1 | initial version |

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.

2 | No.2 Revision |

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
```

3 | No.3 Revision |

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.

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