Solution set to a matrix equation

Question

I am trying to find the solution set to the equation

$Bx^T = 0^T$

where

$B = \begin{bmatrix} 1 &0 &-1 &0 &0 \\ 0 &1 &1 &-1 &-2 \\ 0 &0 &0 &0 &0 \\ 0 &0 &0 &0 &0 \end{bmatrix}$

I have the columns with the leading ones expressed as the columns with the leading variables ($x, y, z$) and have come up with the solution set of

$\{x, -x + y + 2z, x, y, z\}$

It is a multiple choice question and the 5 answers to choose from are:

• $span([1,0,1,1,0],[0,0,0,2,-1])$

• $\{[x,y,x,x + y + 2z,-z] : x,y,z \in R\}$

• $\{[0,0,0,2t,-1] : t \in R\}$

• $\{[0,t,0,t,0] : t \in R\}$

• $span( [1, 0, 1, 1, 0] )$

Am I correct in thinking, by the process of elimination, that it must be either of the span ones (1st and last options)?

Answer 1

Using a process of elimination is a reasonable way to proceed here, but be sure that you eliminate the right things. How did you come to the conclusion that the other choices are wrong?

Hint: What is the dimension of the kernel (null space) of this matrix? How does that compare to the dimensions of the possible answers?

Answer 2

The rank of this matrix is $2$, hence the space of solutions has dimension $3$ since you work in a space of dimension $5$.

What can you conclude?

Answer 3

You say you came up with solution set {x, -x+ y+ 2z, x, y, z}. That can be written as {x, -x, x, 0, 0}+ {0, y, 0, y, 0}+ {0, 2z, 0, 0, z}= x{1, -1, 1, 0, 0}+ y{0, 1, 0, 1, 0}+ z{0, 2, 0, 0, 1}. You should be able to see what a basis for the solution space is.