# [SOLVED]T20 Suppose that A is a square matrix of size n and .....

#### karush

##### Well-known member
example solution

$\tiny{4.288.T20}$
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... #### Country Boy

##### Well-known member
MHB Math Helper
The problem is asking you to prove this for any positive integer, n. You cannot just show it for n= 5. I would probably use "induction on n". When n= 1 this "n by n matrix" is just a number, a. Then $det(\alpha A)= \alpha a= \alpha^1 det(A)$.

Now suppose that for A any "k by k" matrix, it is true that $det(\alpha A)= \alpha^k det(A)$ and consider B, an arbitrary k+1 by k+ 1 matrix. Calculate $det(\alpha A)$ by "expansion on the first row. You get a sum of k+ 1 terms, each a product of a number, which will be multiplied by $\alpha$, times the determinant of a k by k matrix.

Last edited:
• karush

#### karush

##### Well-known member
thanks that helped a lot....

I always have a  with proofs