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. :)


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.

  • $\begingroup$ Thanks so much! Finally a procedure; I know what exactly I have to do. $\endgroup$ Sep 14 '13 at 9:11
  • $\begingroup$ @mikhailcazi: You’re very welcome. $\endgroup$ Sep 14 '13 at 9:12
  • $\begingroup$ It might be important to know why this method works. It's a rather simple and basic reason connected with what "elementary operations by rows" actually are. $\endgroup$
    – DonAntonio
    Sep 14 '13 at 9:34
  • 1
    $\begingroup$ @mikhailcazi: You might take a look at this question and its answers; I believe that one of the answers or comments also has a link to an explanation. $\endgroup$ Sep 14 '13 at 9:46
  • 3
    $\begingroup$ Every row elementary operation is a multiplication by an elementary matrix $\,E_i\;$ from the left, @mikhailcazi, so following the method given by Brian you get $$E_k\cdot E_{k-1}\cdot\ldots\cdot E_2 A=I$$ which means $\,E_kE_{k-1}\cdot\ldots\cdot E_1=A^{-1}\;$ ... $\endgroup$
    – DonAntonio
    Sep 14 '13 at 9:53

I give you an example of a similar approach to Brian's one. Start from the matrix $$A=\left[ \begin{array}{ccc}1&2&0\\2&0&-1\\2&3&-1\end{array}\right]$$ First step is to write the identity matrix on right side of the matrix A. $$\left[ \begin{array}{cccccc}1&2&0&1&0&0\\2&0&-1&0&1&0\\2&3&-1&0&0&1\end{array}\right]\begin{array}{c}a\\b\\c\end{array}$$ Now, you have to start to apply elementary transformations to the whole matrix, since on the left side the identity matrix appears. $$\left[ \begin{array}{cccccc}1&2&0&1&0&0\\0&3&0&0&-1&1\\2&3&-1&0&0&1\end{array}\right]\begin{array}{c}a \rightarrow a\\-b+c \rightarrow b\\c\rightarrow c\end{array}$$ $$\left[ \begin{array}{cccccc}1&2&0&1&0&0\\0&1&0&0&-1/3&1/3\\0&1&1&2&0&-1\end{array}\right]\begin{array}{c}a \rightarrow a\\b/3 \rightarrow b\\-c+2a\rightarrow c\end{array}$$ $$\left[ \begin{array}{cccccc}1&0&0&1&2/3&-2/3\\0&1&0&0&-1/3&1/3\\0&0&1&2&1/3&-4/3\end{array}\right]\begin{array}{c}a-2b \rightarrow a\\b \rightarrow b\\c-b\rightarrow c\end{array}$$

At this point, the matrix on the right side is the inverse of $A$! Namely: $$A^{-1}=\left[ \begin{array}{ccc}1&2/3&-2/3\\0&-1/3&1/3\\2&1/3&-4/3\end{array}\right]$$

  • $\begingroup$ This doesn't exactly answer my question. As Brian says, one way to do reduce it is making the element $a_{ii}$ 1, then making all elements of column $i$ 0. Then move onto $a_{i+1, i+1}$. I was asking for a procedure which could be followed. What procedure have you followed in your answer? $\endgroup$ Sep 14 '13 at 11:39
  • $\begingroup$ well, I managed to get all $a_{ii} = 1$ on the left hand side. I achieved this by using elementary operation reported on the rightmost side of the whole matrix. This is a method to obtain on the right hand side the inverse of a matrix. $\endgroup$ Sep 14 '13 at 17:11

Consider a square matrix A of order 3. Now by property of matrices if |A|$\neq$ 0 only then it has an inverse i.e. A should be a non singular matrix. Assuming A to be a non singular matrix, by property of matrices
$$(A)(A^{-1})=I$$ where I is identity matrix. $$ $$ Now substitute the value of A and I. Use transformations as suggested above to get identity matrix in place of A and simultaneously applying those elementary transformation to the Identity matrix on RHS would give you the value of $A^{-1}$ .

So basically the equation changes to $$(I)(A^{-1})=A^{-1}$$

  • $\begingroup$ In case you need me to add the steps of transformation do comment. $\endgroup$ Nov 11 '16 at 4:29

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