Why does $D \det(I)B=\frac{d}{d t}\det(I+t B)|_{t=0}$ hold? The following problem is from Taylor's PDE I. I do not get why $D \det(I)B=\frac{d}{d t}\det(I+t B)|_{t=0}$ hold. Since $\det(I)=1$ and $D$ is $n\times n$ matrix, the left side seems to be a matrix, while the right side is clearly a scalar. Could anyone tell me how I should correctly interpret $\det(I)$ in this case?


suppose the eigenvalues of $B$ are $\lambda_1, \lambda_2, \cdots, \lambda_n.$  then  the eigenvalues of $I + tB$ are $1 + t\lambda_1, 1 + t\lambda_2, \cdots, 1 + t\lambda_n.$  we will use the fact that the determinant of a matrix is the product of its eigenvalues.
now we have $$det(I + tB) = (1 + t\lambda_1) (1 + t\lambda_2) \cdots (1 + t\lambda_n) = det(I)+ t(\lambda_1 + \lambda_2 \cdots +\lambda_n) + \cdots  $$
therefore $$\frac{d}{dt}det(I + tB)\big|_{t = 0}  = trace(B).$$
