MHB Calculating the Inverse Matrix for a 3x3 Matrix

[karush]

$\tiny{311.2.2.31}$
$A=\left[\begin{array}{rrrrr}
1&0&-2\\-3&1&4\\2&-3&4
\end{array}\right]$
RREF with augmented matrix
$\left[ \begin{array}{ccc|ccc}
1&0&-2&1&0&0 \\&&&\\-3&1&4&0&1&0 \\&&&\\ 2&-3&4&0&0&1\end{array}\right]
\sim
\left[ \begin{array}{ccc|ccc}1&0&0&8&3&1 \\&&&\\0&1&0&10&4&1 \\&&&\\ 0&0&1&\dfrac{7}{2}&\dfrac{3}{2}&\dfrac{1}{2}
\end{array}\right]
\quad \therefore A^{-1}=\left[
\begin{array}{ccc}8 & 3 & 1 \\\\ 10 & 4 & 1 \\\\ \dfrac{7}{2} & \dfrac{3}{2} & \dfrac{1}{2} \end{array} \right]$

ok I left out the row reduction steps
but I tried to use the desmos matrix calculator to check this
but after you put in the matrix didn't see how to run it.

 

[topsquark]

This is why I prefer W|A.

-Dan

 

[karush]

when all else fails there is W|A

 

[HOI]

Do you really have to use some kind of calculator do the arithmetic for you?

Surely it is not that hard to do
$\begin{bmatrix}1 & 0 & -2 \\ -3 & 1 & 4 \\2 & 3 & 4 \end{bmatrix}\begin{bmatrix}8 & 3 & 1 \\ 10 & 4 6 & 1 \\ \frac{7}{2} & \frac{3}{2} & \frac{1}{2}\end{bmatrix}= \begin{bmatrix}8- 7 & 3- 3 & 1- 1\\ -24+ 10+ 14 & -9+ 4+ 6 & -3+ 1+ 2 \\ 16- 30+ 14 & 6- 12+ 6 & 2- 3+ 2 \end{bmatrix}= \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
and
$\begin{bmatrix}8 & 3 & 1 \\ 10 & 4 6 & 1 \\ \frac{7}{2} & \frac{3}{2} & \frac{1}{2}\end{bmatrix}$$\begin{bmatrix}8 & 3 & 1 \\ 10 & 4 6 & 1 \\ \frac{7}{2} & \frac{3}{2} & \frac{1}{2}\end{bmatrix}$$= \begin{bmatrix}8- 9+ 2 & 3- 3 & -16+ 12+ 4 \\ 10- 12+ 2 & 4- 3 & -20+ 16+ 4 \\ \frac{7}{2}- \frac{9}{2}+ 1 & \frac{3}{2}- \frac{3}{2} & -7+ 6+ 2 \end{bmatrix}=$$ \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.

(I'm just too old!)

 

[skeeter]

Any operation with matrices larger than 2 by 2 isn't meant to be done by hand, especially if one is prone to arithmetic errors. ;)

 

[karush]

its kinda like a bingo game