possible values of the determinant of the matrix

Question

What are the possible values of the determinant of the matrix A of n-degree, if

a) $A^2 = 8A^{-1}$

b) $A^T = 4A^{-1}$

what I already have is

a) L = det $(A^2) = det(A) * det(A) = det(A)^2 $ - from Cauchy theorem

R = det $(8A)^{-1} = 8[det(1)]^{-1}$

b) L = det $(A^T) = det (A)$

R = det $(4A)^{-1} = 4[det(A)]^{-1}$

Answer 1

For $A_{n \times n}$ $$A^2=8A^{-1} \implies A^3 =8 I \implies |A|^3= |8 I| \implies |A|^3=8^n \implies |A|=8^{n/3}.$$ Next, $$ A^{T}=4 A^{-1} \implies |A^T|=|4 A^{-1}| \implies |A|=4^n |A|^{-1} \implies |A|=2^n$$

Answer 2

(a) From $A^2=8A^{-1}$,

taking determinant on both sides, we have

$$\det(A)^2=\frac{8^n }{\det(A)}$$

Hence $$\det(A)^3=8^n$$

$$\det(A)=2^n$$

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(b) $A^T=4A^{-1}$

$$\det(A)=\frac{4^n}{\det(A)}$$

$$\det(A)^2=4^n$$

$$\det(A)=\pm2^n$$

Remark about your attempt: Remember to raise the power as you factor out a constant from the determinant, i.e. $\det(cA)=c^n \det(A)$.