Matrices Do Commute

Both $A$ & $B$ are symmetrical. Let us denote them as the following with positive integers $a$ to $f$ : $A = \begin{bmatrix} a & b \\ b & c \end{bmatrix}$

$B = \begin{bmatrix} d & e \\ e & f \end{bmatrix}$

Therefore, $AB = BA = \begin{bmatrix} ad+be & ae+bf \\ bd+ce & be+cf \end{bmatrix}$

Thus, we can set up system of equations as:

$(ad+be)-(be+cf) = ad - cf = 54-17 = 37$

$ae+bf = bd+ce = 37$ or $e(a-c) = b(d-f)$

Given that all the elements are positive integers, each determinant will also be an integer, and we can apply the product determinant rule:

$\det(AB) = \det(A) \cdot \det(B) = 54\cdot 17 - 37^2 = -451 = -41\times 11 = -11\times 41$

Neither $A$ nor $B$ can not have a determinant exceeding $9^2 - 1^2 = 80$ , so the determinant value of $451$ is not possible. Then, for generality, there are two possible scenarios for $A$ : $$ $\det(A) = 41$ or $\det(A) = -41$ $$ Considering the former case, $41 = ac - b^2$ .

Then $b^2 = ac - 41$ . By $ac$ must exceed $41$ , the selection of digits can start from $6\cdot 7 = 42$ .

By continuing the higher digit multiplication, there are only two applicable products:

$1^2 = 6\cdot 7 - 41$ , which leads to two possible matrices: $A = \begin{bmatrix} 7 & 1 \\ 1 & 6 \end{bmatrix}$ or
$A = \begin{bmatrix} 6 & 1 \\ 1 & 7 \end{bmatrix}$

$2^2 = 5\cdot 9 - 41$ , which leads to two possible matrices: $A = \begin{bmatrix} 9 & 2 \\ 2 & 5 \end{bmatrix}$ or
$A = \begin{bmatrix} 5 & 2 \\ 2 & 9 \end{bmatrix}$

Now we plug in the values for the equation $ad - cf = 37$ and $e(a-c) = b(d-f)$

For $(a, b, c) = (7,1,6)$ , $7d-6f = 37$ . $f= \dfrac{7d-37}{6}$ . Only $d=7$ results in $f=2$ , but $a=7$ . The matrices can't share the element, according to the question.

For $(a, b, c) = (6,1,7)$ , $6d-7f = 37$ . $f= \dfrac{6d-37}{7}$ . No digit for $d$ can result in a digit for $f$ .

For $(a, b, c) = (9,2,5)$ , $9d-5f = 37$ . $f= \dfrac{9d-37}{5}$ . Only $d=8$ results in $f=7$ , but $a=7$ . The matrices can't share the element, according to the question.

For $(a, b, c) = (5,2,9)$ , $5d-9f = 37$ . $f= \dfrac{5d-37}{9}$ . No digit for $d$ can result in a digit for $f$ .

$$ Thus, $\det(A) = -41$ and $\det(B) = 11$

Then $b^2 = ac + 41$ . The only squares exceeding $41$ are $7^2$ , $8^2$ , and $9^2$ .

If $b=7$ , then $ac = 49 - 41 = 8 = 1\cdot 8 = 2\cdot 4$ .

If $b=8$ , then $ac = 64 - 41 = 23$ . No digit multiplication can apply.

If $b=9$ , then $ac = 81 - 41 = 40 = 5\cdot 8 = 8\cdot 5$ .

Therefore, there are six possible matrices:

$A = \begin{bmatrix} 1 & 7 \\ 7 & 8 \end{bmatrix}$ , for this case $d-8f = 37$ . $d-37 = 8f$ . No digit $d$ can make $f$ positive, so it's not the solution. $$ $A = \begin{bmatrix} 8 & 7 \\ 7 & 1 \end{bmatrix}$ , for this case $8d-f = 37$ . $f = 8d-37$ . Only $d=5$ results in $f=3$ . Then $e(a-c) = b(d-f)$ . $e(8-1) = 7(5-3)$ . $e=2$ . This seems to work out for all constraints, but let us check the other possibilities first. $$

$A = \begin{bmatrix} 2 & 7 \\ 7 & 4 \end{bmatrix}$ , for this case, $e(2-4) = 7(d-f)$ . That means $7|e$ or $e=7$ , but but $b=7$ . The matrices can't share the element, according to the question. $$

$A = \begin{bmatrix} 4 & 7 \\ 7 & 2 \end{bmatrix}$ , for this case, $e(4-2) = 7(d-f)$ . That means $7|e$ or $e=7$ , but but $b=7$ . The matrices can't share the element, according to the question. $$

$A = \begin{bmatrix} 8 & 9 \\ 9 & 5 \end{bmatrix}$ , for this case, $8d-5f = 37$ . $f= \dfrac{8d-37}{5}$ . Only $d=9$ results in $f=7$ , but $b=9$ . The matrices can't share the element, according to the question. $$

$A = \begin{bmatrix} 5 & 9 \\ 9 & 8 \end{bmatrix}$ , for this case, $5d-8f = 37$ . $f= \dfrac{5d-37}{8}$ . Only $d=9$ results in $f=1$ , but $b=9$ . The matrices can't share the element, according to the question. $$

Finally, $A = \begin{bmatrix} 8 & 7 \\ 7 & 1 \end{bmatrix}$ , and $B = \begin{bmatrix} 5 & 2 \\ 2 & 3 \end{bmatrix}$ $$

Checking the multiplication:

$AB=BA=\begin{bmatrix} 8 & 7 \\ 7 & 1 \end{bmatrix} \begin{bmatrix} 5 & 2 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} 5 & 2 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 8 & 7 \\ 7 & 1 \end{bmatrix} = \begin{bmatrix} 54 & 37 \\ 37 & 17 \end{bmatrix}$

Thus, $A+B = \begin{bmatrix} 13 & 9 \\ 9 & 4 \end{bmatrix}$

Therefore, $|\det(A+B)| = |13\cdot 4 - 9^2| = \boxed{29}$ .