- Thread starter
- #1

#### karush

##### Well-known member

- Jan 31, 2012

- 3,066

$\tiny{4.288.T20}$

Suppose that A is a square matrix of size n and $\alpha \in \CC$ is $\alpha$ scalar.

Prove that $\det{\alpha A} = \alpha^n\det{A}$.

Using $\alpha=5$

$\det{5A}=\det\left(5\left[

\begin{array}{rrr}

1&2\\3&4

\end{array} \right]\right)

=\det\left[

\begin{array}{rrr}

5&10\\15&20

\end{array} \right]=-50

$

$5^2\det{A}=5^2\det\left[

\begin{array}{rrr}

1&2\\3&4

\end{array} \right]

=\left[\begin{array}{cc} 25 & 50 \\ 75 & 100 \end{array} \right]=(25)(-50)$

Solution: $aA$ can be obtained from A by elementary row operations %of type II.

$\alpha A = E_1 \cdots E_n A$

where E, is the corresponding elementary matrix that multiplies the i-th row by the constant a.

It follows that

$\det{\alpha A}= \det{E_i}\cdots \det{E_n} \det(A)=\alpha^n\det{A} $

ok I obviouly tried to follow the example above (link) but not quite sure I got the message on it...