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Consider an $n \times n$ matrix of the form $$ A = \begin{pmatrix} M & P \\ 0 & N \end{pmatrix} $$ with $M \in \Bbb K^{r \times r}$, $P \in \Bbb K^{r \times s}$, and $0 \in \Bbb K^{s \times r}$ is the zero $s \times r$ matrix, $n = s+r$. Show that $$\det(A) = \det(M) \cdot \det(N)$$

I know that I have to approach this question by changing $M$ and $N$ into upper triangular matrix, but how do I do this? I am only allowed to use row operations. Thanks in advance.

Similar questions aren't showing the row operations method.

  • @LinAlg The post you referred didn't show how to transform it into the upper triangular form. I am stuck in transforming A into its upper triangular form. –  Mar 15 '21 at 15:43

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Let's agree to perform only row operations that multiply the determinant by $\pm 1$ and to record the product of the minus ones. If $|N|=0,$ then reduce the final $r$ rows until the last row is zero. Therefore, $|A|$ is also zero. Hence, $|N|$ is nonzero and we can reduce the final $r$ rows so that $N$ becomes a diagonal matrix with nonzero entries--which you can use to turn $P$ into a zero matrix! You can finish the proof.

David
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