# Matrices and Vectors?

Assuming that X is a three dimensional vector and:

AX = ( 1 )

( 2 )

( 0 )

Use A^ -1 to find X

Update:

Sorry, the A Matrix is:

[ 1 0 2 ]

A = [ -1 3 4 ]

[ 0 1 0 ]

and

[ 2/3 -1/3 1 ]

A^-1 = [ 0 0 1 ]

[ 1/6 1/6 -1/2 ]

Relevance

Let S be that solution vector.

So AX = S.

Then (A^-1)AX = (A^-1)S

Which gives X = (A^-1)S.

Since we don't know what A or A^-1 are, we can't actually compute X.

• Yea, treat A like a number so that AX=S, then X = S/A,...except A is a matrix, so you can't divide a 3-D vector S by a matrix A. Hence we do A^(-1) S.

Be careful. The multiplication is a pre-multiplication on the left of S. And if A is not invertible (that means you cannot find the inverse of A), then you might still have a solution but it depends on whether S is in the column space generated by the matrix A. If it does, then you need elementary row operations to find the solutions.

Source(s): Elementary Linear Algebra, by DAvid Lay
• What is Matrix A? It must be a 3x3 matrix of course!