Question

Calculați determinanți de ordinul 2 ​

Answer 1

$\displaystyle a)~\left|\begin{array}{ccc}2&5\\1&3\end{array}\right|=2 \cdot 3-5 \cdot 1=6-5=1\\\\ b)~\left|\begin{array}{ccc}5&-3\\2&-1\end{array}\right|=5 \cdot(-1)-(-3) \cdot 2=-5+6=1\\\\ c)~\left|\begin{array}{ccc}1&2\\3&6\end{array}\right|=1 \cdot 6-2 \cdot 3=6-6=0$

$\displaystyle d)~\left|\begin{array}{ccc}\sqrt{2} &1\\4&\sqrt{8} \end{array}\right|=\sqrt{2} \cdot \sqrt{8} -1 \cdot 4=\sqrt{16} -4=4-4=0\\\\ e)~\left|\begin{array}{ccc}\cfrac{1}{2} &-1\\\cfrac{1}{4} &\cfrac{1}{3} \end{array}\right|=\frac{1}{2} \cdot \frac{1}{3} -(-1) \cdot \frac{1}{4} =\frac{1}{6} +\frac{1}{4} =\frac{2+3}{12} =\frac{5}{12}$

$\displaystyle f)~\left|\begin{array}{ccc}i&2i\\-i&i\end{array}\right|=i \cdot i -2i \cdot (-i)=i^2+2i^2=-1+2 \cdot (-1)=-1-2=-3$