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).

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