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Find the Inverse? \[ A=\left[\begin{array}{lll} 2 & 1 & 1 \\ 3 & 2 & 1 \\ 2 & 1 & 2 \end{array}\right] \quad A^{-1}=? \] |
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Answer» We have A = IA \(\begin{bmatrix}2&1&1\\3&2&1\\2&1&2\end{bmatrix}=\) \(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)A Applying R3 → R3 — R1 R2 → 2R2 — 3R1 \(\begin{bmatrix}2&1&1\\0&1&-1\\0&0&1\end{bmatrix}=\)\(\begin{bmatrix}1&0&0\\-3&2&0\\-1&0&1\end{bmatrix}\)A Applying R2 → R2 + R3 \(\begin{bmatrix}2&1&1\\0&1&0\\0&0&1\end{bmatrix}=\)\(\begin{bmatrix}1&0&0\\-4&2&1\\-1&0&1\end{bmatrix}\)A Applying R1 → R1 — R2 — R3 \(\begin{bmatrix}2&0&0\\0&1&0\\0&0&1\end{bmatrix}=\) \(\begin{bmatrix}6&-2&-2\\-4&2&1\\-1&0&1\end{bmatrix}\)A Applying R1 → \(\frac{R_1}2\) \(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\)\(\begin{bmatrix}3&-1&-1\\-4&2&1\\-1&0&1\end{bmatrix}\)A Hence A-1 = \(\begin{bmatrix}3&-1&-1\\-4&2&1\\-1&0&1\end{bmatrix}\) |
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