Inverse of a Nonsingular Matrix by Elementary Transformation

By the definition of inverse, if $A^{-1}$ exists then $AA^{-1} = A^{-1}A = I$. Consider the equation $AA^{-1}=I$: here $A$ is the given matrix of order $m$, $I$ is the identity matrix of order $m$, and the only unknown is $A^{-1}$. So to find $A^{-1}$, the strategy is to convert $A$ into $I$ using elementary transformations -- and whatever happens to $A$ must happen, in step, to the matrix standing in for $A^{-1}$.

Why the same row transformation can be applied to both sides. Whenever an elementary row transformation is applied to the product $AB=C$ of two matrices, it is enough to apply it only to the prefactor $A$ -- $B$ stays unchanged -- and to apply the identical transformation to $C$ as well; the equation remains true. For example, if $A=\begin{bmatrix}1&2\\3&4\end{bmatrix}$ and $B=\begin{bmatrix}-1&0\\1&5\end{bmatrix}$, then $AB=\begin{bmatrix}1&10\\1&20\end{bmatrix}=C$. Transforming $C$ by $R_1\leftrightarrow R_2$ gives $C\sim\begin{bmatrix}1&20\\1&10\end{bmatrix}$. Applying the same transformation to $A$ alone (leaving $B$ unchanged) gives $A\sim\begin{bmatrix}3&4\\1&2\end{bmatrix}$, and now $\begin{bmatrix}3&4\\1&2\end{bmatrix}\begin{bmatrix}-1&0\\1&5\end{bmatrix}=\begin{bmatrix}1&20\\1&10\end{bmatrix}$ -- exactly the already-transformed $C$, confirming the shortcut is valid.

Hence the equation $AA^{-1}=I$ can be transformed into $IA^{-1}=B$ by applying the same series of row transformations to both sides of the equation. Symbolically:

If, instead, one starts from the equally-valid equation $A^{-1}A=I$, the transformations used must be column transformations, applied to the postfactor $A$ and to the right-hand $I$, while the prefactor $A^{-1}$-slot stays put:

A row-only derivation and a column-only derivation must never be mixed inside a single computation of $A^{-1}$ -- pick one system and stay in it throughout.

Standard pivoting order for a $3\times3$ matrix. For $A = \begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}$, reducing to $I_3=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$ by row transformations typically proceeds: (1) reduce $a_{11}$ to $1$; (2) then reduce $a_{21}$ and $a_{31}$ to $0$; (3) reduce $a_{22}$ to $1$; (4) then reduce $a_{12}$ and $a_{32}$ to $0$; (5) reduce $a_{33}$ to $1$; (6) then reduce $a_{13}$ and $a_{23}$ to $0$. A similar (but not identical) working order is used for column transformations. This is a convenient default order, not a rigid law -- any valid sequence of elementary transformations that reaches the identity is an acceptable derivation.

Worked Example (checking invertibility first). Before computing an inverse it is standard to confirm $|A|\neq0$. For $A=\begin{bmatrix}2&1\\4&2\end{bmatrix}$: $|A|=2(2)-4(1)=0$, so $A$ is singular and not invertible. For $B=\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}$: $|B|=\cos^2\theta+\sin^2\theta=1\neq0$, so $B$ is non-singular and invertible. For $C=\begin{bmatrix}1&3&2\\3&1&2\\1&2&3\end{bmatrix}$: expanding along row 1, $|C|=1(1\cdot3-2\cdot2)-3(3\cdot3-2\cdot1)+2(3\cdot2-1\cdot1)=1(-1)-3(7)+2(5)=-1-21+10=-12\neq0$, so $C$ is non-singular and invertible.

Worked Example -- inverse of a $2\times2$ matrix by row transformations. For $A=\begin{bmatrix}1&2\\3&4\end{bmatrix}$: $|A|=1(4)-3(2)=-2\neq0$, so $A^{-1}$ exists. Starting from $AA^{-1}=I$ (row transformations only): $\begin{bmatrix}1&2\\3&4\end{bmatrix}A^{-1}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$. Using $R_2\to R_2-3R_1$: $\begin{bmatrix}1&2\\0&-2\end{bmatrix}A^{-1}=\begin{bmatrix}1&0\\-3&1\end{bmatrix}$. Using $R_2\to -\tfrac12R_2$: $\begin{bmatrix}1&2\\0&1\end{bmatrix}A^{-1}=\begin{bmatrix}1&0\\\tfrac32&-\tfrac12\end{bmatrix}$. Using $R_1\to R_1-2R_2$: $\begin{bmatrix}1&0\\0&1\end{bmatrix}A^{-1}=\begin{bmatrix}-2&1\\\tfrac32&-\tfrac12\end{bmatrix}$. Hence $A^{-1}=\begin{bmatrix}-2&1\\\tfrac32&-\tfrac12\end{bmatrix}$. …