Matrices: Reduced Row Echelon Form 1
The matrices are just arrays of numbers that are shorthand for this system of equations.
We can make a coefficient matrix! Let's see this matrix is independent or not. We will transform this matrix to Reduced Row Echelon form(RREF (A)).
The variables that you associate with your pivot entries, we call these pivot variables
(x1, x3). And the variables that are not associated with the pivot entry, we call them free variables
(x2, x4).
We can represent vector X like this. Our solution set is essentially in $R^4$.
Linear combinations of a and b will construct a plane that contains the position vector or contains the point (2, 0, 5, 0).
Solving linear systems with matrices
Let's solve a linear system using an augmented matrix, and put it in RREF.
Matrices: Reduced row echelon form 3
But when we make a RREF matrix, we can see some problems.
0
can never equal with -4
! This means that it is impossible to find a solution that satisfies all three equations at the intersection of three systems of equations.
There are three cases when we make a RREF matrix.
The first case is no solution
and the second case is unique solution.
In here, green columns are free columns and red things are pivot entries. So then we have unlimited solutions, or no unique solutions.
Only if you have whole zero values and you have free variables, then you have an infinite number of solutions.
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