More Problems on Linear Systems and Matrices

Let us answer (1). Consider the augmented matrix of the system

$$\left(\begin{array}{lcr\vert c} 1 & 3 & -1 & a \\ 1 & 2 & 0 &b \\ 3 & 7 & -1 & c \end{array} \right) \;.$$

Let us do some row elementary operations. First take second Row minus first row and third row minus 3 times first row, that is

$$\mbox{\bf Row}_2 - \mbox{\bf Row}_1 \;\;\mbox{and}\;\; \mbox{\bf Row}_3 - 3\;\mbox{\bf Row}_1$$

to get

$$\left(\begin{array}{rrr\vert c} 1 & 3 & -1 & a \\ 0 & -1 & 1 &b-a \\ 0 & -2 & 2 & c-3a \end{array} \right) \;.$$

Next take $\mbox{\bf Row}_3 - 2\; \mbox{\bf Row}_2$ to get

$$\left(\begin{array}{rrr\vert c} 1 & 3 & -1 & a \\ 0 & -1 & 1 &b-a \\ 0 & 0 & 0 & c-3a -2(b-a) \end{array} \right) \;.$$

So the original system is consistent if and only if

$$c-3a -2(b-a) = c - a -2b = 0\;.$$

Next we answer (2). If $(a,b,c) = (1,1,3)$ then we have

$$c - a -2b = 3-1-2 = 0$$

so the condition is satisfied. Therefore the system is consistent.
If $(a,b,c) = (1,0,-1)$, then we have

$$c - a -2b = -1-1-0 = -2 \neq 0$$

so the system in this case is not consistent.

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