Mixing
Using the property $Amathrm{adj}(A) = det(A)I_n$, prove that $det(mathrm{adj}(A^3)) = (det(A))^{3n-3}$
$begingroup$
Suppose that $A$ is invertible $n times n$ matrices. Then using the property $A~mathrm{adj}(A) = det(A)I_n$, prove that $det(mathrm{adj}(A^3)) = (det(A))^{3n-3}$
I started with.
$$ A~mathrm{adj}(A) = det(A)I_n $$
I took $det()$ of both sides, then raised both side to power of $3$, then I multiplied both sides by $1/det(A^3)$, then I use the rule $(mathrm{adj}(A))^n = mathrm{adj}(A^n)$ on the left side, and combine exponents on the right side to complete the proof that
$$ det(mathrm{adj}(A^3)) = (det(A))^{3n-3} $$
the only thing I wanted to confirm if i did this right is if $(mathrm{adj}(A))^n = mathrm{adj}(A^n)$?
linear-algebra adjoint-operators
edited Feb 3 at 12:29
Community♦
1
asked Oct 12 '14 at 2:07
J LJ L
634102654
$endgroup$
add a comment |
$begingroup$
Suppose that $A$ is invertible $n times n$ matrices. Then using the property $A~mathrm{adj}(A) = det(A)I_n$, prove that $det(mathrm{adj}(A^3)) = (det(A))^{3n-3}$
I started with.
$$ A~mathrm{adj}(A) = det(A)I_n $$
I took $det()$ of both sides, then raised both side to power of $3$, then I multiplied both sides by $1/det(A^3)$, then I use the rule $(mathrm{adj}(A))^n = mathrm{adj}(A^n)$ on the left side, and combine exponents on the right side to complete the proof that
$$ det(mathrm{adj}(A^3)) = (det(A))^{3n-3} $$
the only thing I wanted to confirm if i did this right is if $(mathrm{adj}(A))^n = mathrm{adj}(A^n)$?
linear-algebra adjoint-operators
edited Feb 3 at 12:29
Community♦
1
asked Oct 12 '14 at 2:07
J LJ L
634102654
$endgroup$
add a comment |
$begingroup$
Suppose that $A$ is invertible $n times n$ matrices. Then using the property $A~mathrm{adj}(A) = det(A)I_n$, prove that $det(mathrm{adj}(A^3)) = (det(A))^{3n-3}$
I started with.
$$ A~mathrm{adj}(A) = det(A)I_n $$
I took $det()$ of both sides, then raised both side to power of $3$, then I multiplied both sides by $1/det(A^3)$, then I use the rule $(mathrm{adj}(A))^n = mathrm{adj}(A^n)$ on the left side, and combine exponents on the right side to complete the proof that
$$ det(mathrm{adj}(A^3)) = (det(A))^{3n-3} $$
the only thing I wanted to confirm if i did this right is if $(mathrm{adj}(A))^n = mathrm{adj}(A^n)$?
linear-algebra adjoint-operators
edited Feb 3 at 12:29
Community♦
1
asked Oct 12 '14 at 2:07
J LJ L
634102654
$endgroup$
Suppose that $A$ is invertible $n times n$ matrices. Then using the property $A~mathrm{adj}(A) = det(A)I_n$, prove that $det(mathrm{adj}(A^3)) = (det(A))^{3n-3}$
I started with.
$$ A~mathrm{adj}(A) = det(A)I_n $$
I took $det()$ of both sides, then raised both side to power of $3$, then I multiplied both sides by $1/det(A^3)$, then I use the rule $(mathrm{adj}(A))^n = mathrm{adj}(A^n)$ on the left side, and combine exponents on the right side to complete the proof that
$$ det(mathrm{adj}(A^3)) = (det(A))^{3n-3} $$
the only thing I wanted to confirm if i did this right is if $(mathrm{adj}(A))^n = mathrm{adj}(A^n)$?
linear-algebra adjoint-operators
linear-algebra adjoint-operators
edited Feb 3 at 12:29
Community♦
1
asked Oct 12 '14 at 2:07
J LJ L
634102654
edited Feb 3 at 12:29
Community♦
1
asked Oct 12 '14 at 2:07
J LJ L
634102654
edited Feb 3 at 12:29
Community♦
1
edited Feb 3 at 12:29
Community♦
1
edited Feb 3 at 12:29
Community♦
1
1
asked Oct 12 '14 at 2:07
J LJ L
634102654
asked Oct 12 '14 at 2:07
J LJ L
634102654
asked Oct 12 '14 at 2:07
J LJ L
634102654
634102654
add a comment |
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1 Answer
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$begingroup$
Yes (link). Also, as $A^3$ is also invertible, we can start with:$$A^3{rm adj(A^3)}={rm det}(A^3)I_n.$$
Then, taking determinant on both sides:
$$ {rm det}(A^3){rm det}({rm adj(A^3)})= ({rm det}(A^3))^n.$$
answered Oct 12 '14 at 2:14
ir7ir7
4,19811115
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$begingroup$
Yes (link). Also, as $A^3$ is also invertible, we can start with:$$A^3{rm adj(A^3)}={rm det}(A^3)I_n.$$
Then, taking determinant on both sides:
$$ {rm det}(A^3){rm det}({rm adj(A^3)})= ({rm det}(A^3))^n.$$
answered Oct 12 '14 at 2:14
ir7ir7
4,19811115
$endgroup$
add a comment |
$begingroup$
Yes (link). Also, as $A^3$ is also invertible, we can start with:$$A^3{rm adj(A^3)}={rm det}(A^3)I_n.$$
Then, taking determinant on both sides:
$$ {rm det}(A^3){rm det}({rm adj(A^3)})= ({rm det}(A^3))^n.$$
answered Oct 12 '14 at 2:14
ir7ir7
4,19811115
$endgroup$
add a comment |
$begingroup$
Yes (link). Also, as $A^3$ is also invertible, we can start with:$$A^3{rm adj(A^3)}={rm det}(A^3)I_n.$$
Then, taking determinant on both sides:
$$ {rm det}(A^3){rm det}({rm adj(A^3)})= ({rm det}(A^3))^n.$$
answered Oct 12 '14 at 2:14
ir7ir7
4,19811115
$endgroup$
Yes (link). Also, as $A^3$ is also invertible, we can start with:$$A^3{rm adj(A^3)}={rm det}(A^3)I_n.$$
Then, taking determinant on both sides:
$$ {rm det}(A^3){rm det}({rm adj(A^3)})= ({rm det}(A^3))^n.$$
answered Oct 12 '14 at 2:14
ir7ir7
4,19811115
edited Oct 12 '14 at 2:19
answered Oct 12 '14 at 2:14
ir7ir7
4,19811115
answered Oct 12 '14 at 2:14
ir7ir7
4,19811115
answered Oct 12 '14 at 2:14
ir7ir7
4,19811115
4,19811115
add a comment |
add a comment |
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