Here is a solution to the hint:
First of all, note that since all the ideals in question contain $\mathfrak a$, we may replace
$A$ by $A/\mathfrak a$, and so assume that $\mathfrak a = 0$; this simplifies the notation somewhat.
Next, the condition that $\mathfrak b_1 \cap \cdots \cap \mathfrak b_r = 0$ is equivalent
to the requirement that the natural map $A \to A/\mathfrak b_1 \times \cdot \times
A/\mathfrak b_r$ (the product ot the natural quotient maps) is injective, while the
condition that $\mathfrak b_i$ is irreducible is equivalent to the statement that
if $I$ and $J$ are non-zero ideals in $A/\mathfrak b_i$, then $I \cap J \neq 0$ also.
Now suppose given our two irreducible decompositions of $0$. Choose $i$ as in the hint,
and set $I_j := \mathfrak b_1 \cap \cdots \cap \mathfrak b_{i-1} \cap \mathfrak c_j\cap \mathfrak b_{i+1}
\cap \cdots \cap \mathfrak b_r,$ for each $j =1,\ldots,s$.
Then $I_1\cap \cdots \cap I_s = 0$ (since it is contained in the intersection
of the $\mathfrak c_j$, which already vanishes).
Now we recall that $A$ embeds into the product of the $A/\mathfrak b_{i'}$.
Note that $I_j$ is contained in $\mathfrak b_{i'}$ for $i'\neq i$. Thus,
if we let $J_j$ denote the image of $I_j$ in $A/\mathfrak b_i$, then
we see that the image of $I_j$ under the embedding
$A \hookrightarrow A/\mathfrak b_1\times\cdots\times A/\mathfrak b_i \times \cdots \times A/\mathfrak b_r$
is equal to $0 \times \cdots \times J_j \times\cdots \times 0$.
Thus the intersection of the images of the $I_j$, which is the image
of the intersection of the $I_j$ (since we looking at images under an embedding)
is equal to $0\times \cdots \times (\bigcap J_j) \times \cdots \times 0.$
Thus, since the intersection of the $I_j$ is equal to $0$, we see
that $\bigcap J_j = 0.$ But $\mathfrak b_i$ is irreducible, and so
one of the $J_j = 0$. Equivalently, the corresponding $I_j = 0.$
This proves the hint.
(I think the exercise should be a fairly easy deduction from the hint. The
statement that $r = s$ at least follows directly, using the hint together with
minimality of the two decompositions.)
I can show that there is no universal formula. More precisely, let $k$ be an algebraically closed field and let $R = k[w,x,y,z,\Delta]/(\Delta^2-wz+xy)$. I claim that there do not exist $2 \times 2$ matrices $A_1$, $A_2$, ..., $A_{2t}$ with entries in $R$, such that every matrix occurs an even number of times in the list up to transpose, and $\left( \begin{smallmatrix} w & x \\ y & z \end{smallmatrix} \right) = A_1 A_2 \cdots A_{2t}$. In fact, we are going to show something much stronger: It is impossible to write $\left( \begin{smallmatrix} w & x \\ y & z \end{smallmatrix} \right)$ as $MN$ for two noninvertible $2 \times 2$ matrices with entries in $R$.
Since $R$ is a graded integral domain, $\det M$ and $\det N$ would have to be of pure degree. If $\det M$ has degree $0$ then $M$ is invertible. So we must have $\det M$ and $\det N$ of degree $1$.
Write $M=M_0 + M_1 + \cdots$ where $M_k$ is pure of degree $k$, and likewise for $N$. So $$\left( \begin{smallmatrix} w & x \\ y & z \end{smallmatrix} \right) = M_0 N_1 + M_1 N_0.$$
Since $\det M$ has no constant term, we have $\det M_0 =0$ and, similarly, $\det N_0 =0$. Let $\left( \begin{smallmatrix} u_1 & u_2 \end{smallmatrix} \right) \in k^2$ be in the left kernel of $M_0$ and let $\left( \begin{smallmatrix} v_1 \\ v_2 \end{smallmatrix} \right) \in k^2$ be in the right kernel of $N_0$. So the above equation shows that $\left( \begin{smallmatrix} u_1 & u_2 \end{smallmatrix} \right) \left( \begin{smallmatrix} w & x \\ y & z \end{smallmatrix} \right) \left( \begin{smallmatrix} v_1 \\ v_2 \end{smallmatrix} \right) =0$.
