# Augmented matrix explanation

1. May 14, 2015

### rajeshmarndi

I just couldn't understand how does augmented matrix deduce inverse of a matrix. I mean what is it in the row operation because of which we get the inverse of a matrix. I just don't want to learn the steps but to understand why it works.

Thank you.

2. May 14, 2015

### HallsofIvy

Staff Emeritus
I am not certain what you mean by "augmented matrix". I think of the augmented matrix as meaning that, in order to solve, say the equations ax+ by= c, dx+ ey= f, which we could write as the matrix equation $\begin{bmatrix}a & b \\ d & e \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}c \\ f\end{bmatrix}$. To solve that we "augment" the matrix $\begin{bmatrix}a & b \\ d & e \end{bmatrix}$ by the single column $\begin{bmatrix}c \\ d\end{bmatrix}$ to get $\begin{bmatrix}a & b & c \\ d & e & f\end{bmatrix}$.

But that has nothing, directly, to do with finding the inverse. I think you are talking about a related problem- "augment" the matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ with the indentity matrix to get $\begin{bmatrix}a & b & 1 & 0 \\ c & d & 0 & 1\end{bmatrix}$. Then do whatever "row-operations" are necessary to reduce the first two columns to the identity while also applying them to the last two columnst to get $\begin{bmatrix}1 & 0 & p & q \\ 0 & 1 & r & s\end{bmatrix}$ and the new matrix $\begin{bmatrix}p & q \\ r & s\end{bmatrix}$ is the inverse matrix to the first matrix.

That works because every "row operation" corresponds to an "elementary" matrix- the matrix we get by applying that particular row operation to the identity matrix. One row operation is "add a multiple of one row to another". For example, "add -2 times the first row to the second row". If we do that to the identity matrix $\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$ we get $\begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix}$. And look what happens when we multiply a matrix by that (on the left): $\begin{bmatrix}1 & 0 \\ -2 & 1\end{bmatrix}\begin{bmatrix}a & b \\ c & d \end{bmatrix}= \begin{bmatrix}a & b \\ c- 2a & d- 2b\end{bmatrix}$. That is precisely the result of applying the row operation to that matrix!

Now, suppose we have a matrix, A, and there is a sequence of row operations, $R_1, R_2, \cdot\cdot\cdot, R_n$, in that order, that reduce A to the identity. Those row-operations correspond to a sequence of elementary operations, $E_1, E_2, \cdot\cdot\cdot, E_n$ such that $E_nE_{n-1}\cdot\cdot\cdot E_2E_1A= (E_nE_{n-1}\cdot\cdot\cdot E_2E_1)A= I$ which means that $E_nE_{n-1}\cdot\cdot\cdot E_2E_1$ is the inverse matrix to A. But, of course $E_nE_{n-1}\cdot\cdot\cdot E_2E_1$ times I is the same as $E_nE_{n-1}\cdot\cdot\cdot E_2E_1$ and, again, those multiplications are the same thing as applying the row operations to the identity matrix so:
Applying the same series of row operations that make A the identity matrix to the identity matrix results in the inverse matrix to A.

Writing the matrix A and the identity matrix side by side and then row-reducing both is a convenient way of applying the same row operations to A and the indentity matrix, reducing A to the identity matrix and changing the identity matrix to A inverse.

Last edited by a moderator: May 14, 2015
3. Jun 11, 2015

### rajeshmarndi

Thanks I do now understand how the row operation works i.e [A-1] [A] = identity matrix. But still couldn't understand how the other way works i.e [A] [A-1]= identity matrix. Seems really difficult, the former was easy to comprehend.

4. Jun 12, 2015

### Hawkeye18

There are two theorems to help.

The first one is that if you have a square matrix $A$ and it is left invertible (i.e. have a left inverse $B$, $BA=I$) then it is invertible.
And the second one is that if a matrix $A$ is invertible, then its left inverse $B$ is unique and is also a right inverse, i.e. if $BA=I$ then $AB=I$; the same of course goes for the right inverse.

5. Jun 13, 2015

### rajeshmarndi

I didn't understand what is the relation of the coefficient of the equations and their row operation(inverse) to get identity matrix.

6. Jun 13, 2015

### HallsofIvy

Staff Emeritus
I'm not certain what your question is. but perhaps this will help. If we have the two equations ax+ by= e, cx+ dy= f we can write that as the matrix equation
$$\begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}e \\ f\end{bmatrix}$$
Do you understand that part? That is essentially saying that "Ax= b" where A, x, and b are those three matrices. One way to solve that equation is to find the multiplicative inverse of $A$, $A^{-1}$, and multiply both sides of the equation by that inverse: $A^{1}Ax= x= A^{-1}b$

The "augmented matrix" is just a short way of writing that matrix equation.
$$\begin{bmatrix}a & b & e \\ c & d & f\end{bmatrix}$$

Now, "row-operations" are one way to find the inverse of a matrix. The key is that every row operation corresponds to multiplying by a matrix- and we can get that matrix by applying the row operation to the identity matrix. In terms of 3 by 3 matrices, the row operation "add three times the second row to the first row" is
$$\begin{bmatrix} 1 & 3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$$
See what happens when we multiply that matrix by a general matrix:
$$\begin{bmatrix}1 & 3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}= \begin{bmatrix}a+ 3d & b+ 3e & c+ 3f \\ d & e & f \\ g & h & i\end{bmatrix}$$
3 times the second row has been added to the first row!

So when you apply a series of row operations that reduce matrix A to the identity matrix that is the same as multiplying a series of matrices whose product is $A^{-1}$. When you apply those same row operations to the last column of the augmented matrix, the "b" in Ax+ b, you are multiplying b by $A^{-1}$

7. Jun 14, 2015

### rajeshmarndi

I do understand the meaning of A-1 A = identity matrix i.e row operation just applied to A.

But my question is, why and how we get the identity matrix , also when A A-1 = identity matrix?

8. Jun 14, 2015

### symbolipoint

Not so much because of LINEAR ALGEBRA, but more important is the simple Algebra concept of INVERSE. Start with a entity and do an operation on it using another but different identity, and a result entity is the result. How to you take this result entity and operate on it with an entity which gives you back the entitiy originally used?

This can be done using real numbers, and this can be done with matrices.