More Problems on Linear Systems and Matrices

Recall that the vector $B =
\left(\begin{array}{rrrr}
a \\
b \\
c \\
\end{array} \right)$ is in the range of $A$ if and only if there exists a vector $X = \left(\begin{array}{rrrr}
x \\
y \\
z \\
t \\
\end{array} \right)$ such that $AX=B$. This clearly implies that $B$ is in the range of $A$ if and only if the linear system $AX=B$ is consistent. To discuss this let us consider the augmented matrix

\begin{displaymath}\left(\begin{array}{rrrr\vert c}
1 & 1 & 3 & -2 & a \\
2 & 1 & -1 & 2 & b \\
4 & 3 & 5 & -2& c
\end{array} \right) \;.\end{displaymath}

Let us do some row elementary operations. First take $\mbox{\bf Row}_2 - 2\; \mbox{\bf Row}_1$ and $\mbox{\bf Row}_3 - 4\;
\mbox{\bf Row}_1$ we will get

\begin{displaymath}\left(\begin{array}{rrrr\vert c}
1 & 1 & 3 & -2 & a \\
0 & -...
...7 & 6 & b-2a \\
0 & -1 & -7 & 6 & c-4a
\end{array} \right) \;.\end{displaymath}

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

\begin{displaymath}\left(\begin{array}{rrrr\vert c}
1 & 1 & 3 & -2 & a \\
0 & -...
...& b-2a \\
0 & 0 & 0 & 0 & c-4a -(b-2a)
\end{array} \right) \;.\end{displaymath}

So the system $AX=B$ is consistent if and only if

\begin{displaymath}c-4a -(b-2a) = c - b -2a = 0\end{displaymath}

So the vector $B$ is in the range of $A$ if and only if $c - b -2a
= 0$.


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