Finding $ A^{-1}$

Construct the augmented matrix $ [A \hspace{0.05in} I_n]$ for the $ n\times n$ matrix $ A$ of interest with $ n\times n$ identity matrix $ I_n$, then produce a reduced echelon form (REF).

CASE I: The REF is given in the form  $ [I_n \hspace{0.05in} B]$ with $ n\times n$ identity matrix $ I_n$ in the left side. Then $ rank A = n$, and the $ n\times n$ matrix $ B$ in the right side gives the inverse $ A^{-1}$.

CASE II: Otherwise, the size $ k$ of pivot positions in the $ n$ leftmost columns must be less than $ n$. For example,

$\displaystyle \begin{bmatrix}
1 & 0 & \alpha_1 & 0 & b_{11} & b_{12} & b_{13} &...
...3} & b_{34} \\
0 & 0 & 0 & 0 & b_{41} & b_{42} & b_{43} & b_{44}
\end{bmatrix}$

where $ k = 3$ and $ n = 4$ (that is, rank$ A = 3$). Then rank$ A < n$, and $ A$ is not invertible. Thus, $ A^{-1}$ does not exist.

EXAMPLE 4. Find the inverse of the matrix $ A =
\begin{bmatrix}
0 & 1 & 2 \\
1 & 0 & 3 \\
4 &-3 & 8
\end{bmatrix}$, if it exists.

EXAMPLE 5 Find the inverse of the matrix $ A =
\begin{bmatrix}
1 &-2 &-1 \\
-1 & 5 & 6 \\
5 &-4 & 5
\end{bmatrix}$, if it exists.

EXAMPLE 6 Decide whether $ A =
\begin{bmatrix}
1 & 0 &-2 \\
3 & 1 &-2 \\
-5 &-1 & 9
\end{bmatrix}$ is invertible or not.



Department of Mathematics
Last modified: 2005-10-21