Let A = [abcd]\begin{bmatrix}a & b\\c & d\end{bmatrix}[acbd]
i.e [abcd]\begin{bmatrix}a & b\\c & d\end{bmatrix}[acbd] [012−1]\begin{bmatrix}0 & 1\\2 & -1\end{bmatrix}[021−1] = [2−110]\begin{bmatrix}2 & -1\\1 & 0\end{bmatrix}[21−10]
→[a(0)+b(2)a(1)+b(−1)c(0)+d(2)c(1)+d(−1)]\rightarrow\begin{bmatrix}a(0)+b(2) & a(1)+b(-1)\\c(0)+d(2) & c(1)+d(-1)\end{bmatrix}→[a(0)+b(2)c(0)+d(2)a(1)+b(−1)c(1)+d(−1)]
→[2ba−b2dc−d]\rightarrow\begin{bmatrix}2b & a-b\\2d & c-d\end{bmatrix}→[2b2da−bc−d]
By comparing
∴The matrice A = [01frac1212]\begin{bmatrix}0 & 1\\frac{1}{2} & \frac{1}{2}\end{bmatrix}[0frac12121]
