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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.

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