-1

Prove by mathematical induction on the size of matrix A that

$\begin{equation} \begin{aligned} &\operatorname{If} M \in M_{n \times n}(K) \text { and } A \text { y } B \text { are square } M=\left[\begin{array}{cc} A & 0 \\ 0 & B \end{array}\right] \text { then }\\ &\det (M)=\det(A)\det(B) \end{aligned} \end{equation}$

I try this

If A is a Matrix $1\times1$ then $\det(M)=\det(\begin{bmatrix} A & 0 \\ 0 & B \end{bmatrix})=a_1\det(B)=\det(A)*\det(B)$

If A is a Matrix $2\times2$ then $\det(M)=\det(\begin{bmatrix} A & 0 \\ 0 & B \end{bmatrix})=a_1\det(\begin{bmatrix} A_{11} & 0 \\ 0 & B \end{bmatrix})+(-1)^{3}a_2\det(\begin{bmatrix} A_{12} & 0 \\ 0 & B \end{bmatrix})=a_1a_4\det(B)-a_2a_3\det(B)=\det(B)(a_1a_4-a_2a_3)=\det(B)\det(A)$

Then I suppose is true for a matrix $n\times n$ with a matrix $A_{s\times s}$ and this is where I'm stuck to prove it is valid for a matrix $A_{s+1\times s+1}$

Bernard
  • 175,478
Kevin Duran
  • 353

2 Answers2

1

If $A$ is $n_1\times n_1$ and $B$ is $n_2\times n_2$ then $\begin{pmatrix}A&0\\0&B\end{pmatrix}=\begin{pmatrix}A&0\\0&I_{n_2}\end{pmatrix}\begin{pmatrix}I_{n_1}&0\\0&B\end{pmatrix}$

What you showed for $1\times 1$ matrix just extends to $I_n$ of any size (very straightforward induction). So the determinant of the first matrix is $\det(A)$ and for the second one $\det(B)$.

Then use $\det(XY)=\det(X)\det(Y)$ property.

zwim
  • 28,563
0

Hint: Expand along the first row.

markvs
  • 19,653
  • The user who voted to delete this answer cannot expand along the first row? – markvs Sep 05 '20 at 00:20
  • I did this $\begin{equation} det(M)=\sum_{j=1}^{s+m+1}(-1)^{1+j} a_{1 j} det(M_{1j}) \end{equation} $ = $\begin{equation} det(M)=\sum_{j=1}^{s+m+1}(-1)^{1+j} a_{1 j} det(A_{1j}det(B)) \end{equation}$ = $detAdetB$ is this right? – Kevin Duran Sep 06 '20 at 00:02
  • Yes, that is right. – markvs Sep 06 '20 at 00:04
  • There is a ")" missing. And before the last "=" you should take $\det B$ outside the $\Sigma$. – markvs Sep 06 '20 at 00:16