finding constants on the right side of a linear system

Question

I am trying to find all values of a, b, and c so that this:

$$\begin{cases} &x &- &3y &= a \\ &2y &- &z &= b \\ &x &- &5y &+ &z &= c \end{cases} $$

linear system has solutions. But I am finding that I end up with no solutions when following elimination, basing off of coefficient vs augmented matrix rank.

Based off of how the question was worded, I feel that I am expected to find solutions. Am I missing something here?

my elimination looks like:

$$ \begin{bmatrix} 1 & -3 & 0 &| &a \\ 0 & 2 & -1 &| &b \\ 1 & -5 & 1 &| &c \\ \end{bmatrix} $$ -R1 + R3 --> R3 $$ \begin{bmatrix} 1 & -3 & 0 &| &a \\ 0 & 2 & -1 &| &b \\ 0 & -2 & 1 &| &c-a \\ \end{bmatrix} $$ R2 + R3 --> R3 $$ \begin{bmatrix} 1 & -3 & 0 &| &a \\ 0 & 2 & -1 &| &b \\ 0 & 0 & 0 &| &-a+b+c \\ \end{bmatrix} $$

Coefficient matrix rank: 2 Augmented matrix rank: 3

Answer 1

HINT

It is a correct way to solve.

Now observe that the system has solution if and only if $$-a+b+c=0\implies c=a-b$$

With this condition the system becomes

\begin{bmatrix} 1 & -3 & 0 &| &a \\ 0 & 2 & -1 &| &b \\ \end{bmatrix}

which has (infinite) solutions for every $a,b$ (can you see why?).

Answer 2

The vector equation $A\mathbf x=\mathbf b$ has a solution iff $\mathbf b$ lies in the column space of the matrix $A$. If you compute that column space for your system of linear equations, you will end up with constraints on the constants $a$, $b$ and $c$.