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Find the Inverse?\[A=\left[\begin{array}{lll}2 & 1 & 1 \\3 & 2 & 1 \\2 & 1 & 2\end{array}\right] \quad A^{-1}=?\]

Monjur Alahi

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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Salim Ahmed Kawsar

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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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this is the solution!!!! 

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