Help with Matrices

Any help is appreciated, thanks.

 

A matrix is written with ( ) or [ ] brackets. Do not confuse a matrix with a determinant which uses vertical bars | |. A matrix is a pattern of numbers; a determinant gives us a single number. The size of a matrix is written: rows × columns. A 3 × 3 matrix. A matrix with the same number of rows and columns is called a square matrix. The elements in a matrix A are denoted by aij, where i is the row number and j is the column number. A matrix which is simultaneously upper and lower triangular is diagonal. The identity matrix is the only matrix which is both upper and lower unitriangular. The determinant of a triangular matrix equals the product of the diagonal entries. Since for any triangular matrix A the matrix xI − A, whose determinant is the characteristic polynomial of A, is also triangular, the diagonal entries of A in fact give the multiset of eigenvalues of A. The variable L is commonly used for lower triangular matrix, standing for lower/left, while the variable U or R is commonly used for upper triangular matrix, standing for upper/right.

 

Thanks but I all ready know that stuff,

I just got the (a) out then

 

So what part are you actually struggling with?

Do you need help with the entire worksheet or are there just concepts that you need refined?

 

the rref(A) comes out to be the same as the upper triangular matrix.

 

Thanks, found that the answer for that is 2.

 

#3.

You found rref(a) =
1 2 0 1
0 0 1 2
0 0 0 0

Thus, the augmented matrix of the homogeneous system is:
1 2 0 1 0
0 0 1 2 0
0 0 0 0 0

So, if we use w, x, y, and z as our variables, the system is:
w + 2x + z = 0
y + 2z = 0

So, w and y are fixed variables, and x and z are free variables.
w = -2x - z
x
y = -2z
z

The solutions to this can therefore be expressed as a linear combination (which I'll write horizontally, just because it's too difficult to express vertically).

[ w x y z ] = [-2 1 0 0]x + [-1 0 -2 1]z

So:

-2
1
0
0

and

-1
0
-2
1

form the basis of the null space.

 

l0l0l0 i hate mat bro D:

 

are those igcse papers?

 

Why are we able to assume the system is homogeneous...? Sorry if this is an obvious answer but I haven't dabbled in matrices except for the occasional eigenvector/value here and there :x

 

It's just part of the definition of what he's trying to find: the null space is the set of solutions to the homogenous system. Accordingly, the basis of the null space is just the minimum vectors whose linear combinations form the entire set of solutions to the homogenous system.