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(iii) \( \left[\begin{array}{cc}4 & -2 \\ 3 & 1\end{array}\right] \)

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\(A=\begin{bmatrix}4&-2\\3&1\end{bmatrix}\)

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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