A 2-digit number has two properties: The digits sum to 11, and if the number is written with digits reversed, and subtracted from the original number, the result is 45.

Find the number.

Solution.

The key to this problem is noticing that our 2-digit number can be written as $10A + B$, where $A, B$ are the 10s and 1s digit respectively.

The digits summing to 11 then yields the equality $A + B = 11$.

The number with digits reversed is $10B + A$, and so the second property yields the equation $10A + B – (10B + A) = 45$. Simplifying, we have the system of equations
\begin{align*}
A+B &= 11\\
9A-9B &= 45
\end{align*}

To solve this system, we create the augmented matrix and use elementary row operations to put it into reduced row-echelon form:
\begin{align*}
\left[\begin{array}{rr|r} 1 & 1 & 11 \\ 9 & -9 & 45 \end{array} \right] \xrightarrow{ R_2 – 9 R_1 } \left[\begin{array}{rr|r} 1 & 1 & 11 \\ 0 & -18 & -54 \end{array} \right]\\[6pt] \xrightarrow{ \frac{-1}{18} R_2 } \left[\begin{array}{rr|r} 1 & 1 & 11 \\ 0 & 1 & 3 \end{array} \right] \xrightarrow{R_1 – R_2} \left[\begin{array}{rr|r} 1 & 0 & 8 \\ 0 & 1 & 3 \end{array} \right].
\end{align*}
With this final augmented matrix, we can read off the solution $A = 8$ and $B = 3$.

Thus our number is $10A + B = 83$.

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