Answer to Question #289136 in Linear Algebra for rakesh kumar

Question

find the inverse of the following matrix A and rank of a 2. 1. 2

1. 3. 0

-1. 1. 2

Answer 1

Augmenting A with a $3×3$ identity matrix

$=\begin{pmatrix} 2 & 1&2| &1&0&0\\ 1&3&0| & 0&1&0\\ -1&1&2|&0&0&1 \end{pmatrix}$

$\frac{1}{2}R_1\to\>R_1$

$-R_3\to\>R_3$

$\begin{pmatrix} 1&\frac{1}{2}&1&|& \frac{1}{2}&0&0 \\ 1&3&0&|& 0&1&0\\ 1&-1&-2&|&0&0&-1 \end{pmatrix}$

$R_2-R_1\to\>R_2$

$R_3-R_1\to\>R_3$

$\begin{pmatrix} 1&\frac{1}{2}&1&|&\frac{1}{2}&0&0 \\ 0&\frac{5}{2}&-1&|&\frac{-1}{2}&1&0 \\ 0&\frac{-3}{2}&-3&|&\frac{-1}{2}&0&-1 \end{pmatrix}$

$\frac{2}{5}R_2\to\>R_2$

$\frac{-2}{3}R_3\to\>R_3$

$\begin{pmatrix} 1&\frac{1}{2}&1&|&\frac{1}{2}&0&0 \\ 0&1&\frac{-2}{5}&|&\frac{-1}{5}&\frac{2}{5}&0 \\ 0&1&2&|&\frac{1}{3}&0&\frac{2}{3} \end{pmatrix}$

$\frac{5}{12}(R_3-R_2)\to\>R_3$

$R_2+\frac{2}{5}R_3\to\>R_2$

$\begin{pmatrix} 1&\frac{1}{2}&1&|&\frac{1}{2}&0&0\\ 0&1&0&|&\frac{-1}{9} & \frac{1}{3}&\frac{1}{9}\\ 0&0&1&|&\frac{2}{9}&\frac{-1}{6}&\frac{5}{18} \end{pmatrix}$

$R_1-R_3\to\>R_3$

$\begin{pmatrix} 1&\frac{1}{2}&0&|&\frac{5}{18}&\frac{1}{6}&\frac{-5}{18} \\ 0&1&0&|&\frac{-1}{9} & \frac{1}{3}&\frac{1}{9}\\ 0&0&1&|&\frac{2}{9}&\frac{-1}{6}&\frac{5}{18} \end{pmatrix}$

$R_1-\frac{1}{2}R_2\to\>R_1$

$=\begin{pmatrix} 1&0&0| & \frac{1}{3}&0&\frac{-1}{3}\\ 0&1&0| & \frac{-1}{9}&\frac{1}{3}&\frac{1}{9}\\ 0&0&1|&\frac{2}{9}&\frac{-1}{6}&\frac{5}{18} \end{pmatrix}$

Inverse is matrix on the right of augmented matrix

$\therefore\>A^{-1}=\begin{pmatrix} \frac{1}{3}& 0&\frac{-1}{3} \\ \frac{-1}{9}&\frac{1}{3} & \frac{1}{9}\\ \frac{2}{9}&\frac{-1}{6}&\frac{5}{18} \end{pmatrix}$

Rank is number of non-zeros row on the left of the augmented matrix

$\therefore$ Rank $=3$