While using the elementary transformation method to find the inverse of a matrix, our goal is to convert the given matrix into an identity matrix.
We can use three transformations:-
1) Multiplying a row by a constant
2) Adding a multiple of another row
3) Swapping two rows
The thing is, I can't seem to figure out what to do to achieve that identity matrix. There are so many steps which I can start off with, but how do I know which one to do? I think of one step to get a certain position to a $1$ or a $0$, and then get a new matrix. Now again there are so many options, it's boggling.
Is there some specific procedure to be followed? Like, first convert the top row into:
\begin{bmatrix}
1&0&0\\
a_{21}&a_{22}&a_{23}\\
a_{31}&a_{32}&a_{33}
\end{bmatrix}
Then do the second row and then the third?
What do I start off with? I hope I've made my question clear enough.
Thanks to @Brian M. Scott.
$P.S:$ Does anyone have any other methods? Brian's works perfectly, but it's always great to know more than one method. 🙂
Best Answer
First get a non-zero entry in the upper lefthand corner by swapping to rows if necessary. If that entry is $a_{11}\ne 0$, multiply the first row by $a_{11}^{-1}$ to get a $1$ in the upper lefthand corner. Now use operation (2) to get $0$’s in the rest of the first column.
Now get a non-zero entry in the $a_{22}$ position, the second entry in the second row, by swapping the second row with one of the lower rows if necessary, and multiply the (possibly new) second row by $a_{22}^{-1}$ to get a $1$ in the $a_{22}$ position. Then use operation (2) to get $0$’s in the rest of the second column; notice that since $a_{21}$, the first element in the second row, is $0$, this will not affect anything in the first column.
At this point your matrix looks like this:
$$\begin{bmatrix} 1&0&a_{13}&\dots&a_{1n}\\ 0&1&a_{23}&\dots&a_{2n}\\ 0&0&a_{33}&\dots&a_{3n}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&a_{n3}&\dots&a_{nn} \end{bmatrix}$$
Continue in the same fashion: get a non-zero entry in the $a_{33}$ position by swapping row $3$ with a lower row if necessary, multiply row $3$ by a suitable constant to make $a_{33}=1$, and use operation (2) to $0$ out the rest of the third column.
If at any point the necessary operation is impossible, your original matrix was not invertible.