Determinant of a matrix multiplied by a scalar

In summary, the conversation is discussing how to prove the relation det(QA) = Q^n det(A) for a nxn matrix A and scalar Q, using induction and the definition of the determinant. The conversation also touches on the use of sub-determinants and the alternating positive and negative multiplication.
  • #1
karnten07
213
0

Homework Statement



it is problem 1 on the attachment

Homework Equations


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



The Attempt at a Solution


ah i see the first part i think, but I am nto sure I am 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.
 

Attachments

  • problems2.pdf
    21.9 KB · Views: 1,065
Last edited:
Physics news on Phys.org
  • #2
In general, for a nxn matrix A, and any scalar [itex]\alpha[/itex], det([itex]\alpha[/itex]A) = [itex]\alpha^n[/itex] 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
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 I've wrote sum of, this is sigma with an n on the top and j=1 on the bottom.
 
  • #4
im unsure how to write out the induction for the (n-1)x(n-1) matrix. I think I am right in sayign that the form given in the question of the determinant is the laplace way of writing it? many thanks
 
  • #5
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
radou said:
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.

Sorry I am 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
OK, for a nxn matrix A, we have det(A)=[itex]\sum_{j=1}^n (-1)^{i+j} a_{ij} det A_{ij}[/itex], where [itex]A_{ij}[/itex] 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([itex]\alpha[/itex]A) = [itex]\alpha^2[/itex] 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([itex]\alpha[/itex]A) = [itex]\sum_{j=1}^n (-1)^{i+j} a_{ij} det(\alpha A_{ij})[/itex]. Do you know how to proceed now?
 

What is a determinant of a matrix?

A determinant is a scalar value associated with a square matrix. It provides essential information about the properties of the matrix, such as whether it is invertible (non-singular) or singular and whether the matrix's linear transformation scales volumes.

How is the determinant of a matrix calculated?

The determinant of a square matrix can be calculated using various methods, such as cofactor expansion or row reduction. The most common method for a 2x2 matrix [a b; c d] is ad - bc, and for larger matrices, it involves expanding along rows or columns using a recursive formula.

What happens to the determinant of a matrix when it is multiplied by a scalar?

When a matrix is multiplied by a scalar (a constant), the determinant of the resulting matrix is equal to the determinant of the original matrix multiplied by that scalar. In other words, if you have a matrix A and you multiply it by a scalar k to create a new matrix B (B = kA), then det(B) = k * det(A).

Is there a formula to express the determinant of a scaled matrix?

Yes, you can express the determinant of a scaled matrix as follows: \[det(kA) = k^n * det(A)\] Where: - \(det(kA)\) is the determinant of the scaled matrix kA. - \(k\) is the scalar by which the original matrix A is multiplied. - \(n\) is the order (size) of the square matrix A.

Can you provide an example of finding the determinant of a scaled matrix?

Sure! Let's say you have the 3x3 matrix A: \[A = \begin{bmatrix} 2 & 1 & 0 \\ 0 & -3 & 4 \\ 5 & 2 & 1 \end{bmatrix}\] And you want to find the determinant of 2A (i.e., a scaled matrix of A by a factor of 2). Using the formula, you can calculate it as: \[det(2A) = 2^3 * det(A) = 8 * det(A)\] So, you first find the determinant of matrix A using your preferred method, and then multiply that result by 8 to find the determinant of 2A.

Why is finding the determinant of a scaled matrix important in linear algebra?

Finding the determinant of a scaled matrix is important in linear algebra because it helps determine whether the matrix transformation scales volumes. If the determinant is zero, the transformation collapses volumes (and is singular), whereas a non-zero determinant indicates that volumes are preserved (and the transformation is invertible).

Are there any special cases or considerations when calculating the determinant of scaled matrices?

One consideration is that the scalar factor (k) should not be zero, as it would make the entire determinant equal to zero. Additionally, when scaling matrices in practice, you may need to find the original matrix's determinant first before scaling it.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
264
  • Calculus and Beyond Homework Help
Replies
1
Views
722
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
343
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
702
  • Calculus and Beyond Homework Help
Replies
1
Views
845
Back
Top