For real numbers $a,\ b,\ c$, on the coordinate plane with the origin $O$, let $T$ be a linear transformation expressed by a matrix $\left(
\begin{array}{cc}
a & 1 \\
b & c
\end{array}
\right)$.

Suppose that the transformation $T$ satisfies the following two conditions:

(i) $(1,\ 2)$ is mapped to $(1,\ 2)$

(ii) Denote $A,\ B$ the images of $(1,\ 0),\ (0,\ 1)$ by $T$ respectively, the area of $\triangle{OAB}$ is $\frac 12$.

Find the values of $a,\ b,\ c$.

[i]2011 Kyoto University entrance exam/Science, Problem 2[/i]
1. Given the linear transformation matrix \( T \) as:
   \[
   T = \begin{pmatrix} a & 1 \\ b & c \end{pmatrix}
   \]
   and the conditions:
   - (i) \((1, 2)\) is mapped to \((1, 2)\)
   - (ii) The area of \(\triangle OAB\) is \(\frac{1}{2}\), where \(A\) and \(B\) are the images of \((1, 0)\) and \((0, 1)\) respectively.

2. From condition (i), we have:
   \[
   T \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}
   \]
   This gives us the equation:
   \[
   \begin{pmatrix} a & 1 \\ b & c \end{pmatrix} \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} a + 2 \\ b + 2c \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}
   \]
   Therefore, we have:
   \[
   a + 2 = 1 \quad \text{and} \quad b + 2c = 2
   \]
   Solving these equations, we get:
   \[
   a = 1 - 2 = -1
   \]
   \[
   b + 2c = 2
   \]

3. From condition (ii), we need to find the area of \(\triangle OAB\). The images of \((1, 0)\) and \((0, 1)\) under \(T\) are:
   \[
   T \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} -1 \\ b \end{pmatrix}
   \]
   \[
   T \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ c \end{pmatrix}
   \]
   The area of \(\triangle OAB\) is given by:
   \[
   \text{Area} = \frac{1}{2} \left| \det \begin{pmatrix} -1 & 1 \\ b & c \end{pmatrix} \right| = \frac{1}{2} \left| (-1)c - (1)b \right| = \frac{1}{2} \left| -c - b \right|
   \]
   Given that the area is \(\frac{1}{2}\), we have:
   \[
   \frac{1}{2} \left| -c - b \right| = \frac{1}{2}
   \]
   Simplifying, we get:
   \[
   \left| -c - b \right| = 1
   \]
   This gives us two cases:
   \[
   -c - b = 1 \quad \text{or} \quad -c - b = -1
   \]

4. Solving the equations:
   - For \( -c - b = 1 \):
     \[
     b = -c - 1
     \]
     Substituting into \( b + 2c = 2 \):
     \[
     -c - 1 + 2c = 2
     \]
     \[
     c - 1 = 2
     \]
     \[
     c = 3
     \]
     \[
     b = -3 - 1 = -4
     \]
     Thus, one solution set is \( (a, b, c) = (-1, -4, 3) \).

   - For \( -c - b = -1 \):
     \[
     b = -c + 1
     \]
     Substituting into \( b + 2c = 2 \):
     \[
     -c + 1 + 2c = 2
     \]
     \[
     c + 1 = 2
     \]
     \[
     c = 1
     \]
     \[
     b = -1 + 1 = 0
     \]
     Thus, another solution set is \( (a, b, c) = (-1, 0, 1) \).

The final answer is \(\boxed{(-1, 0, 1), (-1, -4, 3)}\).