More Problems on Linear Systems and Matrices

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

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

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

\begin{displaymath}\mbox{\bf Row}_2 - \mbox{\bf Row}_1 \;\;\mbox{and}\;\;
\mbox{\bf Row}_3 - 3\;\mbox{\bf Row}_1\end{displaymath}

to get

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

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

\begin{displaymath}\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) \;.\end{displaymath}

So the original system is consistent if and only if

\begin{displaymath}c-3a -2(b-a) = c - a -2b = 0\;.\end{displaymath}

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

\begin{displaymath}c - a -2b = 3-1-2 = 0\end{displaymath}

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

\begin{displaymath}c - a -2b = -1-1-0 = -2 \neq 0\end{displaymath}

so the system in this case is not consistent.


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