But there are no nonzero vectors in $k^2$ for which $\left( \begin{smallmatrix} u_1 & u_2 \end{smallmatrix} \right) \left( \begin{smallmatrix} w & x \\ y & z \end{smallmatrix} \right) \left( \begin{smallmatrix} v_1 \\ v_2 \end{smallmatrix} \right) =0$, a contradiction.
UPDATE: Pushing this argument further, I can show that $\begin{pmatrix} 0 & u & v & w \\ -u & 0 & x & y \\ -v & -x & 0 & z \\ -w & -y & -z & 0 \end{pmatrix}$ cannot be expressed as $MN$ with $\det M$ and $\det N$ nonconstant. Proof: Abbreviate the above matrix to $S$. As before, deduce that we can write $M_0 N_0 =0$ and $M_0 N_1 + M_1 N_0 = S$ with $\det M_0 = \det N_0 = 0$. So $\mathrm{rank} M_0 + \mathrm{rank} N_0 \leq 4$. Without loss of generality, suppose that $M_0$ has nullity at least $2$ and $N_0$ has nullity at least $1$. So, as before, we can find linearly independent constant vectors $p$ and $q$ with $p^T M_0=q^T M_0 =0$ and a nonzero vector $r$ with $N_0 r=0$. We have $p^T S r = q^T S r = 0$ as before.
Let $p=(p_1, p_2, p_3, p_4)$ and $r=(r_1, r_2, r_3, r_4)$. Looking at the coefficient of $u$ in $p^T S r$, we get $p_1 r_2 = p_2 r_1$. Looking at $v$, $w$, $x$, $y$ and $z$, we see that $p_i r_ = p_j r_i$ for all $(i,j)$, so $p$ and $r$ are proportional. Similarly, $q$ and $r$ are proportional, so $p$ and $q$ are proportional, contradicting that we choose them linearly independent. QED.
I have no $2 \times 2$ counter-example.
Best Answer
Let M be the $n\times m$ matrix representing the injection $A^m \to A^n$. Define Di(M) to be the ideal generated by the determinants of all i-by-i minors of M. Let r be the largest possible integer such that Dr(M) has no annihilator (i.e. there is no nonzero element a∈A such that aDr(M)=0); I think r is usually called the McCoy rank of M.
Assume that $m>n$. We shall get a contradiction.
Choose a nonzero a∈A such that aDr+1(M)=0. By assumption, $a$ does not annihilate Dr(M), so there is some r-by-r minor that is not killed by $a$; we may assume it is the upper-left r-by-r minor. Thus, $r \leq n$, so that $r+1 \leq n+1\leq m$. Let M1, ..., Mr+1 be the cofactors of the upper-left (r+1)-by-(r+1) minor obtained by expanding along the bottom row. (This is well-defined even if the $r+1$-th row of $M$ does not exist, because these cofactors use only the first $r$ rows and the first $r+1$ columns of $A$, and $A$ has both since $r \leq n$ and $r+1 \leq m$.) Note, in particular, we know Mr+1 is the determinant of the upper-left r-by-r minor, so aMr+1≠0.
The claim is that the vector (aM1,...,aMr+1,0,0,...) (which we've already shown is non-zero) is in the kernel of M. To see that, note that when you dot this vector with the k-th row of M, you get $a$ times the determinant of the matrix obtained by replacing the bottom row of the upper-left (r+1)-by-(r+1) minor with the first r+1 entries in the k-th row. If k≤r, this determinant is zero because a row is repeated, and if k>r, this determinant is the determinant of some (r+1)-by-(r+1) minor, so it is annihilated by $a$. But since A is injective, the kernel of M is 0, and we have a contradiction.