The code to question 35658 gives a wrong answer i.e. for the matrix
A=matrix([[4,1,2],[0,2,-4],[0,1,6]])
Where can I find an improvement?
![]() | 1 | initial version |
The code to question 35658 gives a wrong answer i.e. for the matrix
A=matrix([[4,1,2],[0,2,-4],[0,1,6]])
Where can I find an improvement?
![]() | 2 | None |
The code to question 35658 gives a wrong answer i.e. for the matrix
A=matrix([[4,1,2],[0,2,-4],[0,1,6]])
Where can I find an improvement?
@crisman The power of a matrix A satisfies by definition the recursion Ak+1=A∗Ak and the initial condition A0=I. Substitution of k=0 and k=1 should therefore give the identity matrix I resp. the matrix A. The following live code does not give the expected answer.
Btw: One of the M-programs gives the correct answer
![]() | 3 | None |
The code to question 35658 gives a wrong answer i.e. for the matrix
A=matrix([[4,1,2],[0,2,-4],[0,1,6]])
Where can I find an improvement?
@crisman
The power of a matrix A satisfies by definition the recursion Ak+1=A∗Ak and the initial condition A0=I.
Substitution of k=0 and k=1 should therefore give the identity matrix I resp. the matrix A. The following following
live code does not give the expected answer.
Btw: One of the M-programs gives the correct answer
![]() | 4 | None |
The code to question 35658 gives a wrong answer i.e. for the matrix
A=matrix([[4,1,2],[0,2,-4],[0,1,6]])
Where can I find an improvement?
@crisman
The power of a matrix A satisfies by definition the recursion Ak+1=A∗Ak and the initial condition A0=I.
Substitution of k=0 and k=1 should therefore give the identity matrix I resp. the matrix A. The following
live code following \href{live code}{http://sagecell.sagemath.org/?z=eJzLVrBVKEss0lDPVtfk5XK0zU0sKcqs0IiONtEx1DGK1Yk20DHS0TUBMwx1zGJjgaoK8ktsHeOyebmKM_LLNYA8oBiCo1dcmlSskW1roKmJKWioqQkA_MQgTQ==&lang=sage} does not give the expected answer.
Btw: One of the M-programs gives the correct answer
![]() | 5 | None |
The code to question 35658 gives a wrong answer i.e. for the matrix
A=matrix([[4,1,2],[0,2,-4],[0,1,6]])
Where can I find an improvement?
@crisman
The power of a matrix A satisfies by definition the recursion Ak+1=A∗Ak and the initial condition A0=I.
Substitution of k=0 and k=1 should therefore give the identity matrix I resp. the matrix A. The following
\href{live code}{http://sagecell.sagemath.org/?z=eJzLVrBVKEss0lDPVtfk5XK0zU0sKcqs0IiONtEx1DGK1Yk20DHS0TUBMwx1zGJjgaoK8ktsHeOyebmKM_LLNYA8oBiCo1dcmlSskW1roKmJKWioqQkA_MQgTQ==&lang=sage} \href{lhttp://sagecell.sagemath.org/?z=eJzLVrBVKEss0lDPVtfk5XK0zU0sKcqs0IiONtEx1DGK1Yk20DHS0TUBMwx1zGJjgaoK8ktsHeOyebmKM_LLNYA8oBiCo1dcmlSskW1roKmJKWioqQkA_MQgTQ==&lang=sage}{live code} does not give the expected answer.
Btw: One of the M-programs gives the correct answer
![]() | 6 | None |
The code to question 35658 gives a wrong answer i.e. for the matrix
A=matrix([[4,1,2],[0,2,-4],[0,1,6]])
Where can I find an improvement?
@crisman
The power of a matrix A satisfies by definition the recursion Ak+1=A∗Ak and the initial condition A0=I.
Substitution of k=0 and k=1 should therefore give the identity matrix I resp. the matrix A. The following
\href{lhttp://sagecell.sagemath.org/?z=eJzLVrBVKEss0lDPVtfk5XK0zU0sKcqs0IiONtEx1DGK1Yk20DHS0TUBMwx1zGJjgaoK8ktsHeOyebmKM_LLNYA8oBiCo1dcmlSskW1roKmJKWioqQkA_MQgTQ==&lang=sage}{live code} live code
sagecell.sagemath.org/?z=eJzLVrBVKEss0lDPVtfk5XK0zU0sKcqs0IiONtEx1DGK1Yk20DHS0TUBMwx1zGJjgaoK8ktsHeOyebmKM_LLNYA8oBiCo1dcmlSskW1roKmJKWioqQkA_MQgTQ==&lang=sage
does not {\bf not} give the expected answer.
