To find the inverse of the matrix $A = [121; 011; 311]$ by elementary transformation method.

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Solution

$\begin{matrix} A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 1 0 & 1 & 1 3 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦ A = A I ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 1 0 & 1 & 1 3 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦ = A ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ R_{3} \to R_{3} - 3 R_{1} ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 1 0 & 1 & 1 0 & - 5 & - 2 \end{matrix} ⎤ ⎥ ⎦ = A ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 - 3 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ R_{3} \to R_{3} - 5 R_{2} R_{1} \to R_{1} - 2 R_{2} ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & - 2 0 & 1 & 1 0 & 0 & - 7 \end{matrix} ⎤ ⎥ ⎦ = A ⎡ ⎢ ⎣ \begin{matrix} 1 & - 2 & 0 0 & 1 & 0 - 8 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ R_{3} \to \frac{- 1}{7} R_{3} ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & - 2 0 & 1 & 1 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ = A ⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 2 & 0 0 & 1 & 0 \frac{8}{7} & 0 & \frac{- 1}{7} \end{matrix} ⎤ ⎥ ⎥ ⎦ R_{2} \to R_{2} - R_{3} R_{1} \to R_{1} + 2 R_{3} ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ = A ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 + \frac{16}{7} & - 2 & \frac{- 2}{7} \frac{8}{7} & 1 & \frac{- 1}{7} \frac{8}{7} & 0 & \frac{- 1}{7} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ A^{- 1} = ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{25}{7} & - 2 & \frac{- 2}{7} \frac{8}{7} & 1 & \frac{- 1}{7} \frac{8}{7} & 0 & \frac{- 1}{7} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ \end{matrix}$

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