# Michael Artin’s Algebra Ch.1 M.1

(I have started reading Michael Artin’s Algebra and I think uploading solutions of some problems, especially some of the starred ones, might be useful to others like me who are currently self-studying mathematics. If I have obtained the solution from some source, I will mention the source with the solution. Pointing out errors is always very welcome.)

Problem Statement: $M= \begin{bmatrix} A & B \\ C & D \end{bmatrix}$ where each block is an $n \times n$ matrix. Suppose that $A$ is invertible and that $AC=CA$ . Use block multiplication to prove that $det M=det(AD-CB)$.

Solution:

We note that $G=\begin{bmatrix} 0 & I_n \\ A & -C \end{bmatrix}$ is invertible, and by expanding the determinant along its first row, we see that $det G = det A$.

Then $GM= \begin{bmatrix} A & B \\ 0 & AD-CB \end{bmatrix}$.

Take $H=\begin{bmatrix} A^{-1} & 0 \\ 0 & I_n \end{bmatrix}$.

Then $HGM = \begin{bmatrix} I_n & A^{-1}B \\ 0 & AD-CB \end{bmatrix}$.

Clearly, $det M = det (HGM)$. Expanding $HGM$ along its first column, we get $det (HGM) = det (AD - CB)$.