# Revision history [back]

One could use Sage to explore the problem.

Define the matrix $A$:

sage: A = matrix(SR, 4, [0, 1, 1, 1, 1, 0, 2^x, 2^x, 1, 2^x, 0, 2^x, 1, 2^x, 2^x, 0])
sage: A
[  0   1   1   1]
[  1   0 2^x 2^x]
[  1 2^x   0 2^x]
[  1 2^x 2^x   0]


Compute its determinant:

sage: det(A)
-3*2^(2*x)


The sum of its entries:

sage: sum(A.list())
6*2^x + 6


Its inverse:

sage: A^-1
[-2/3*2^x      1/3      1/3      1/3]
[     1/3 -2/3/2^x  1/3/2^x  1/3/2^x]
[     1/3  1/3/2^x -2/3/2^x  1/3/2^x]
[     1/3  1/3/2^x  1/3/2^x -2/3/2^x]


The sum of the entries of its inverse:

sage: sum((A^-1).list())
-2/3*2^x + 2


The rest (solving $\det(A) = 0$ and $\lVert A^-1 \rVert = 0$) is left as an exercise.

One could use Sage to explore the problem.

Define the matrix $A$:

sage: A = matrix(SR, 4, [0, 1, 1, 1, 1, 0, 2^x, 2^x, 1, 2^x, 0, 2^x, 1, 2^x, 2^x, 0])
sage: A
[  0   1   1   1]
[  1   0 2^x 2^x]
[  1 2^x   0 2^x]
[  1 2^x 2^x   0]


Compute its determinant:

sage: det(A)
-3*2^(2*x)


The sum of its entries:

sage: sum(A.list())
6*2^x + 6


Its inverse:

sage: A^-1
~A
[-2/3*2^x      1/3      1/3      1/3]
[     1/3 -2/3/2^x  1/3/2^x  1/3/2^x]
[     1/3  1/3/2^x -2/3/2^x  1/3/2^x]
[     1/3  1/3/2^x  1/3/2^x -2/3/2^x]


The sum of the entries of its inverse:

sage: sum((A^-1).list())
sum((~A).list())
-2/3*2^x + 2


The rest (solving Now solve $\det(A) = 0$ and $\lVert A^-1 A^{-1} \rVert = 0$) is left as an exercise.

0$, extract positive solutions, take the minimum, and compute a numerical approximation. sage: S = solve([det(A), x > 0], x) sage: T = solve([sum((~A).list()), x > 0], x) sage: S [[0 < x, -3*2^(2*x)]] sage: T [[0 < x, -1/3*2^(x + 1) + 2]] sage: S = solve(det(A) == 0, x); S [] sage: T = solve(sum((~A).list()) == 0, x); T [x == (log(6) - log(2))/log(2)] sage: min([e.rhs() for e in S + T if e.rhs() > 0]).n() 1.58496250072116  One could use Sage to explore the problem. Define the matrix$A$: sage: A = matrix(SR, 4, [0, 1, 1, 1, 1, 0, 2^x, 2^x, 1, 2^x, 0, 2^x, 1, 2^x, 2^x, 0]) sage: A [ 0 1 1 1] [ 1 0 2^x 2^x] [ 1 2^x 0 2^x] [ 1 2^x 2^x 0]  Compute its determinant: sage: det(A) -3*2^(2*x)  The sum of its entries: sage: sum(A.list()) 6*2^x + 6  Its inverse: sage: ~A [-2/3*2^x 1/3 1/3 1/3] [ 1/3 -2/3/2^x 1/3/2^x 1/3/2^x] [ 1/3 1/3/2^x -2/3/2^x 1/3/2^x] [ 1/3 1/3/2^x 1/3/2^x -2/3/2^x]  The sum of the entries of its inverse: sage: sum((~A).list()) -2/3*2^x + 2  Now solve$\det(A) = 0$and$\lVert A^{-1} \rVert = 0\$, extract positive solutions, take the minimum, and compute a numerical approximation.

sage: S = solve([det(A), x > 0], x)
sage: T = solve([sum((~A).list()), x > 0], x)
sage: S
[[0 < x, -3*2^(2*x)]]
sage: T
[[0 < x, -1/3*2^(x + 1) + 2]]
sage: S = solve(det(A) == 0, x); S
[]
sage: T = solve(sum((~A).list()) == 0, x); T
[x == (log(6) - log(2))/log(2)]
sage: min([e.rhs() for e in S + T if e.rhs() > 0]).n()
1.58496250072116


This can all be put in a function which, given a matrix A as above, returns this minimum positive solution.