Answer
See below
Work Step by Step
a) We can deduce the following from the formula 3.1.3:
$\det A=\sum \sigma(p_1,p_2,...,p_n)a_{1p_1}a_{2p_2}...a_{np_n}$
When $a_{ik}=0$ for all $i \lt k$, $A$ is lower triangular. So, in the above equation, the only nonzero terms are those with $p_i \leq i \forall i$. Since all of the $p_i$ must be distinct, the only possibility is $p_n \leq n,p_{n-1}\leq n-1,...p_1\leq 1$, which means that $p_i=1; i=1,2,...,n$ is the only option.
Consequently,
$\det(A)=\sigma(p_1,p_2,...p_n)a_{11}a_{22}...a_{nn}\\=a_{11}a_{22}...a_{nn}\Pi^n_{i=1}a_{ii}$
b) From the given matrix we get:
$1.A_{42}(-1),A_{43}=-4,A_{41}=-5\\
2.A_{31}(-3),A_{32}(-2)\\
3. A_{21}(-\frac{13}{16})$
Hence, $B=\begin{pmatrix}
-\frac{13}{8} & 0 & 0 & 0\\
2 & 16 & 0 & 0\\
-1 & -8 & 1 & 0\\
1 & 2 & 0 & 1
\end{pmatrix}$
Since $\det(A)=26\\
\det(B)=-\frac{13}{8}16=26\\
\rightarrow \det(A)=\det(B)$
