(1/(16)) ⇒ k=±(1/4)  so ∣k∣=(1/4)

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

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