Computing Determinant of a Matrix $textrm{det}(A^{101} - A)$











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Let $A$ be a $n$-by-$n$ matrix with real entries such that $A^{-1} = 3A$. What is the determinant of $A^{101} - A$ i.e. $textrm {det}(A^{101} - A)$?



My attempt:

$$A^{-1} = 3A Rightarrow 3A^{2} = I Rightarrow A^{2} = frac{1}{3}I$$
begin{align}
textrm{det}(A^{101} - A) & = textrm{det}[A(A^{100}-I)] \
& = textrm{det}(A)textrm{det}(A^{100}-I) \
& = textrm{det}(A)textrm{det}[(A^{2})^{50}-I] \
& = textrm{det}(A)textrm{det}[(frac{1}{3}I)^{50}-I] \
& = textrm{det}(A)textrm{det}[(frac{1}{3^{50}})I-I] \
& = frac{1}{3^{n}}textrm{det}(A^{-1})textrm{det}[(frac{1}{3^{50}})I-I] \
end{align}
Any hints on how to continue from here will be appreciated.






linear-algebra matrices determinant







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  • 4




    Determinants of diagonal matrices are really very easy to calculate. At your second-to-last line, you have $det(A)$ multiplied by the determinant of something diagonal, so you just need to find $det(A)$. But note that $det(A)^2 = det(A^2), and that's diagonal too, by your first line. .
    – user3482749
    Nov 18 at 11:20















up vote
3
down vote

favorite












Let $A$ be a $n$-by-$n$ matrix with real entries such that $A^{-1} = 3A$. What is the determinant of $A^{101} - A$ i.e. $textrm {det}(A^{101} - A)$?



My attempt:

$$A^{-1} = 3A Rightarrow 3A^{2} = I Rightarrow A^{2} = frac{1}{3}I$$
begin{align}
textrm{det}(A^{101} - A) & = textrm{det}[A(A^{100}-I)] \
& = textrm{det}(A)textrm{det}(A^{100}-I) \
& = textrm{det}(A)textrm{det}[(A^{2})^{50}-I] \
& = textrm{det}(A)textrm{det}[(frac{1}{3}I)^{50}-I] \
& = textrm{det}(A)textrm{det}[(frac{1}{3^{50}})I-I] \
& = frac{1}{3^{n}}textrm{det}(A^{-1})textrm{det}[(frac{1}{3^{50}})I-I] \
end{align}
Any hints on how to continue from here will be appreciated.






linear-algebra matrices determinant







share|cite|improve this question


















  • 4




    Determinants of diagonal matrices are really very easy to calculate. At your second-to-last line, you have $det(A)$ multiplied by the determinant of something diagonal, so you just need to find $det(A)$. But note that $det(A)^2 = det(A^2), and that's diagonal too, by your first line. .
    – user3482749
    Nov 18 at 11:20













up vote
3
down vote

favorite









up vote
3
down vote

favorite











Let $A$ be a $n$-by-$n$ matrix with real entries such that $A^{-1} = 3A$. What is the determinant of $A^{101} - A$ i.e. $textrm {det}(A^{101} - A)$?



My attempt:

$$A^{-1} = 3A Rightarrow 3A^{2} = I Rightarrow A^{2} = frac{1}{3}I$$
begin{align}
textrm{det}(A^{101} - A) & = textrm{det}[A(A^{100}-I)] \
& = textrm{det}(A)textrm{det}(A^{100}-I) \
& = textrm{det}(A)textrm{det}[(A^{2})^{50}-I] \
& = textrm{det}(A)textrm{det}[(frac{1}{3}I)^{50}-I] \
& = textrm{det}(A)textrm{det}[(frac{1}{3^{50}})I-I] \
& = frac{1}{3^{n}}textrm{det}(A^{-1})textrm{det}[(frac{1}{3^{50}})I-I] \
end{align}
Any hints on how to continue from here will be appreciated.






linear-algebra matrices determinant







share|cite|improve this question













Let $A$ be a $n$-by-$n$ matrix with real entries such that $A^{-1} = 3A$. What is the determinant of $A^{101} - A$ i.e. $textrm {det}(A^{101} - A)$?



My attempt:

$$A^{-1} = 3A Rightarrow 3A^{2} = I Rightarrow A^{2} = frac{1}{3}I$$
begin{align}
textrm{det}(A^{101} - A) & = textrm{det}[A(A^{100}-I)] \
& = textrm{det}(A)textrm{det}(A^{100}-I) \
& = textrm{det}(A)textrm{det}[(A^{2})^{50}-I] \
& = textrm{det}(A)textrm{det}[(frac{1}{3}I)^{50}-I] \
& = textrm{det}(A)textrm{det}[(frac{1}{3^{50}})I-I] \
& = frac{1}{3^{n}}textrm{det}(A^{-1})textrm{det}[(frac{1}{3^{50}})I-I] \
end{align}
Any hints on how to continue from here will be appreciated.






