Thread: True or False

A linear system of two equations in three unknowns can not have exactly one solution. True or False?

Think like this, we can find the values of variables if the number of equations and number of variables is same.

rather than give a "formal" answer i'll give an "informal one". one can think of the dimension of a vector space as quantifying its "size" in some sense. a system of two equations in 3 unknowns is equivalent to a 2x3 matrix, which takes your inputs (the 3 unknowns) and spits out two numbers (the two equations evaluated at the inputs).

so we have a mapping from a three-dimensional space to a two-dimensional one. this can't happen without at least one dimension "collapsing" down to 0. the dimension(s) that collapses is some line in the three-dimension space that gets mapped to a single point (the origin of the two-dimensional space). this is a many-to-one operation, which is NOT uniquely reversible.

all of this can be formalized by using the concepts of rank, linear independence and spanning set, but the "idea" is what i just said above.

Originally Posted by Civy71

A linear system of two equations in three unknowns can not have exactly one solution. True or False?

True

xa +yb +zc = b
xa +yb = b-zc
where z can be assigned arbitrarily

a.b,c,d are the columns of the two linear equations. x,y,z are the unknowns

The rank of augmented matrix A:B and A have to be same(=r say) for solution to exist, if r=n unique solution, r<n, infinitely many solutions . In case of 2 equations and 3 variables the rank of augmented matrix and matrix A can be at the most 2 wich is less than 3, so infinitely many solutions exist wich can be found out by putting one unknown equal to a parameter and finding other unknowns in terms of that parameter.

ax + by + cz = d is only valid if d is a LC of a,b and c.

rank is less transparent and more wordy (in my opinion).