If we have C=($A^t$)$^2$BA$^3$B$^-$$^1$A$^-$$^3$ and detA=-2 and detB doesnt equal 0, how do we calculate det C?

I know that the transpose of a matrix does not affect the determinant. Does this mean that ($A^t$)$^2$=(-2)$^2$=4?

And then how is A$^-$$^3$ affected? Does this mean the inverse of A cubed? And how does the inverse affect the determinant? Thanks

  • You need to repeatedly use the determinant properties. – vadim123 May 10 '15 at 14:37
  • @vadim123 by using these, i would get C = 4B(-8) $1/B$ $1/(-8)$ am i able to cancel this down to equal 4? – Lauren Bathers May 10 '15 at 14:44
up vote 1 down vote accepted

One may recall that, if $\det A \neq 0$, then $$ \det A{}^t=\det A, \qquad \det (A{}^{p})=(\det A)^p,\quad p=0,\pm 1,\pm2, \ldots. $$and $$ \det (AB)=(\det A)(\det B)=(\det B)(\det A)=\det (BA) $$ Thus here: $$ \det C=\det ( (A^t)^2BA^3B^{-1}A^{-3})=(\det A)^{2+3-3}(\det B)^{1-1}=(\det A)^{2}=4. $$

  • are you allow to reorder the matrices order like that? thank you this answer is clear – Lauren Bathers May 10 '15 at 14:46
  • @LaurenBathers When is it about determinants, they are real numbers, so you can reorder... Thanks. – Olivier Oloa May 10 '15 at 14:48
  • Ah that makes a lot of sense! thank you! – Lauren Bathers May 10 '15 at 14:50

Remember the following

$$\operatorname{det}(AB)=\operatorname{det}{A}\operatorname{det}{B}$$ $$\operatorname{det}{A^n}=(\operatorname{det}{A})^n$$ $$\operatorname{det}{A^T}=\operatorname{det}{A}$$

So in our case we have $B$ and $B^{-1}$ whose determinant cancel out because they are not $0$ (otherwise $B^{-1}$ would not exist) $A^3$ and $A^{-3}$ whose determinant cancel out and we are left with $(\operatorname{det}{A^T})^2=4$

\begin{equation*} \det(A^k) = \det(A)^k~\text{and}~\det(A^t) = \det(A) \end{equation*}

  • What does the square at the end mean? – Kitegi May 10 '15 at 15:25
  • is there a proof to $det(A^k)=det(A)^k$? – gbox May 10 '15 at 15:35
  • 2
    @gbox yes if you know $det(AB) = det(A)det(B)$ – rackne May 10 '15 at 16:06

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