Prove $\det(1+tA)=1+t\cdot tr(A)+O(t^2)$ I need help proving 
$\det(1+tA)=1+t\cdot \operatorname{tr}(A)+O(t^2)$
I'm not really sure where to start due to the $(1+tA)$, the $1$ is throwing me off.

Assume $A$ is $M_{n\times n}(\mathbb{C}).$ Suppose that $A$ is diagonalizable, with eigenvalues $\lambda_1,\dotsc,\lambda_n.$ Then $1+tA$ is also diagonalizable, with eigenvalues $1+t\lambda_i.$ The determinant of a diagonalizable matrix is the product of its eigenvalues, so we have
$$
\det(1+tA)=(1+t\lambda_1)\cdot\dotsb\cdot(1+t\lambda_n)\\=1+t(\lambda_1+\dotsb+\lambda_n)+\dotsb + t^n(\lambda_1\cdot\dotsb\cdot\lambda_n)\\=1+\text{tr}(A)t+\dotsb+t^n\det(A)=1+\text{tr}(A)t + O(t^2).
$$
If $A$ is not diagonalizable, observe that the diagonalizable matrices are dense in $M_{n\times n}(\mathbb{C}).$ By continuity, the above equation also holds for $A$.
