Answer
(a) $-2$
(b) $-2$
(c) Yes.
Work Step by Step
(a) Expand det(B) by the second row, we have:
$\begin{array}( \\|B|= \\ \\ \end{array}
\begin{vmatrix} 4 &1 &0\\-2 &-1 &1\\4 &0 &3 \end{vmatrix} \begin{array}( \\=-(-2) \\ \\\end{array}
\begin{vmatrix}1 &0\\0 &3 \end{vmatrix} \begin{array}( \\+(-1) \\ \\\end{array}
\begin{vmatrix} 4 &0\\4&3 \end{vmatrix} \begin{array}( \\-(+1) \\ \\\end{array}
\begin{vmatrix} 4 &1\\4 &0 \end{vmatrix}$
$|B|=2(3-0)-(12-0)-(0-4)=-2$
(b) Expand det(B) by the third column, we have:
$\begin{array}( \\|B|=(0)-(1) \\ \\ \end{array}
\begin{vmatrix} 4 &1\\4&0 \end{vmatrix} \begin{array}( \\+(3)\\ \\\end{array}
\begin{vmatrix} 4 &1\\-2 &-1 \end{vmatrix}$
$|B|=-(0-4)+3(-4+2)=-2$
(c) Clearly the results in parts (a) and (b) agree with each other.