Btw: One of the M-programs gives {\bf gives}
http://www.wolframalpha.com/input/?i=MatrixPower%5B%7B%7B4,1,2%7D,%7B0,2,-4%7D,%7B0,1,6%7D%7D,k%5D
the correct answer
![]() | 7 | None |
The code to question 35658 gives a wrong answer i.e. for the matrix
A=matrix([[4,1,2],[0,2,-4],[0,1,6]])
Where can I find an improvement?
@crisman The power of a matrix A satisfies by definition the recursion Ak+1=A∗Ak and the initial condition A0=I. Substitution of k=0 and k=1 should therefore give the identity matrix I resp. the matrix A. The following live code
sagecell.sagemath.org/?z=eJzLVrBVKEss0lDPVtfk5XK0zU0sKcqs0IiONtEx1DGK1Yk20DHS0TUBMwx1zGJjgaoK8ktsHeOyebmKM_LLNYA8oBiCo1dcmlSskW1roKmJKWioqQkA_MQgTQ==&lang=sage
does {\bf not} give the expected answer.
Btw: One of the M-programs {\bf gives}
http://www.wolframalpha.com/input/?i=MatrixPower%5B%7B%7B4,1,2%7D,%7B0,2,-4%7D,%7B0,1,6%7D%7D,k%5D
the correct answer
![]() | 8 | None |
The code to question 35658 gives a wrong answer i.e. for the matrix
A=matrix([[4,1,2],[0,2,-4],[0,1,6]])
Where can I find an improvement?
@crisman The power of a matrix A satisfies by definition the recursion Ak+1=A∗Ak and the initial condition A0=I. Substitution of k=0 and k=1 should therefore give the identity matrix I resp. the matrix A. The following live code
sagecell.sagemath.org/?z=eJzLVrBVKEss0lDPVtfk5XK0zU0sKcqs0IiONtEx1DGK1Yk20DHS0TUBMwx1zGJjgaoK8ktsHeOyebmKM_LLNYA8oBiCo1dcmlSskW1roKmJKWioqQkA_MQgTQ==&lang=sagehttp://sagecell.sagemath.org/?z=eJzLVrBVKEss0lDPVtfk5XK0zU0sKcqs0IiONtEx1DGK1Yk20DHS0TUBMwx1zGJjgaoK8ktsHeOyebmKM_LLNYA8oBiCo1dcmlSskW1roKmJKWioqQkA_MQgTQ==&lang=sage
does {\bf not} not give the expected answer.
Btw: One of the M-programs {\bf gives}gives
http://www.wolframalpha.com/input/?i=MatrixPower%5B%7B%7B4,1,2%7D,%7B0,2,-4%7D,%7B0,1,6%7D%7D,k%5D
the correct answer
![]() | 9 | None |
The code to question 35658 gives a wrong answer i.e. for the matrix
A=matrix([[4,1,2],[0,2,-4],[0,1,6]])
Where can I find an improvement?
@crisman The power of a matrix A satisfies by definition the recursion Ak+1=A∗Ak and the initial condition A0=I. Substitution of k=0 and k=1 should therefore give the identity matrix I resp. the matrix A. The following live code
http://sagecell.sagemath.org/?z=eJzLVrBVKEss0lDPVtfk5XK0zU0sKcqs0IiONtEx1DGK1Yk20DHS0TUBMwx1zGJjgaoK8ktsHeOyebmKM_LLNYA8oBiCo1dcmlSskW1roKmJKWioqQkA_MQgTQ==&lang=sage
does not give the expected answer.
Btw: One of the M-programs gives
http://www.wolframalpha.com/input/?i=MatrixPower%5B%7B%7B4,1,2%7D,%7B0,2,-4%7D,%7B0,1,6%7D%7D,k%5Dhttps://www.wolframalpha.com/input/?i=MatrixPower%5B%7B%7B4,1,2%7D,%7B0,2,-4%7D,%7B0,1,6%7D%7D,k%5D
the correct answer
![]() | 10 | None |
The code to question 35658 gives a wrong answer i.e. for the matrix
A=matrix([[4,1,2],[0,2,-4],[0,1,6]])
Where can I find an improvement?
