More Problems on Linear Systems and Matrices

The parabola will have the form $y = ax^2 + bx +c$. The three points will be on the parabola if and only if

\begin{displaymath}\left\{\begin{array}{rcl} a-b+c &=& 6 \\ a+b+c &=& 4 \\ 4a+2b+c &=& 9 \\ \end{array} \right.\end{displaymath}

The existence of such a parabola translates into the consistency of the above system. Consider the augmented matrix

\begin{displaymath}\left(\begin{array}{rrr\vert c} 1 & -1 & 1 & 6 \\ 1 & 1 & 1 & 4 \\ 4 & 2 & 1 & 9 \end{array} \right) \;.\end{displaymath}

Take $\mbox{\bf Row}_2 - \; \mbox{\bf Row}_1$ and $\mbox{\bf Row}_3 - 4 \; \mbox{\bf Row}_1$ to get

\begin{displaymath}\left(\begin{array}{rrr\vert c} 1 & -1 & 1 & 6 \\ 0 & 2 & 0 & -2 \\ 0 & 6 & -3 & -15 \end{array} \right) \;.\end{displaymath}

Take $\displaystyle \frac{1}{2} \mbox{\bf Row}_2$ and $\displaystyle \frac{1}{3} \mbox{\bf Row}_3$ to get

\begin{displaymath}\left(\begin{array}{rrr\vert c} 1 & -1 & 1 & 6 \\ 0 & 1 & 0 & -1 \\ 0 & 2 & -1 & -5 \end{array} \right) \;.\end{displaymath}

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

\begin{displaymath}\left(\begin{array}{rrr\vert c} 1 & -1 & 1 & 6 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & -1 & -3 \end{array} \right) \;.\end{displaymath}

From here, we can conclude that the parabola does exist. To find it, let us continue with the row elementary operations. Take $-\mbox{\bf Row}_3$ and $\mbox{\bf Row}_2 + \; \mbox{\bf Row}_1$ to get

\begin{displaymath}\left(\begin{array}{rrr\vert c} 1 & 0 & 1 & 5 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 3 \end{array} \right) \;.\end{displaymath}

Finally take $\mbox{\bf Row}_1 - \; \mbox{\bf Row}_3$


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

Hence

\begin{displaymath}(a,b,c) = (2,-1,3)\end{displaymath}

or $y = 2x^2-x+3$.

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