Answer to Question #110429 in Quantitative Methods for Gayatri Yadav

Question

Find the inverse of the following matrix, using Gauss Jordan method.
1 2 3 4
2 3 4 1
3 4 1 2
4 1 2 3

Answer 1

$A=\begin{bmatrix} 1 & 2&3&4 \\ 2&3&4&1\\ 3&4&1&2\\ 4&1&2&3 \end{bmatrix}$

$[A|E]=\begin{bmatrix} 1 & 2&3&4 |1&0&0&0\\ 2&3&4&1|0&1&0&0\\ 3&4&1&2|0&0&1&0\\ 4&1&2&3|0&0&0&1 \end{bmatrix}$

IIr+Ir(-2)

IIIr+Ir(-3)

IIIIr+Ir(-4)

$\begin{bmatrix} 1 & 2&3&4 |1&0&0&0\\ 0&-1&-2&-7|-2&1&0&0\\ 0&-2&-8&-10|-3&0&1&0\\ 0&-7&-10&-13|-4&0&0&1 \end{bmatrix}$

IIr(-1)

IIIr+IIr(-2)

IIIIr+IIr(-7)

$\begin{bmatrix} 1 & 2&3&4 |1&0&0&0\\ 0&1&2&7|2&-1&0&0\\ 0&0&-4&4|1&-2&1&0\\ 0&0&4&36|10&-7&0&1 \end{bmatrix}$

IIIr$\cdot(-\frac{1}{4})$

IIIIr+IIIr

$\begin{bmatrix} 1 & 2&3&4 |1&0&0&0\\ 0&1&2&7|2&-1&0&0\\ 0&0&1&-1|-0.25&0.5&-0.25&0\\ 0&0&0&40|11&-9&1&1 \end{bmatrix}$

IIIIr$\cdot\frac{1}{40}$

IIIr+IIIIr

IIr+IIIr(-2)+IIIIr(-7)

Ir+IIr(-2)+IIIr(-3)+IIIIr(-4)

$\begin{bmatrix} 1 & 0&0&0 |\\&&&&-0.225&0.025&0.025&0.275\\ 0&1&0&0|\\&&&&0.025&0.025&0.275&-0.225\\ 0&0&1&0|\\&&&&0.025&0.275&-0.225&0.025\\ 0&0&0&1|\\&&&&0.275&-0.225&0.025&0.025 \end{bmatrix}$

$A^{-1}==\begin{bmatrix} -0.225&0.025&0.025&0.275\\ 0.025&0.025&0.275&-0.225\\ 0.025&0.275&-0.225&0.025\\ 0.275&-0.225&0.025&0.025 \end{bmatrix}$