@crisman
@krisman
The power of a matrix A satisfies by definition the recursion Ak+1=A∗Ak and the initial condition A0=I.
Substitution of k=0 and k=1 should therefore give the identity matrix I resp. the matrix A. The following
live code
http://sagecell.sagemath.org/?z=eJzLVrBVKEss0lDPVtfk5XK0zU0sKcqs0IiONtEx1DGK1Yk20DHS0TUBMwx1zGJjgaoK8ktsHeOyebmKM_LLNYA8oBiCo1dcmlSskW1roKmJKWioqQkA_MQgTQ==&lang=sage
does not give the expected answer.
Btw: One of the M-programs gives
https://www.wolframalpha.com/input/?i=MatrixPower%5B%7B%7B4,1,2%7D,%7B0,2,-4%7D,%7B0,1,6%7D%7D,k%5D
the correct answer
![]() | 11 | None |
The code to question 35658 gives a wrong answer i.e. for the matrix
A=matrix([[4,1,2],[0,2,-4],[0,1,6]])
Where can I find an improvement?
@krisman
@kcrisman
The power of a matrix A satisfies by definition the recursion Ak+1=A∗Ak and the initial condition A0=I.
Substitution of k=0 and k=1 should therefore give the identity matrix I resp. the matrix A. The following
live code
http://sagecell.sagemath.org/?z=eJzLVrBVKEss0lDPVtfk5XK0zU0sKcqs0IiONtEx1DGK1Yk20DHS0TUBMwx1zGJjgaoK8ktsHeOyebmKM_LLNYA8oBiCo1dcmlSskW1roKmJKWioqQkA_MQgTQ==&lang=sage
does not give the expected answer.
Btw: One of the M-programs gives
https://www.wolframalpha.com/input/?i=MatrixPower%5B%7B%7B4,1,2%7D,%7B0,2,-4%7D,%7B0,1,6%7D%7D,k%5D
the correct answer
![]() | 12 | None |
The code to question 35658 gives a wrong answer i.e. for the matrix
A=matrix([[4,1,2],[0,2,-4],[0,1,6]])
Where can I find an improvement?
@kcrisman The power of a matrix A satisfies by definition the recursion Ak+1=A∗Ak and the initial condition A0=I. Substitution of k=0 and k=1 should therefore give the identity matrix I resp. the matrix A. The following live code
http://sagecell.sagemath.org/?z=eJzLVrBVKEss0lDPVtfk5XK0zU0sKcqs0IiONtEx1DGK1Yk20DHS0TUBMwx1zGJjgaoK8ktsHeOyebmKM_LLNYA8oBiCo1dcmlSskW1roKmJKWioqQkA_MQgTQ==&lang=sage
does not give the expected answer.
Btw: One of the M-programs gives
https://www.wolframalpha.com/input/?i=MatrixPower%5B%7B%7B4,1,2%7D,%7B0,2,-4%7D,%7B0,1,6%7D%7D,k%5D
the correct answer
![]() | 13 | None |
The code to question 35658 gives a wrong answer i.e. for the matrix
A=matrix([[4,1,2],[0,2,-4],[0,1,6]])
Where can I find an improvement?
@kcrisman
The For k∈N the k-th power of a matrix A A∈Rn×n satisfies by definition the recursion Ak+1=A∗Ak and the initial condition A0=I.
Substitution of k=0 and k=1 should therefore give the identity matrix I resp. the matrix A. The following
live code
http://sagecell.sagemath.org/?z=eJzLVrBVKEss0lDPVtfk5XK0zU0sKcqs0IiONtEx1DGK1Yk20DHS0TUBMwx1zGJjgaoK8ktsHeOyebmKM_LLNYA8oBiCo1dcmlSskW1roKmJKWioqQkA_MQgTQ==&lang=sage
does not give the expected answer.
Btw: One of the M-programs gives
https://www.wolframalpha.com/input/?i=MatrixPower%5B%7B%7B4,1,2%7D,%7B0,2,-4%7D,%7B0,1,6%7D%7D,k%5D
the correct answeranswer.
![]() | 14 | None |
The code to question 35658 gives a wrong answer i.e. for the matrix
A=matrix([[4,1,2],[0,2,-4],[0,1,6]])
Where can I find an improvement?
