# Thread: finding inverse of a matrix through row transformation

1. ## finding inverse of a matrix through row transformation

Sir,

The question was to find A-1 through row transformation for the given matrix A.

1 2 3 0 1 0

0 3 5 = 1 1 0 x A

0 -5 -8 0 -3 1

Can I say R1 -> R1 + R2 + R3

Or I have two take only any one or two rows for applying row transformation?

kindly guide me.

I saw the preview of the post. I am not able to put it in a matrix form. It is 3 x 3 matrix resembles

D = C x A

All D, C and A are of 3 x 3 matrices.

with warm regards,

Aranga

2. ## Re: finding inverse of a matrix through row transformation

Each transformation uses at most two rows, but the transformation you give can be achieved in two steps. (R1 -> R1+R2, R1->R1+R3).

3. ## Re: finding inverse of a matrix through row transformation

I was under the belief any number of rows can be used for row transformation. Thanks.

4. ## Re: finding inverse of a matrix through row transformation

There are three kinds of row operations:
1) Swap two rows
2) Multiply every number in a row by a fixed number
3) Add a multiple of one row to another.

The simplest way to use row operations to get the inverse matrix of a given matrix is to write the given matrix with the identity matrix right next to it:
$\begin{bmatrix}1 & 2 & 3 \\ 0 & 3 & 5 \\ 0 & -5 & -8\end{bmatrix}\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$

Now apply row operations to convert the left matrix to the identity matrix, applying those same row operations to the right matrix, working one column at a time. To start with I see that the left column of the matrix, $\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}$, is already the left column of the identity matrix so we don't need to do anything to that. Moving to the middle column, we want "1" in the middle row. Now there is a "3" there so divide the entire middle row by 3:
$\begin{bmatrix}1 & 2 & 3 \\ 0 & 1 & \frac{5}{3} \\ 0 & -\frac{5}{3} & -\frac{8}{3}\end{bmatrix}\begin{bmatrix}1 & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & 1\end{bmatrix}$

There is a "2" in the top row of the middle column and we want "0" so add -2 times the middle row:
$\begin{bmatrix}1 & 0 & 3 \\ 0 & 1 & \frac{5}{3} \\ 0 & -\frac{5}{3} & -\frac{8}{3}\end{bmatrix}\begin{bmatrix}1 & -\frac{2}{3} & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & 1\end{bmatrix}$

There is a "-5" in the bottom row of the middle column and we want "0" so add 5 times the middle row:
$\begin{bmatrix}1 & 0 & 3 \\ 0 & 1 & \frac{5}{3} \\ 0 & 0 & \frac{1}{3}\end{bmatrix}\begin{bmatrix}1 & -\frac{2}{3} & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & \frac{5}{3} & 1\end{bmatrix}$

(With a little practice you can do those three operations on one copy of the matrix)

That finishes the third column, now go on to the third column. The bottom row of the third column has $\frac{1}{3}$ so multiply the third row by 3:
$\begin{bmatrix}1 & 0 & 3 \\ 0 & 1 & \frac{5}{3} \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}1 & -\frac{2}{3} & 0 \\ 0 & 1 & 0 \\ 0 & 5 & 3\end{bmatrix}$

The second row, third column is, $\frac{5}{3}$ so add $-\frac{5}{3}$ times the third row to the second row:
$\begin{bmatrix}1 & 0 & 3 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}1 & -\frac{2}{3} & 0 \\ 0 & 1 & -8 \\ 0 & -8 & 1\end{bmatrix}$

Finally, the first row, third column, is $-\frac{1}{3}$ so add $\frac{1}{3}$ times the third row to the first row:
$\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}1 & 1 & 1 \\ 0 & -8 & -5 \\ 0 & 5 & 3\end{bmatrix}$

Of course, you should check that $\begin{bmatrix}1 &2 & 3 \\ 0 & 3 & 5 \\ 0 & -5 & -8\end{bmatrix}\begin{bmatrix}1 & 1 & 1 \\ 0 & -8 & -5 \\ 0 & 5 & 3 \end{bmatrix}= \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

and that
$\begin{bmatrix}1 & 1 & 1 \\ 0 & -8 & -5 \\ 0 & 5 & 3 \end{bmatrix}\begin{bmatrix}1 &2 & 3 \\ 0 & 3 & 5 \\ 0 & -5 & -8\end{bmatrix}=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

This works because every "row operation" corresponds to multiplying the matrix by the "elementary matrix" we get by applying that row operation to the identity matrix. That is "swap the first and third rows" is the same as multiplying the matrix by $\begin{bmatrix}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}$, "multiply every number in the second row by 3" is the same as multiplying the matrix by $\begin{bmatrix}1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 1\end{bmatrix}$, and "add two times the first row to the second row" is the same as multiplying the matrix by $\begin{bmatrix}1 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$.

Reducing the matrix X to the identity matrix is the same as determining the elementary matrices, $F_1$, $F_2$, ..., $F_k$ that will reduce X to the identity: $(F_1F_2\cdot\cdot\cdot F_k)A= I$ so that $A^{-1}$ is simply $A^{-1}= F_1F_2\cdot\cdot\cdot F_k$ while applying them to the identity matrix is just $(F_1F_2\cdot\cdot\cdot F_n)I= F_1F_2\cdot\cdot\cdot F_n= A^{-1}$.

5. ## Re: finding inverse of a matrix through row transformation

Thanks. The detailed explanation is very useful.