Q.Using elementary transformations, find the inverse of the matrix: $A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix}$

elementary row operations to reduce [A|I] to [I|A⁻¹]

$A=\begin{bmatrix}0&1&2\\1&2&3\\3&1&1\end{bmatrix}$. Write $A=IA$:

$\begin{bmatrix}0&1&2\\1&2&3\\3&1&1\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}A$

$R_1\leftrightarrow R_2$:

$\begin{bmatrix}1&2&3\\0&1&2\\3&1&1\end{bmatrix}=\begin{bmatrix}0&1&0\\1&0&0\\0&0&1\end{bmatrix}A$

$R_3\to R_3-3R_1$:

$\begin{bmatrix}1&2&3\\0&1&2\\0&-5&-8\end{bmatrix}=\begin{bmatrix}0&1&0\\1&0&0\\0&-3&1\end{bmatrix}A$

$R_1\to R_1-2R_2$, $R_3\to R_3+5R_2$:

$\begin{bmatrix}1&0&-1\\0&1&2\\0&0&2\end{bmatrix}=\begin{bmatrix}-2&1&0\\1&0&0\\5&-3&1\end{bmatrix}A$

$R_3\to \dfrac12R_3$:

$\begin{bmatrix}1&0&-1\\0&1&2\\0&0&1\end{bmatrix}=\begin{bmatrix}-2&1&0\\1&0&0\\\frac52&-\frac32&\frac12\end{bmatrix}A$

$R_1\to R_1+R_3$, $R_2\to R_2-2R_3$:

$\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}-2+\frac52&1-\frac32&0+\frac12\\1-5&0+3&0-1\\\frac52&-\frac32&\frac12\end{bmatrix}A=\begin{bmatrix}\frac12&-\frac12&\frac12\\-4&3&-1\\\frac52&-\frac32&\frac12\end{bmatrix}A$

Hmm — let me re-verify this elimination arithmetic directly by instead computing $A^{-1}$ via the adjugate method as a check, since the row-reduction has many steps where an arithmetic slip is easy.

Cross-check via cofactors. $|A|=0(2\cdot1-3\cdot1)-1(1\cdot1-3\cdot3)+2(1\cdot1-2\cdot3)=0-1(1-9)+2(1-6)=0+8-10=-2$.

Cofactors: $C_{11}=2(1)-3(1)=-1$; $C_{12}=-(1(1)-3(3))=-(1-9)=8$; $C_{13}=1(1)-2(3)=1-6=-5$;

$C_{21}=-(1(1)-2(1))=-(1-2)=1$; $C_{22}=0(1)-2(3)=-6$; $C_{23}=-(0(1)-1(3))=3$;

$C_{31}=1(3)-2(2)=3-4=-1$; $C_{32}=-(0(3)-2(1))=2$; $C_{33}=0(2)-1(1)=-1$.

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