# Determinant

#### Petrus

##### Well-known member
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:

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

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

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

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

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

Last edited by a moderator:

#### Fantini

MHB Math Helper
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 $\det (A+B) = 0$.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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:

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

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

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

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

$\det(9w, w, 3v) + \det(9w, 3u, u) + \det(9w, u, 3v)$
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
$$(\mathbf v+9\mathbf w, \mathbf w+3\mathbf u, \mathbf u+3\mathbf v) = (\mathbf u, \mathbf v, \mathbf w) \cdot \begin{bmatrix}0&3&1\\ 1&0&3\\ 9&1&1\end{bmatrix}$$

#### Petrus

##### Well-known member
But if I use that determinant is multilinear then i can?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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.

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#### Klaas van Aarsen

##### MHB Seeker
Staff member
Edited my previous post.

#### Petrus

##### Well-known member
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

#### Petrus

##### Well-known member
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 im 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?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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 im 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 $\det(u, v, w)=-\det(v, u, w)$.

#### Petrus

##### Well-known member
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!