Find $\det(A+B)$ given $\det(A)=4$

  • MHB
  • Thread starter Petrus
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In summary: You are a great explainer so thank you!In summary, when finding the determinant of a matrix, it is important to remember that $\det(A+B) \neq \det(A)+\det(B)$ and that $\det(A \cdot B)= \det(A) \cdot \det(B)$. In addition, when calculating the determinant of a matrix, it is important to consider properties such as multilinear and alternating forms, and the effects of swapping rows or columns on the determinant. By applying these concepts, the determinant of $\det(v+9w, w+3u, u+3v)$ can be found.
  • #1
Petrus
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Let $u,v,w$ be the three columns in a 3 x 3 - matrix $A$. Determinant to matrix $A$ can then be considered as a function of $u,v,w$. Assume that $\det(u,v,w)= \det(A)=4$ then find $\det(v+9w, w+3u, u+3v)$.

My progress:

I start with:

$\det(A+B)= \det(A)+\det(B)$

$\det(v+9w, w+3u, u+3v)=\det(v, w+3u,u+3v)+\det(9w, w+3u, u+3v) =$

$\det( v, w, u+3v) + \det(v, 3u, u+3v) + \det(9w, w, u+3v) + \det(9w, 3u, u+3v) = $

$\det(v, w, u) + \det(v, w , 3v) + \det(v, 3u, u) + \det(v, 3u, 3v) + \det(9w, w, u) + $

$\det(9w, w, 3v) + \det(9w, 3u, u) + \det(9w, u, 3v)$
 
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  • #2
It's best to point out that $\det (A+B) = \det A + \det B$ is invalid! Take $A = I_2$ and $B = - I_2$. Then $A+B = 0_2$ but $\det A = 1$ and $\det B = 1$, whereas $\det (A+B) = 0$.
 
  • #3
Petrus said:
Let $u,v,w$ be the three columns in a 3 x 3 - matrix $A$. Determinant to matrix $A$ can then be considered as a function of $u,v,w$. Assume that $\det(u,v,w)= \det(A)=4$ then find $\det(v+9w, w+3u, u+3v)$.

My progress:

I start with:

$\det(A+B)= \det(A)+\det(B)$

$\det(v+9w, w+3u, u+3v)=\det(v, w+3u,u+3v)+\det(9w, w+3u, u+3v) =$

$\det( v, w, u+3v) + \det(v, 3u, u+3v) + \det(9w, w, u+3v) + \det(9w, 3u, u+3v) = $

$\det(v, w, u) + \det(v, w , 3v) + \det(v, 3u, u) + \det(v, 3u, 3v) + \det(9w, w, u) + $

$\det(9w, w, 3v) + \det(9w, 3u, u) + \det(9w, u, 3v)$

As Fantini already stated, $\det(A+B)= \det(A)+\det(B)$ is not true.

You can use $\det(A \cdot B)= \det(A) \cdot \det(B)$.
And consider that
$$(\mathbf v+9\mathbf w, \mathbf w+3\mathbf u, \mathbf u+3\mathbf v) = (\mathbf u, \mathbf v, \mathbf w) \cdot \begin{bmatrix}0&3&1\\ 1&0&3\\ 9&1&1\end{bmatrix}$$
 
  • #4
But if I use that determinant is multilinear then i can?
 
  • #5
Petrus said:
But if I use that determinant is multilinear then i can?

Your method of calculation does work.
It's just your statement about $\det(A+B) = \det(A)+\det(B)$ that is not true.
It should be:
$$\det(\lambda\mathbf a + \mu\mathbf b, \mathbf v, \mathbf w) = \lambda\det(\mathbf a, \mathbf v, \mathbf w) + \mu\det(\mathbf b, \mathbf v, \mathbf w)$$

But what is your question then?
Seems as if you need to continue applying determinant calculation rules.
See properties of the determinant on wiki.
 
Last edited:
  • #6
Edited my previous post.
 
  • #7
I like Serena said:
As Fantini already stated, $\det(A+B)= \det(A)+\det(B)$ is not true.

