Question
Evaluate $\begin{bmatrix}\sqrt{3}&\sqrt{2}\\-1&2\sqrt{3}\end{bmatrix}$ is:
  1. $6-3\sqrt{2}$
  2. $6-\sqrt{2}$
  3. $6+3\sqrt{2}$
  4. $6+\sqrt{2}$

Answer

  1. $6+\sqrt{2}$

Solution:

$\triangle=\begin{bmatrix}\sqrt{3}&\sqrt{2}\\-1&2\sqrt{3}\end{bmatrix}$

$\triangle=(\sqrt{3}\times2\sqrt{3})+\sqrt{2}$

$\triangle=6+\sqrt{2}$

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