More Problems on Linear Systems and Matrices

Write $N=xyz$. By definition we have $N =
z + 10y + 100x$. The first condition translates into

\begin{displaymath}N = 15 (x+y+z)\end{displaymath}

Since the new number with reversed digits is $zyx = x + 10y +
100z$, the second condition translates into

\begin{displaymath}x + 10y + 100z = 15 (x+y+z) + 396\end{displaymath}

And finally the last condition gives

\begin{displaymath}z = x+y + 1\end{displaymath}

Putting all the above equations together we get

\begin{displaymath}\left\{\begin{array}{lcl}
z + 10y + 100x &=&15 (x+y+z) \\
...
... &=&15 (x+y+z) + 396 \\
z &=& x+y + 1 \\
\end{array} \right.\end{displaymath}

or

\begin{displaymath}\left\{\begin{array}{lcl}
85x - 5y -14z &=&0 \\
-14 x -5y + 85z &=&396 \\
-x-y +z &=& 1 \\
\end{array} \right.\end{displaymath}

Consider the augmented matrix of this system

\begin{displaymath}\left(\begin{array}{rrr\vert c}
85 & -5 & -14 & 0 \\
-14 & -5 & 85 & 396 \\
-1 & -1 & 1 &1
\end{array} \right) \;.\end{displaymath}

Let us do some elementary row operations to solve it. First we interchange the third row with the first one and multiply the new first row with $-1$ to get

\begin{displaymath}\left(\begin{array}{rrr\vert c}
1 & 1 & -1 & -1\\
-14 & -5 & 85 & 396 \\
85 & -5 & -14 & 0
\end{array} \right) \;.\end{displaymath}

Next take the $\mbox{\bf Row}_2 + 14 \; \mbox{\bf Row}_1$ and $\mbox{\bf Row}_3 - 85 \; \mbox{\bf Row}_1$ to get

\begin{displaymath}\left(\begin{array}{rrr\vert c}
1 & 1 & -1 & -1\\
0 & 9 & 71 & 382 \\
0 & -90 & 71 & 85
\end{array} \right) \;.\end{displaymath}

Next take $\mbox{\bf Row}_3 + 10 \; \mbox{\bf Row}_2$ to get

\begin{displaymath}\left(\begin{array}{rrr\vert c}
1 & 1 & -1 & -1\\
0 & 9 & 71 & 382 \\
0 & 0 & 781 & 3905
\end{array} \right) \;.\end{displaymath}

Divide the third row with 781 to get

\begin{displaymath}\left(\begin{array}{rrr\vert c}
1 & 1 & -1 & -1\\
0 & 9 & 71 & 382 \\
0 & 0 & 1 & 5
\end{array} \right) \;.\end{displaymath}

Next take $\mbox{\bf Row}_2 - 71 \; \mbox{\bf Row}_3$ and $\mbox{\bf Row}_1 + \; \mbox{\bf Row}_3$ to get

\begin{displaymath}\left(\begin{array}{rrr\vert c}
1 & 1 & 0 & 4\\
0 & 9 & 0 & 27 \\
0 & 0 & 1 & 5
\end{array} \right) \;.\end{displaymath}

Divide the second row by 9 to get $\mbox{\bf Row}_1 + \; \mbox{\bf Row}_3$ to get

\begin{displaymath}\left(\begin{array}{rrr\vert c}
1 & 1 & 0 & 4\\
0 & 1 & 0 & 3 \\
0 & 0 & 1 & 5
\end{array} \right) \;.\end{displaymath}

Finally take $\mbox{\bf Row}_1 - \; \mbox{\bf Row}_2$ to get

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

So we have

\begin{displaymath}(x,y,z) = (1,3,5)\end{displaymath}

or $N = 135$. Check that the original conditions are satisfied.


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