\(\frac{1}{5}\) \(\begin{vmatrix} 2 & 1 \\ 3 & 4\end{vmatrix}\)
\(\frac{1}{5}\) \(\begin{vmatrix} 4 & -3 \\ -1 & 2\end{vmatrix}\)
\(\frac{1}{5}\) \(\begin{vmatrix} 2 & -1 \\ -3 & 4\end{vmatrix}\)
\(\frac{1}{5}\) \(\begin{vmatrix} 4 & 3 \\ 1 & 2\end{vmatrix}\)
Correct answer is B
N = [2 3]
N-1 = \(\frac{adj N}{|N|}\)
adj N = \(\begin{vmatrix} 4 & -3 \\ -1 & 2 \end{vmatrix}\)
|N| = (2 x4) - (1 x 3)
= 8 - 3
=5
N-1 = \(\frac {1}{5}\) \(\begin{vmatrix} 4 & -3 \\ -1 & 2 \end{vmatrix}\)
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