linear-algebra matrices determinant




linear-algebra matrices determinant






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asked Nov 18 at 11:14









Zach

211




211








  • 4




    Determinants of diagonal matrices are really very easy to calculate. At your second-to-last line, you have $det(A)$ multiplied by the determinant of something diagonal, so you just need to find $det(A)$. But note that $det(A)^2 = det(A^2), and that's diagonal too, by your first line. .
    – user3482749
    Nov 18 at 11:20














  • 4




    Determinants of diagonal matrices are really very easy to calculate. At your second-to-last line, you have $det(A)$ multiplied by the determinant of something diagonal, so you just need to find $det(A)$. But note that $det(A)^2 = det(A^2), and that's diagonal too, by your first line. .
    – user3482749
    Nov 18 at 11:20








4




4




Determinants of diagonal matrices are really very easy to calculate. At your second-to-last line, you have $det(A)$ multiplied by the determinant of something diagonal, so you just need to find $det(A)$. But note that $det(A)^2 = det(A^2), and that's diagonal too, by your first line. .
– user3482749
Nov 18 at 11:20




Determinants of diagonal matrices are really very easy to calculate. At your second-to-last line, you have $det(A)$ multiplied by the determinant of something diagonal, so you just need to find $det(A)$. But note that $det(A)^2 = det(A^2), and that's diagonal too, by your first line. .
– user3482749
Nov 18 at 11:20










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Use:
$$det (A^2)=det(Acdot A)=det(A)cdot det(A)=det(frac13I) Rightarrow det(A)=pmfrac1{3^{n/2}} $$
Hence:
$$textrm{det}(A)textrm{det}[(frac{1}{3^{50}})I-I] =pmfrac1{3^{n/2}}cdot left(frac1{3^{50}}-1right)^n.
$$
Checking with $A_{1times 1}=left(-frac{1}{sqrt{3}}right)$:
$$A^{101}-A=left(-frac{1}{sqrt{3}}right)^{101}-left(-frac{1}{sqrt{3}}right)=left(-frac{1}{sqrt{3}}right)left(frac{1}{3^{50}}-1right).$$






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    Use:
    $$det (A^2)=det(Acdot A)=det(A)cdot det(A)=det(frac13I) Rightarrow det(A)=pmfrac1{3^{n/2}} $$
    Hence:
    $$textrm{det}(A)textrm{det}[(frac{1}{3^{50}})I-I] =pmfrac1{3^{n/2}}cdot left(frac1{3^{50}}-1right)^n.
    $$
    Checking with $A_{1times 1}=left(-frac{1}{sqrt{3}}right)$:
    $$A^{101}-A=left(-frac{1}{sqrt{3}}right)^{101}-left(-frac{1}{sqrt{3}}right)=left(-frac{1}{sqrt{3}}right)left(frac{1}{3^{50}}-1right).$$






    share|cite|improve this answer

























      up vote
      1
      down vote













      Use:
      $$det (A^2)=det(Acdot A)=det(A)cdot det(A)=det(frac13I) Rightarrow det(A)=pmfrac1{3^{n/2}} $$
      Hence:
      $$textrm{det}(A)textrm{det}[(frac{1}{3^{50}})I-I] =pmfrac1{3^{n/2}}cdot left(frac1{3^{50}}-1right)^n.
      $$
      Checking with $A_{1times 1}=left(-frac{1}{sqrt{3}}right)$:
      $$A^{101}-A=left(-frac{1}{sqrt{3}}right)^{101}-left(-frac{1}{sqrt{3}}right)=left(-frac{1}{sqrt{3}}right)left(frac{1}{3^{50}}-1right).$$






      share|cite|improve this answer























        up vote
        1
        down vote










        up vote
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        down vote









        Use:
        $$det (A^2)=det(Acdot A)=det(A)cdot det(A)=det(frac13I) Rightarrow det(A)=pmfrac1{3^{n/2}} $$
        Hence:
        $$textrm{det}(A)textrm{det}[(frac{1}{3^{50}})I-I] =pmfrac1{3^{n/2}}cdot left(frac1{3^{50}}-1right)^n.
        $$
        Checking with $A_{1times 1}=left(-frac{1}{sqrt{3}}right)$:
        $$A^{101}-A=left(-frac{1}{sqrt{3}}right)^{101}-left(-frac{1}{sqrt{3}}right)=left(-frac{1}{sqrt{3}}right)left(frac{1}{3^{50}}-1right).$$






        share|cite|improve this answer












        Use:
        $$det (A^2)=det(Acdot A)=det(A)cdot det(A)=det(frac13I) Rightarrow det(A)=pmfrac1{3^{n/2}} $$
        Hence:
        $$textrm{det}(A)textrm{det}[(frac{1}{3^{50}})I-I] =pmfrac1{3^{n/2}}cdot left(frac1{3^{50}}-1right)^n.
        $$
        Checking with $A_{1times 1}=left(-frac{1}{sqrt{3}}right)$:
        $$A^{101}-A=left(-frac{1}{sqrt{3}}right)^{101}-left(-frac{1}{sqrt{3}}right)=left(-frac{1}{sqrt{3}}right)left(frac{1}{3^{50}}-1right).$$







        share|cite|improve this answer












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        share|cite|improve this answer










        answered Nov 18 at 13:06









        farruhota

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