@kcrisman For k∈N the k-th power of a matrix A∈Rn×n satisfies by definition the recursion Ak+1=A∗Ak and the initial condition A0=I. Substitution of k=0 and k=1 should therefore give the identity matrix I resp. the matrix A. The following live code
http://sagecell.sagemath.org/?z=eJzLVrBVKEss0lDPVtfk5XK0zU0sKcqs0IiONtEx1DGK1Yk20DHS0TUBMwx1zGJjgaoK8ktsHeOyebmKM_LLNYA8oBiCo1dcmlSskW1roKmJKWioqQkA_MQgTQ==&lang=sage
does not give the expected answer.
Btw: One of the M-programs gives
https://www.wolframalpha.com/input/?i=MatrixPower%5B%7B%7B4,1,2%7D,%7B0,2,-4%7D,%7B0,1,6%7D%7D,k%5D
the correct answer.
![]() | 15 | retagged |
The code to question 35658 gives a wrong answer i.e. for the matrix
A=matrix([[4,1,2],[0,2,-4],[0,1,6]])
Where can I find an improvement?
@kcrisman For k∈N the k-th power of a matrix A∈Rn×n satisfies by definition the recursion Ak+1=A∗Ak and the initial condition A0=I. Substitution of k=0 and k=1 should therefore give the identity matrix I resp. the matrix A. The following live code
http://sagecell.sagemath.org/?z=eJzLVrBVKEss0lDPVtfk5XK0zU0sKcqs0IiONtEx1DGK1Yk20DHS0TUBMwx1zGJjgaoK8ktsHeOyebmKM_LLNYA8oBiCo1dcmlSskW1roKmJKWioqQkA_MQgTQ==&lang=sage
does not give the expected answer.
Btw: One of the M-programs gives
https://www.wolframalpha.com/input/?i=MatrixPower%5B%7B%7B4,1,2%7D,%7B0,2,-4%7D,%7B0,1,6%7D%7D,k%5D
the correct answer.
![]() | 16 | None |
The code to question 35658 35658 gives a wrong answer i.e. for the matrix
A=matrix([[4,1,2],[0,2,-4],[0,1,6]])
Where can I find an improvement?
@kcrisman For k∈N the k-th power of a matrix A∈Rn×n satisfies by definition the recursion Ak+1=A∗Ak and the initial condition A0=I. Substitution of k=0 and k=1 should therefore give the identity matrix I resp. the matrix A. The following live code
http://sagecell.sagemath.org/?z=eJzLVrBVKEss0lDPVtfk5XK0zU0sKcqs0IiONtEx1DGK1Yk20DHS0TUBMwx1zGJjgaoK8ktsHeOyebmKM_LLNYA8oBiCo1dcmlSskW1roKmJKWioqQkA_MQgTQ==&lang=sage
does not give the expected answer.
Btw: One of the M-programs gives
https://www.wolframalpha.com/input/?i=MatrixPower%5B%7B%7B4,1,2%7D,%7B0,2,-4%7D,%7B0,1,6%7D%7D,k%5D
the correct answer.
![]() | 17 | None |
The code to question 35658 gives a wrong answer i.e. for the matrix
A=matrix([[4,1,2],[0,2,-4],[0,1,6]])
Where can I find an improvement?
@kcrisman For k∈N the k-th power of a matrix A∈Rn×n satisfies by definition the recursion Ak+1=A∗Ak and the initial condition A0=I. Substitution of k=0 and k=1 should therefore give the identity matrix I resp. the matrix A. The following live code
live code
http://sagecell.sagemath.org/?z=eJzLVrBVKEss0lDPVtfk5XK0zU0sKcqs0IiONtEx1DGK1Yk20DHS0TUBMwx1zGJjgaoK8ktsHeOyebmKM_LLNYA8oBiCo1dcmlSskW1roKmJKWioqQkA_MQgTQ==&lang=sage
does not give the expected answer.
Btw: One of the M-programs givesM-programs
https://www.wolframalpha.com/input/?i=MatrixPower%5B%7B%7B4,1,2%7D,%7B0,2,-4%7D,%7B0,1,6%7D%7D,k%5D
gives the correct answer.
![]() | 18 | None |
The code to question 35658 gives a wrong answer i.e. for the matrix
A=matrix([[4,1,2],[0,2,-4],[0,1,6]])
Where can I find an improvement?
