Mathematics - Continuity - Differentiability Question with Solution | TestHub

$\begin{array}{rl}& \left|\begin{array}{ccc}1& 2& 3\\ -4& 3& 6\\ 2& -7& 9\end{array}\right|=1\left|\begin{array}{cc}3& 6\\ -7& 9\end{array}\right|-2\left|\begin{array}{cc}-4& 6\\ 2& 9\end{array}\right|+3\\ & \left|\begin{array}{cc}-4& 3\\ 2& -7\end{array}\right|\\ & \end{array}$ $\begin{array}{rl}& =1(3\times 9-6(-7))-2(-4\times 9-2\times 6)+3\\ & ((-4)(-7)-3\times 2)\\ & =(27+42)-2(-36-12)+3(28-6)=231\end{array}$