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(iii) \( \left[\begin{array}{cc}4 & -2 \\ 3 & 1\end{array}\right] \) |
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Answer» \(A=\begin{bmatrix}4&-2\\3&1\end{bmatrix}\) We have A = IA \(\begin{bmatrix}4&-2\\3&1\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}A\) ⇒ \(\begin{bmatrix}4&-2\\0&10\end{bmatrix}=\begin{bmatrix}1&0\\-3&4\end{bmatrix}A\) (By applying R2 → 4R2 - 3R1) ⇒ \(\begin{bmatrix}1&-1/2\\0&1\end{bmatrix}=\begin{bmatrix}1/4&0\\-3/10&2/5\end{bmatrix}A\) (By applying R1 → R1/4 and R2 → R2/10) ⇒ \(\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}1/10&1/5\\-3/10&2/5\end{bmatrix}A\) (By applying R1 → R1 + 1/2 R2) ⇒ \(\begin{bmatrix}1&0\\0&1\end{bmatrix}=\frac1{10}\begin{bmatrix}1&2\\-3&4\end{bmatrix}A\) We get I = A-1A ∴ A-1 \(=\frac1{10}\begin{bmatrix}1&2\\-3&4\end{bmatrix}.\) |
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