Find values of x for which \(|\begin{smallmatrix}3&x\\
x&1\end{smallmatrix}|=|\begin{smallmatrix}3&2\\
4&1\end{smallmatrix}|\)
Solution We have \(|\begin{smallmatrix}3&x\\
x&1\end{smallmatrix}|=|\begin{smallmatrix}3&2\\
4&1\end{smallmatrix}|\) i.e. \(3-x^{2}=3-8\) i.e. \(x^{2}=8\) Hence \(x=\pm2\sqrt{2}\)