Does linearity of cofactor imply linearity of determinant for 3×3 symmetric matrices? $\DeclareMathOperator\cof{cof}$If we suppose that a finite family (greater than 3) of $3\times3$ symmetric matrices $A_i$ and positive reals $a_i$ such that $\sum_ia_i=1$ satisfies
$\cof(\sum_i a_i A_i)=\sum_ia_i\cof A_i$,
does this imply that the same happens for determinant?
Is it true that
$\det(\sum_i a_i A_i)=\sum_ia_i\det A_i$?
If not, what is a counterexample?
 A: Here's a counterexample to the OP's literal question:  Consider the following four symmetric (in fact, diagonal) $3$-by-$3$ matrices:  $A_1 = \mathrm{diag}(0,0,0)$, $A_2 = \mathrm{diag}(\frac12,3,3)$, $A_3 = \mathrm{diag}(0,11,1)$, $A_4 = \mathrm{diag}(1,6,0)$, and let $a_1=a_2=a_3=a_4=\frac14$.  Then the cofactor equation holds, but the determinant equation does not.
Meanwhile, if, instead one asks that the symmetric matrices $A_i$ have the property that the cofactor equation holds for all choices of $a_i$ for which $\sum_i a_i = 1$ and $a_i>0$, then, indeed, the determinant equation holds also for all choices of $a_i$ for which $\sum_i a_i = 1$ and $a_i>0$.
To see this, note first that the cofactor function on $3$-by-$3$ matrices $A\mapsto \mathrm{cof}(A)$ is a quadratic function (i.e., the mapping $\beta(A,B) = \mathrm{cof}(A+B) - \mathrm{cof}(A)-\mathrm{cof}(B)$ is (symmetric and) bilinear in its arguments. In particular, it is homogeneous of degre $2$.
Almost by definition, $\mathrm{cof}(A)=0$ if and only if $A$ has rank at most $1$, i.e. $A = pq^\mathsf{T}$ for some column vectors $p$ and $q$.  In particular, when $A$ is also symmetric, it must be a multiple of $pp^\mathsf{T}$ for some column vector $p$.
Now, suppose that one has $3$-by-$3$ matrices $A_1,\ldots, A_n$ such that $\mathrm{cof}(a_1\,A_1 + \cdots + a_n\,A_n) = a_1\,\mathrm{cof}(A_1)+\cdots+a_n\,\mathrm{cof}(A_n)$ for all $a_i>0$ with $a_1+\cdots+a_n=1$. Writing $a_i = b_i/(b_1+\cdots+b_n)$ for $b_i>0$, substituting and clearing denominators, one sees that
$$
\mathrm{cof}(b_1\,A_1 {+} \cdots {+} b_n\,A_n) = (b_1{+}\cdots{+}b_n)\bigl(b_1\,\mathrm{cof}(A_1)+\cdots+b_n\,\mathrm{cof}(A_n)\bigr)
$$
for all $b_i>0$.  Since both sides of this equation are quadratic polynomials in the $b_i$, it follows that this equation must hold for all $b_i$, not just when the $b_i$ are positive.  In particular, we must have
$$
\mathrm{cof}(b_1\,A_1 {+} \cdots {+} b_n\,A_n)=0
$$
when $b_1+\cdots+b_n=0$.  In particular, $\mathrm{cof}(A_i-A_j) = 0$ for all $i$ and $j$, so that $A_i-A_j$ has rank at most $1$.  Moreover,
$$
\mathrm{cof}\bigl((A_i{-}A_j)- (A_k{-}A_l)\bigr)=0
$$
for all $i$, $j$, $k$ and $l$.  It follows easily that there must be a column vector $p$ and constants $c_i$ such that $A_i = A_0 + c_i\,pp^\mathsf{T}$, where $A_0$ is the average of the $A_i$ for $1\le i\le n$ and $c_1+\cdots+c_n=0$.  Conversely, for $A_i$ of this form, we have
$$
\mathrm{cof}(A_i) = \mathrm{cof}(A_0) + c_i\,\beta(A_0,pp^\mathsf{T}),
$$
and the equation
$$
\mathrm{cof}(a_1\,A_1 + \cdots + a_n\,A_n) = a_1\,\mathrm{cof}(A_1)+\cdots+a_n\,\mathrm{cof}(A_n)
$$
for $a_1+\cdots+a_n=1$ follows immediately from these formulae.  Moreover, one now easily checks that
$$
\det(A_i) = \det(A_0) + c_i\gamma(A_0,A_0,pp^\mathsf{T})
$$
where $\gamma$ is a certain trilinear function on matrices, and the desired determinant identity
$$
\mathrm{det}(a_1\,A_1 + \cdots + a_n\,A_n) = a_1\,\mathrm{det}(A_1)+\cdots+a_n\,\mathrm{det}(A_n)
$$
for $a_1+\cdots+a_n=1$ follows immediately from these formulae.
