3x3 matrix: finding inverse without a calculator
- Thread starter me....
- Start date
D
Deleted member 4993
Guest
Re: 3x3 matrix inverse
What methods have you been taught in your class? What does your textbook say?
There are several methods - try googling "inverse matrix" - you'll find many references.
me.... said:
What methods have you been taught in your class? What does your textbook say?
There are several methods - try googling "inverse matrix" - you'll find many references.
D
Deleted member 4993
Guest
me.... said:my text book say nothing about 3x3s but i will try to google it....thank you
What does it say about inverses of 2 x 2 or 4 x 4 matrices?
You can use elementary row ops.
The idea is to switch the two sides, so that the identity matrix is on the left and the inverse is on the right.
Start with finding the inverse of:
\(\displaystyle \L\\A=\left[\begin{array}1&2&3\\2&5&3\\1&0&8\end{array}\right]\)
Write as:
\(\displaystyle \L\\\left[\begin{array}{ccc|ccc}1&2&3&1&0&0\\2&5&3&0&1&0\\1&0&8&0&0&1\end{array}\right]\)
Now, add -2 times the first row to the second and -1 times the first row to the third:
\(\displaystyle \L\\\left[\begin{array}{ccc|ccc}1&2&3&1&0&0\\0&1&-3&-2&1&0\\0&-2&5&-1&0&1\end{array}\right]\)
Add 2 times the second row to the third:
\(\displaystyle \L\\\left[\begin{array}{ccc|ccc}1&2&3&1&0&0\\0&1&-3&-2&1&0\\0&0&1&-5&2&1\end{array}\right]\)
Multiply the third row by -1:
\(\displaystyle \L\\\left[\begin{array}{ccc|ccc}1&2&3&1&0&0\\0&1&-3&-2&1&0\\0&0&1&5&-1&-1\end{array}\right]\)
Add 3 times the third row to the second and -3 times the third row to the first:
\(\displaystyle \L\\\left[\begin{array}{ccc|ccc}1&2&0&-14&6&3\\0&1&0&13&-5&-3\\0&0&1&5&-2&-2\end{array}\right]\)
Add -2 times the second row to the first:
\(\displaystyle \L\\\left[\begin{array}{ccc|ccc}1&0&0&-40&16&9\\0&1&0&13&-5&-3\\0&0&1&5&-2&-1\end{array}\right]\)
There. Your inverse is the right part:
\(\displaystyle \L\\A^{-1}=\begin{bmatrix}-40&16&9\\13&-5&-3\\5&-2&-1\end{bmatrix}\)
The idea is to switch the two sides, so that the identity matrix is on the left and the inverse is on the right.
Start with finding the inverse of:
\(\displaystyle \L\\A=\left[\begin{array}1&2&3\\2&5&3\\1&0&8\end{array}\right]\)
Write as:
\(\displaystyle \L\\\left[\begin{array}{ccc|ccc}1&2&3&1&0&0\\2&5&3&0&1&0\\1&0&8&0&0&1\end{array}\right]\)
Now, add -2 times the first row to the second and -1 times the first row to the third:
\(\displaystyle \L\\\left[\begin{array}{ccc|ccc}1&2&3&1&0&0\\0&1&-3&-2&1&0\\0&-2&5&-1&0&1\end{array}\right]\)
Add 2 times the second row to the third:
\(\displaystyle \L\\\left[\begin{array}{ccc|ccc}1&2&3&1&0&0\\0&1&-3&-2&1&0\\0&0&1&-5&2&1\end{array}\right]\)
Multiply the third row by -1:
\(\displaystyle \L\\\left[\begin{array}{ccc|ccc}1&2&3&1&0&0\\0&1&-3&-2&1&0\\0&0&1&5&-1&-1\end{array}\right]\)
Add 3 times the third row to the second and -3 times the third row to the first:
\(\displaystyle \L\\\left[\begin{array}{ccc|ccc}1&2&0&-14&6&3\\0&1&0&13&-5&-3\\0&0&1&5&-2&-2\end{array}\right]\)
Add -2 times the second row to the first:
\(\displaystyle \L\\\left[\begin{array}{ccc|ccc}1&0&0&-40&16&9\\0&1&0&13&-5&-3\\0&0&1&5&-2&-1\end{array}\right]\)
There. Your inverse is the right part:
\(\displaystyle \L\\A^{-1}=\begin{bmatrix}-40&16&9\\13&-5&-3\\5&-2&-1\end{bmatrix}\)