Inverse and the Gauss-Jordan Method

We first give a consequence of Theorem 2.5.8 and then use it to find the inverse of an invertible matrix.

COROLLARY 2.5.13   Let $A$ be an invertible $n \times n$ matrix. Suppose that a sequence of elementary row-operations reduces $A$ to the identity matrix. Then the same sequence of elementary row-operations when applied to the identity matrix yields $A^{-1}.$

Proof. Let $A$ be a square matrix of order $n.$ Also, let $E_1, E_2, \ldots, E_k$ be a sequence of elementary row operations such that $E_1 E_2 \cdots E_k A = I_n.$ Then $E_1 E_2 \cdots E_k I_n = A^{-1}.$ This implies $A^{-1} = E_1 E_2 \cdots E_k.$ height6pt width 6pt depth 0pt

Summary: Let $A$ be an $n \times n$ matrix. Apply the Gauss-Jordan method to the matrix $[A \; \; I_n].$ Suppose the row reduced echelon form of the matrix $[A \;\; I_n]$ is $[B \;\; C].$ If $B = I_n,$ then $A^{-1} = C$ or else $A$ is not invertible.

EXAMPLE 2.5.14   Find the inverse of the matrix $\begin{bmatrix}2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{bmatrix}$ using the Gauss-Jordan method.

Solution: Consider the matrix $\begin{bmatrix} 2 & 1 & 1 & 1 & 0 & 0 \\ 1 & 2 & 1 & 0 & 1 & 0 \\ 1 & 1 & 2 & 0 & 0 & 1 \end{bmatrix}.$ A sequence of steps in the Gauss-Jordan method are:

1. $\begin{bmatrix}2 & 1 & 1 & 1 & 0 & 0 \\ 1 & 2 & 1 & 0 & 1 & 0 \\ 1 & 1 & 2 & 0... ...c{1}{2} & 0 & 0 \\ 1 & 2 & 1 & 0 & 1 & 0 \\ 1 & 1 & 2 & 0 & 0 & 1 \end{bmatrix}$
2. $\begin{bmatrix}1 & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & 0 & 0 \\ 1 & 2 & ... ...} & 1 & 0 \\ 0 & \frac{1}{2} & \frac{3}{2} & -\frac{1}{2} & 0 & 1 \end{bmatrix}$
3. $\begin{bmatrix}1 & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & 0 & 0 \\ 0 & \fra... ...2}{3} & 0 \\ 0 & \frac{1}{2} & \frac{3}{2} & -\frac{1}{2} & 0 & 1 \end{bmatrix}$
4. $\begin{bmatrix}1 & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 1 & ... ...}{3} & 0 \\ 0 & 0 & \frac{4}{3} & -\frac{1}{3} & -\frac{1}{3} & 1 \end{bmatrix}$
5. $\begin{bmatrix}1 & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 1 & ... ...}{3} & 0 \\ 0 & 0 & 1 & -\frac{1}{4} & -\frac{1}{4} & \frac{3}{4} \end{bmatrix}$
6. $\begin{bmatrix}1 & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 1 & ... ...c{-1}{4} \\ 0 & 0 & 1 & \frac{-1}{4} & \frac{-1}{4} & \frac{3}{4} \end{bmatrix}$
7. $\begin{bmatrix}1 & \frac{1}{2} & 0 & \frac{5}{8} & \frac{1}{8} & \frac{-3}{8} ... ...{-1}{4} \\ 0 & 0 & 1 & \frac{-1}{4} & \frac{-1}{4} & \frac{3}{4} \end{bmatrix}.$
8. Thus, the inverse of the given matrix is $\begin{bmatrix} 3/4 & -1/4 & -1/4 \\ -1/4 & 3/4 & -1/4 \\ -1/4 & -1/4 & 3/4 \end{bmatrix}.$

EXERCISE 2.5.15   Find the inverse of the following matrices using the Gauss-Jordan method.
$(i)\; \begin{bmatrix}1 & 2 & 3\\ 1 &3&2 \\ 2 & 4&7 \end{bmatrix},$ $\;\;(ii)\; \begin{bmatrix}1 & 3 & 3\\ 2 &3&2 \\ 2 & 4&7 \end{bmatrix},$ $\;\;(iii)\; \begin{bmatrix}2 & -1 & 3\\ -1 &3&-2 \\ 2 & 4&1 \end{bmatrix}.$