Let V be a 5 dimensional vector space

[laminatedevildoll]

Let V be a 5 dimensional vector space, and let $$\Delta$$ be a determinant formon V. Given $$\Delta$$(b1, b2, b3, b4, b5)= -3

How do I find $$\Delta$$(b4, b3, b5, b1, b2)?

 

[mathwonk]

use properties of determinnats, i.e. interchanging two rows changes the determinant by a minus sign. did you know this? if not, go back and review what properties determinants have.

or let me just tell you:

a determinant is an
1) alternating
2) multilinear
3) normalized

function of n variables, where n = dim(V).

normalized means that if there is some preferred basis, like e1,...,en, then det(e1,...,en) = 1.

more intrinsically, if it is only defined for endomorphisms, then det(Id) = 1.

 

[M98Ranger]

To find \Delta(b4, b3, b5, b1, b2), we can use the properties of determinants. Firstly, we know that the determinant of a matrix remains the same if we swap two rows or columns. So, we can rewrite \Delta(b4, b3, b5, b1, b2) as \Delta(b1, b2, b3, b4, b5), which is the given value of -3.

Next, we can use the property that the determinant of a matrix is equal to the negative of the determinant of its transpose. So, we can take the transpose of the matrix \Delta(b1, b2, b3, b4, b5) and find its determinant, which will give us the desired value of \Delta(b4, b3, b5, b1, b2).

The transpose of a matrix is obtained by interchanging the rows and columns. So, the transpose of \Delta(b1, b2, b3, b4, b5) will be \Delta(b1, b4, b3, b5, b2).

Therefore, we can write \Delta(b4, b3, b5, b1, b2) = -\Delta(b1, b4, b3, b5, b2).

Substituting the value of \Delta(b1, b2, b3, b4, b5) = -3, we get \Delta(b4, b3, b5, b1, b2) = -(-3) = 3.

In conclusion, we have found the value of \Delta(b4, b3, b5, b1, b2) to be 3 by using the properties of determinants.