# Determinant of a matrix multiplied by a scalar

1. Oct 29, 2007

### karnten07

1. The problem statement, all variables and given/known data

it is problem 1 on the attachment

2. Relevant equations
detA = ad - bc for a 2 x 2 matrix, where the first column is a c and the 2nd column is b d

3. The attempt at a solution
ah i see the first part i think, but im nto sure im writing my answer in the most correct way, say from first principles (not sure that is the correct eay to describe it). So when i multiply the 2 x 2 matrix by the scalar and work out the determinant by ad - bc, i get the scalar term squared times the determinant of A as the answer. therefore satisfying the first part.

#### Attached Files:

• ###### problems2.pdf
File size:
21.9 KB
Views:
552
Last edited: Oct 29, 2007
2. Oct 29, 2007

In general, for a nxn matrix A, and any scalar $\alpha$, det($\alpha$A) = $\alpha^n$ det(A).

The attachment is still pending approval, so you could post the exact text of the problem, if it's not too long.

3. Oct 29, 2007

### karnten07

If A is an (n x n) matrix and Q is a scalar, prove det(QA) = Q^n det(A)

Directly from the definition of the determinant;

det(A) = Sum of (-1)^(i+j) aij det(A(ij)) n>2
a11a22 - a12a21 n=2

Hint: use induction ie. show for n = 2 first, then show that the statement is true if one assumes it is true for (n-1)(n-1) matrices.

ive tried to write it out as int he question but where ive wrote sum of, this is sigma with an n on the top and j=1 on the bottom.

4. Oct 29, 2007

### karnten07

im unsure how to write out the induction for the (n-1)x(n-1) matrix. I think im right in sayign that the form given in the question of the determinant is the laplace way of writing it? many thanks

5. Oct 29, 2007

You have shown it is true for n = 2. Now assume it is true for some n = k, and write your "definition" of the determinant. Hint: of what order are the sub-determinants appearing in the sum?

Btw, I would not like to cause confusion (so you can ignore this comment after reading it), but what you're using is not exactly the definition of the determinant, that's why I put it under quotes. By using the real definition, one can almost directly see that this relation holds.

6. Oct 29, 2007

### karnten07

Sorry im not very fluent in matrices work, i understand what you mean by sub determinants, i know when working out the determinant the sub det's are alternately positive and negatively multiplied.

So for an n-1 by n-1 matrix, how do i write out the determinant of it using this general definition (given by what my notes call a row expansion?) to show that the relation works?
Im just very unsure on how to go about writing the proof and where to start, if you could give me any more hints and start me off id be most grateful. thanks

7. Oct 30, 2007

OK, for a nxn matrix A, we have det(A)=$\sum_{j=1}^n (-1)^{i+j} a_{ij} det A_{ij}$, where $A_{ij}$ is the matrix we get after removing the i-th row and j-th column from the original matrix (hence, the order of this matrix is n-1). Now, you did prove that, for a 2x2 matrix, det($\alpha$A) = $\alpha^2$ det(A). This is the basis of induction. Now, assume this relation holds for a kxk matrix. We want to show that the relation holds for a (k+1)x(k+1) matrix too, right? Okay, so let A be some (k+1)x(k+1) matrix. Then, det($\alpha$A) = $\sum_{j=1}^n (-1)^{i+j} a_{ij} det(\alpha A_{ij})$. Do you know how to proceed now?