Answer
$a.\qquad\left[\begin{array}{ll}
0.01 & 0.99\\
0 & 1
\end{array}\right]$
$b.\qquad\left[\begin{array}{ll}
0.0001 & 0.9999\\
0 & 1
\end{array}\right]$
$c.\qquad\approx\left[\begin{array}{ll}
0 & 1\\
0 & 1
\end{array}\right]$
$d.\qquad\approx\left[\begin{array}{ll}
0 & 1\\
0 & 1
\end{array}\right]$
Work Step by Step
$a.$
$P^{2}=\left[\begin{array}{ll}
0.1 & 0.9\\
0 &
\end{array}\right]\left[\begin{array}{ll}
0.1 & 0.9\\
0 &
\end{array}\right]$
$=\left[\begin{array}{ccc}{0.1 \cdot 0.1+0.9 \cdot 0}&{0.1\cdot 0.9+0.9\cdot 1}\\{0\cdot 0.1+1\cdot 0}&{0\cdot 0.9+1\cdot 1}\end{array}\right]$
$=\left[\begin{array}{ll}
0.01 & 0.99\\
0 & 1
\end{array}\right]$
$b.$
$P^{4}=\left[\begin{array}{ll}
0.01 & 0.99\\
0 & 1
\end{array}\right]\left[\begin{array}{ll}
0.01 & 0.99\\
0 & 1
\end{array}\right]$
$=\left[\begin{array}{cc}{0.01 \cdot 0.01+0.99 \cdot 0}&{0.01\cdot 0.99+0.99\cdot 1}\\{0\cdot 0.01+1\cdot 0}&{0\cdot 0.99+1\cdot 1}\end{array}\right]$
$=\left[\begin{array}{ll}
0.0001 & 0.9999\\
0 & 1
\end{array}\right]$
$c.$
$P^{8}=\left[\begin{array}{ll}
0.0001 & 0.9999\\
0 & 1
\end{array}\right]$
$=\left[\begin{array}{cc}{0.0001 \cdot 0.0001+0.9999 \cdot 0}&{0.0001\cdot 0.9999+0.9999\cdot 1}\\{0\cdot 0.0001+1\cdot 0}&{0\cdot 0.9999+1\cdot 1}\end{array}\right]$
$=\left[\begin{array}{cc}{0.00000001}&{0.99999999}\\{0}&{1}\end{array}\right]\approx\left[\begin{array}{ll}
0 & 1\\
0 & 1
\end{array}\right]$
$d.$
With the observed pattern, $P^{1000}\approx\left[\begin{array}{ll}
0 & 1\\
0 & 1
\end{array}\right]$
