More Problems on Linear Systems and Matrices

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

$$N = 15 (x+y+z)$$

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

$$x + 10y + 100z = 15 (x+y+z) + 396$$

And finally the last condition gives

$$z = x+y + 1$$

Putting all the above equations together we get

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

or

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

Consider the augmented matrix of this system

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

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

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

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

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

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

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

Divide the third row with 781 to get

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

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

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

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

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

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

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

So we have

$$(x,y,z) = (1,3,5)$$

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

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