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
asked Nov 18 at 11:14
Zach
211
add a comment |
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
asked Nov 18 at 11:14
Zach
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
add a comment |
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
asked Nov 18 at 11:14
Zach
211
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
asked Nov 18 at 11:14
Zach
211
asked Nov 18 at 11:14
Zach
211
asked Nov 18 at 11:14
Zach
211
asked Nov 18 at 11:14
Zach
211
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
add a comment |
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
add a comment |
1 Answer
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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 $A_{1times 1}=left(-frac{1}{sqrt{3}}right)$:
$$A^{101}-A=left(-frac{1}{sqrt{3}}right)^{101}-left(-frac{1}{sqrt{3}}right)=left(-frac{1}{sqrt{3}}right)left(frac{1}{3^{50}}-1right).$$
answered Nov 18 at 13:06
farruhota
17.7k2736
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 $A_{1times 1}=left(-frac{1}{sqrt{3}}right)$:
$$A^{101}-A=left(-frac{1}{sqrt{3}}right)^{101}-left(-frac{1}{sqrt{3}}right)=left(-frac{1}{sqrt{3}}right)left(frac{1}{3^{50}}-1right).$$
answered Nov 18 at 13:06
farruhota
17.7k2736
add a comment |
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 $A_{1times 1}=left(-frac{1}{sqrt{3}}right)$:
$$A^{101}-A=left(-frac{1}{sqrt{3}}right)^{101}-left(-frac{1}{sqrt{3}}right)=left(-frac{1}{sqrt{3}}right)left(frac{1}{3^{50}}-1right).$$
answered Nov 18 at 13:06
farruhota
17.7k2736
add a comment |
up vote
1
down vote
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 $A_{1times 1}=left(-frac{1}{sqrt{3}}right)$:
$$A^{101}-A=left(-frac{1}{sqrt{3}}right)^{101}-left(-frac{1}{sqrt{3}}right)=left(-frac{1}{sqrt{3}}right)left(frac{1}{3^{50}}-1right).$$
answered Nov 18 at 13:06
farruhota
17.7k2736
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 $A_{1times 1}=left(-frac{1}{sqrt{3}}right)$:
$$A^{101}-A=left(-frac{1}{sqrt{3}}right)^{101}-left(-frac{1}{sqrt{3}}right)=left(-frac{1}{sqrt{3}}right)left(frac{1}{3^{50}}-1right).$$
answered Nov 18 at 13:06
farruhota
17.7k2736
answered Nov 18 at 13:06
farruhota
17.7k2736
answered Nov 18 at 13:06
farruhota
17.7k2736
answered Nov 18 at 13:06
farruhota
17.7k2736
17.7k2736
add a comment |
add a comment |
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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