![A=\left[\begin{array}{cc}1&2\\4&5\end{array}\right] \\\\|A|=(5)(1)-(2)(4)\\\\|A|=-3---(1)](/tpl/images/0454/8659/39a5e.png)
![A=\left[\begin{array}{cc}1&2\\4&5\end{array}\right] \\\\|A|=-3\\\\Adj(A)=\left[\begin{array}{cc}5&-2\\-4&1\end{array}\right] \\\\\\A^{-1}=\frac{Adj(A)}{|A|}\\\\A^{-1}=\left[\begin{array}{cc}\frac{5}{-3}&\frac{-2}{-3}\\\frac{-4}{-3}&\frac{1}{-3}\end{array}\right] \\\\\\|A^{-1}|=(\frac{5}{-3})(\frac{1}{-3})-(\frac{-4}{-3})(\frac{-2}{-3})\\\\|A^{-1}|=\frac{5}{9}-\frac{8}{9}\\\\|A^{-1}|=\frac{-1}{3}](/tpl/images/0454/8659/9b101.png)
If a is nonsingular, then det a−1= 1/(det a). verify this
theorem for 2 3 2 matrices....
Mathematics, 14.01.2020 21:31 zozo9987
If a is nonsingular, then det a−1= 1/(det a). verify this
theorem for 2 3 2 matrices.
Answers: 3
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