@kcrisman
For k∈N the k-th power of a matrix A∈Rn×n satisfies by definition the recursion Ak+1=A∗Ak and the initial condition A0=I.
Substitution of k=0 and k=1 should therefore give the identity matrix I resp. the matrix A. The following
live code
live code
http://sagecell.sagemath.org/?z=eJzLVrBVKEss0lDPVtfk5XK0zU0sKcqs0IiONtEx1DGK1Yk20DHS0TUBMwx1zGJjgaoK8ktsHeOyebmKM_LLNYA8oBiCo1dcmlSskW1roKmJKWioqQkA_MQgTQ==&lang=sage
does not give the expected answer.
Btw: One of the M-programs
M-programs
https://www.wolframalpha.com/input/?i=MatrixPower%5B%7B%7B4,1,2%7D,%7B0,2,-4%7D,%7B0,1,6%7D%7D,k%5D
gives the correct answer.
![]() | 19 | None |
The code to question 35658 gives a wrong answer i.e. for the matrix
A=matrix([[4,1,2],[0,2,-4],[0,1,6]])
Where can I find an improvement?
@kcrisman For k∈N the k-th power of a matrix A∈Rn×n satisfies by definition the recursion Ak+1=A∗Ak and the initial condition A0=I. Substitution of k=0 and k=1 should therefore give the identity matrix I resp. the matrix A. The following live code does not give the expected answer.
Btw: One of the M-programs
M-programs
https://www.wolframalpha.com/input/?i=MatrixPower%5B%7B%7B4,1,2%7D,%7B0,2,-4%7D,%7B0,1,6%7D%7D,k%5D
gives the correct answer.
![]() | 20 | None |
The code to question 35658 gives a wrong answer i.e. for the matrix
A=matrix([[4,1,2],[0,2,-4],[0,1,6]])
Where can I find an improvement?
@kcrisman
For k∈N the k-th power of a matrix A∈Rn×n satisfies by definition the recursion Ak+1=A∗Ak and the initial condition A0=I.
Substitution of k=0 and k=1 should therefore give the identity matrix I resp. the matrix A. The following
live code does not give the expected answer.
Btw: One of the M-programs gives the correct answer.
![]() | 21 | None |
The code to question 35658 gives a wrong answer i.e. for the matrix
A=matrix([[4,1,2],[0,2,-4],[0,1,6]])
Where can I find an improvement?
For k∈N the k-th power of a matrix A∈Rn×n satisfies by definition the recursion Ak+1=A∗Ak and the initial condition A0=I. Substitution of k=0 and k=1 should therefore give the identity matrix I resp. the matrix A. The following live code does (currently) not give the expected answer.
Btw: One of the M-programs gives the correct answer.
![]() | 22 | None |
The code to question 35658 gives a wrong answer i.e. for the matrix
A=matrix([[4,1,2],[0,2,-4],[0,1,6]])
A=matrix([[4,1,2],[0,2,-4],[0,1,6]])
Where can I find an improvement?
For k∈N the k-th power of a matrix A∈Rn×n satisfies by definition the recursion Ak+1=A∗Ak and the initial condition A0=I. Substitution of k=0 and k=1 should therefore give the identity matrix I resp. the matrix A. The following live code does (currently) not give the expected answer.
Btw: One of the M-programs gives the correct answer.
![]() | 23 | retagged |
The code to question 35658 gives a wrong answer i.e. for the matrix
A=matrix([[4,1,2],[0,2,-4],[0,1,6]])
Where can I find an improvement?
For k∈N the k-th power of a matrix A∈Rn×n satisfies by definition the recursion Ak+1=A∗Ak and the initial condition A0=I. Substitution of k=0 and k=1 should therefore give the identity matrix I resp. the matrix A. The following live code does (currently) not give the expected answer.
Btw: One of the M-programs gives the correct answer.
![]() | 24 | retagged |
The code to question 35658 gives a wrong answer i.e. for the matrix
A=matrix([[4,1,2],[0,2,-4],[0,1,6]])
Where can I find an improvement?
For k∈N the k-th power of a matrix A∈Rn×n satisfies by definition the recursion Ak+1=A∗Ak and the initial condition A0=I. Substitution of k=0 and k=1 should therefore give the identity matrix I resp. the matrix A. The following live code does (currently) not give the expected answer.
Btw: One of the M-programs gives the correct answer.