Matrix Algebra

Computation of Inverse

Computation of Inverse: Case I. Construct the augmented matrix $[A \hspace{0.05in} I_n]$ for the $n\times n$ matrix

of interest with $n\times n$ identity matrix

, then produce a reduced echelon form (REF). The REF is given in the form $[I_n \hspace{0.05in} B]$ with $n\times n$ identity matrix

in the left side. Then $\mathrm{rank}~A = n$ ; see Rank Theorem for the definition of rank. In this case the $n\times n$ matrix

in the right side gives the inverse $A^{-1}$ .

Computation of Inverse: Case II. Otherwise, the size of pivot positions in the leftmost columns must be less than . For example,

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

where

and

(that is, $\mathrm{rank}~A = 3$ ). Then $\mathrm{rank}~A < n$ , and

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.