Science:Math Exam Resources/Courses/MATH152/April 2015/Question A 20/Solution 1

We compute

$\begin{array}{rlrl}\left|\begin{array}{ccccc}1& 3& 2& 0& 1\\ 2& 7& 1& 0& 2\\ 1& 5& 6& -1& 7\\ 0& 0& 3& 0& 0\\ 1& 1& 4& 0& 4\\ \end{array}\right|& =-3\cdot \left|\begin{array}{cccc}1& 3& 0& 1\\ 2& 7& 0& 2\\ 1& 5& -1& 7\\ 1& 1& 0& 4\\ \end{array}\right|& & \text{expanding row 4}\\ & =-3\cdot -1\cdot \left|\begin{array}{ccc}1& 3& 1\\ 2& 7& 2\\ 1& 1& 4\end{array}\right|& & \text{expanding column 3}\\ & =-3\cdot -1\cdot (\left|\begin{array}{cc}7& 2\\ 1& 4\end{array}\right|-3\cdot \left|\begin{array}{cc}2& 2\\ 1& 4\end{array}\right|+\left|\begin{array}{cc}2& 7\\ 1& 1\end{array}\right|)& & \text{expanding row 1}\\ & =-3\cdot -1\cdot ((7\cdot 4-2\cdot 1)-3(2\cdot 4-2\cdot 1)+(2\cdot 1-7\cdot 1))\\ & =9.\end{array}$