Find $\det(A)$ if $\det(A + A^T)=8, \det(A + 2A^T)=27$ 
Let $A$ be a $2 \times 2$ real matrix such that $\det(A + A^T)=8,
 \det(A + 2A^T)=27$. Find $\det(A)$


I cannot solve this other than making a tedious calculus so a faster solution is appreciated.
 A: Consider the polynomial:
$$
P(x)=\det(A+xA^{T})
$$
Since $A$ is a $2\times 2$ matrix, then the degree of $P$ is 2, and you have that:
$$
P(1)=8,P(2)=27
$$
Our task, reduces to finding $P(0)=\det(A)$ which is the constant term, but the dominant term is $\det{(A^T)}=\det(A)$, so the constant term and the dominating one are equal, hence:
$$
P(x)=a(x^2+1)+bx
$$
We have two equations and two unknowns, hence we deduce that:
$$
a=11,b=-14
$$
And $\det(A)=a=11$
EDIT:
Here, I show that the dominating term is $\det(A^T)$ for a $n\times n$ matrix $A$, we write $A=(a_{ij})$, then recall by the initial definition of the determinant:
$$
\det(A+xA^T)=\sum_{\sigma}\epsilon{(\sigma)}\prod_{i}(a_{\sigma(i),i}+xa_{i,\sigma(i)})
$$
Expanding the product, we observe that the coefficient associated with $x^n$ is:
$$
\sum_{\sigma}\epsilon{(\sigma)}\prod_{i}a_{i,\sigma(i)}=\det(A^T)
$$
A: Let the matrix be
$$\left(\begin{array}{cc} a & b\\ c & d \end{array}\right)$$
Then
$$A+A^T=\left(\begin{array}{cc} 2a & b+c\\ c+b & 2d \end{array}\right)$$
$$\det(A+A^T)=4ad-(b+c)^2=8$$
$$8ad-2(b+c)^2=16$$
$$A+2A^T=\left(\begin{array}{cc} 3a & b+2c\\ c+2b & 3d \end{array}\right)$$
$$\det(A+2A^T)=9ad-(b+2c)(c+2b)=27$$
Subtract,
$$ad-bc = 27-16=11$$
