प्रारंभिक रूपांतरणों का उपयोग करके,निम्नलिखित | vedclass

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(A) माना $A = \left[\begin{array}{cc}1 & 2 \ 2 & -1\end{array}ight]$.हम $A = IA$ लिखते हैं,जहाँ $I = \left[\begin{array}{cc}1 & 0 \ 0 & 1\end{arra

Vedclass · Accepted Answer

$\left[\begin{array}{cc}\frac{1}{5} & \frac{2}{5} \ \frac{2}{5} & -\frac{1}{5}\end{array}ight]$

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(A) माना $A = \left[\begin{array}{cc}1 & 2 \ 2 & -1\end{array}ight]$.हम $A = IA$ लिखते हैं,जहाँ $I = \left[\begin{array}{cc}1 & 0 \ 0 & 1\end{array}ight]$.$\left[\begin{array}{cc}1 & 2 \ 2 & -1\end{array}ight] = \left[\begin{array}{cc}1 & 0 \ 0 & 1\end{array}ight] A$.$R_2 	o R_2 - 2R_1$ लागू करने पर:$\left[\begin{array}{cc}1 & 2 \ 0 & -5\end{array}ight] = \left[\begin{array}{cc}1 & 0 \ -2 & 1\end{array}ight] A$.$R_2 	o -\frac{1}{5} R_2$ लागू करने पर:$\left[\begin{array}{cc}1 & 2 \ 0 & 1\end{array}ight] = \left[\begin{array}{cc}1 & 0 \ \frac{2}{5} & -\frac{1}{5}\end{array}ight] A$.$R_1 	o R_1 - 2R_2$ लागू करने पर:$\left[\begin{array}{cc}1 & 0 \ 0 & 1\end{array}ight] = \left[\begin{array}{cc}1 - 2(\frac{2}{5}) & 0 - 2(-\frac{1}{5}) \ \frac{2}{5} & -\frac{1}{5}\end{array}ight] A$.$\left[\begin{array}{cc}1 & 0 \ 0 & 1\end{array}ight] = \left[\begin{array}{cc}\frac{1}{5} & \frac{2}{5} \ \frac{2}{5} & -\frac{1}{5}\end{array}ight] A$.अतः,$A^{-1} = \left[\begin{array}{cc}\frac{1}{5} & \frac{2}{5} \ \frac{2}{5} & -\frac{1}{5}\end{array}ight]$.

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