Answer
See below
Work Step by Step
Let $\begin{array}{c}{A=\left[\begin{array}{ccc}
1 & 0 \\4 & 1 \end{array}\right] }\end{array},\begin{array}{c}{S=\left[\begin{array}{ccc}
1 & 2 \\0 & 1 \end{array}\right] }\end{array}$ and $B$ be $n × n$ matrices
Obtain $S^{-1}AS=\begin{bmatrix}
1 & -2 \\ 0 & 1
\end{bmatrix}\begin{bmatrix}
1& 0 \\ 4 & 1
\end{bmatrix}\begin{bmatrix}
1 & 2 \\ 0& 1
\end{bmatrix}$ then we have $\det(A_1)=\begin{vmatrix}
0 & 1 \\ 1 & 0
\end{vmatrix}=\begin{array}{c}{A=\left[\begin{array}{ccc}
-7 & =16 \\4 & 9 \end{array}\right] }\end{array}$
Let $\left[\begin{array}{ccc}
-7 & =16 \\4 & 9 \end{array}\right] =B$
Hence, $\det(A)=1-0=1\\
\det(B)=1$
According to propert P9, $\det(S^{-1}AS)\\=\det(B)\\=\det (S^{-1}).\det(A).\det(S)\\=\det (S^{-1}).\det(S).\det(A)\\=\det(I_n).\det(A)\\=\det(A)$
