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

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

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

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

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

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

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

$$c-4a -(b-2a) = c - b -2a = 0$$

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

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