Actually, it's x3 that is the free variable in this system. Notice that both equations in the "simplified" version have x3s in them, so x3 ends up being the thing that we need to use to define the other variables. You could take, for example, x2 as the free variable, but you would have to reduce your system using some other technique so that both of the last equations had x2s in them.
Recall that what the last all-zero row is saying is that "zero = zero". This is always true, not matter what your variable values are. Another way to think of it would be to say that this demonstrates that the two vectors are not scalar multiples of one another.
OOPS! At about 2:40, I said, and wrote, 4 instead of 8!
Very well explained 😍❤😊
Glad you thought so!
Can we take y and z as free variables instead of x and y?
yes! It is free to choose as they are called 'free variables'. You don't need to pay a price for them either!
Actually, it's x3 that is the free variable in this system. Notice that both equations in the "simplified" version have x3s in them, so x3 ends up being the thing that we need to use to define the other variables. You could take, for example, x2 as the free variable, but you would have to reduce your system using some other technique so that both of the last equations had x2s in them.
How do we conclude the 2 basis vectors are linearly independent by reducing to echelon form in the last step?
By definition, a basis vector is linearly independent and will span the vector space.
@@daz7200 Yes, but if by row echelon the last row is all zero's, they are linearly dependent? (verify)
@@markohorn1548 If the last row is all zeros they are linearly INdependent.
Recall that what the last all-zero row is saying is that "zero = zero". This is always true, not matter what your variable values are. Another way to think of it would be to say that this demonstrates that the two vectors are not scalar multiples of one another.