You can use $\det(A \cdot B)= \det(A) \cdot \det(B)$.
And consider that
$$(\mathbf v+9\mathbf w, \mathbf w+3\mathbf u, \mathbf u+3\mathbf v) = (\mathbf u, \mathbf v, \mathbf w) \cdot \begin{bmatrix}0&3&1\\ 1&0&3\\ 9&1&1\end{bmatrix}$$
Ignore this you did edit ur post op so
 
  • #8
progress so far:
$\det(v+9w, w+3u, u+3v) =9\det(v, w+3u, u+3v) + 9\det(w, w+3u, u+3v) =$
$27\det(v, w, u+3v) + 27\det(w, u, u+3v) + 27\det(w, w, u+3v) + 27\det(w, u, u+3v)$
$= 81\det(v, w, u) + 81\det(v, w, v) + 81\det(w, u, u) + 81\det(w, u, v)+$
$81/det(w, w, u) + 81\det(w, w, v) + 81\det(w, u, u) + 81\det(w, u, v)$

The next step I am kinda unsure how to do it, I guess ima swap road or something and try get exemple $\det(w, w, w)$ cause that is equal to zero? I am correct? How do i swap row or column?
 
  • #9
Petrus said:
progress so far:
$\det(v+9w, w+3u, u+3v) =9\det(v, w+3u, u+3v) + 9\det(w, w+3u, u+3v) =$

You have a 9 instead of a 1. It should be:

$\det(1v+9w, w+3u, u+3v) =1\det(v, w+3u, u+3v) + 9\det(w, w+3u, u+3v) =$

$27\det(v, w, u+3v) + 27\det(w, u, u+3v) + 27\det(w, w, u+3v) + 27\det(w, u, u+3v)$

You have repeated the same mistake here.

$= 81\det(v, w, u) + 81\det(v, w, v) + 81\det(w, u, u) + 81\det(w, u, v)+$
$81/det(w, w, u) + 81\det(w, w, v) + 81\det(w, u, u) + 81\det(w, u, v)$

The next step I am kinda unsure how to do it, I guess ima swap road or something and try get exemple $\det(w, w, w)$ cause that is equal to zero? I am correct? How do i swap row or column?

From the wiki section I linked to before:
8. This n-linear function is an alternating form. This means that whenever two columns of a matrix are identical, or more generally some column can be expressed as a linear combination of the other columns (i.e. the columns of the matrix form a linearly dependent set), its determinant is 0.​

So for instance $\det(w,w,u)=0$.Also from that wiki section:
11. Interchanging two columns of a matrix multiplies its determinant by −1. This follows from properties 7 and 8 (it is a general property of multilinear alternating maps). Iterating gives that more generally a permutation of the columns multiplies the determinant by the sign of the permutation. Similarly a permutation of the rows multiplies the determinant by the sign of the permutation.​

So $\det(u, v, w)=-\det(v, u, w)$.
 
  • #10
I like Serena said:
You have a 9 instead of a 1. It should be:

$\det(1v+9w, w+3u, u+3v) =1\det(v, w+3u, u+3v) + 9\det(w, w+3u, u+3v) =$
You have repeated the same mistake here.
From the wiki section I linked to before:
8. This n-linear function is an alternating form. This means that whenever two columns of a matrix are identical, or more generally some column can be expressed as a linear combination of the other columns (i.e. the columns of the matrix form a linearly dependent set), its determinant is 0.​

So for instance $\det(w,w,u)=0$.Also from that wiki section:
11. Interchanging two columns of a matrix multiplies its determinant by −1. This follows from properties 7 and 8 (it is a general property of multilinear alternating maps). Iterating gives that more generally a permutation of the columns multiplies the determinant by the sign of the permutation. Similarly a permutation of the rows multiplies the determinant by the sign of the permutation.​

So $\det(u, v, w)=-\det(v, u, w)$.
Thanks Serena!
I got it correct and now I understand more about this! Thanks for taking your time and giving me great explain!
 

1. How do you find the determinant of the sum of two matrices?

To find the determinant of the sum of two matrices, you simply add the determinants of the two matrices.

2. What is the determinant of the sum of two matrices if one determinant is known?

If you know the determinant of one matrix, you can use that information and the formula for finding the determinant of the sum to find the determinant of the sum of two matrices.

3. Why is the determinant of the sum of two matrices equal to the sum of their determinants?

This is due to the linearity property of determinants, which states that the determinant of a sum of matrices is equal to the sum of their determinants.

4. Can the determinant of the sum of two matrices be negative if both determinants are positive?

No, the determinant of the sum of two matrices cannot be negative if both determinants are positive. The determinant of a matrix is always a positive or negative number, and the sum of two positive numbers will always be positive.

5. Is there a shortcut to finding the determinant of the sum of two matrices if one of the determinants is known?

Yes, there is a shortcut known as the "determinant shortcut" or "matrix shortcut" that can be used to quickly find the determinant of the sum of two matrices if one of the determinants is known. This shortcut involves multiplying the known determinant by the trace of the other matrix and adding it to the trace of the known matrix multiplied by the other determinant.

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