Finding the Inverse of a 2x2 Matrix using Gauss-Jordan Method

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The discussion focuses on finding the inverse of a 2x2 matrix using the Gauss-Jordan method. The original poster attempted to set up an augmented matrix but received feedback that their approach was incorrect, particularly regarding the placement of elements in the matrix. Participants clarified that the correct method involves transforming the left side of the augmented matrix into the identity matrix while applying the same operations to the right side to obtain the inverse. The conversation emphasizes the importance of correctly setting up the augmented matrix and performing row operations accurately. Ultimately, the original poster resolved their issue but acknowledged ongoing challenges with Gaussian elimination.
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Homework Statement


$$
\begin{bmatrix}
a &b \\
c&d
\end{bmatrix}$$

I'm supposed to find the inverse

Homework Equations



Method of Gauss-Jordan

The Attempt at a Solution


So I tried putting zeros in this and I got the following :

$$
\begin{bmatrix}
ad-ac &0 &ad &ad-a \\
0&bc-ad &c &-a
\end{bmatrix}$$

Does my procedure seem to make any sense ?

I know that after that I must have two "1"
 
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Where is your inverse matrix? It should be 2x2.
Just multiply it with the original matrix to check that you got the right result.
 
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Physicaa said:

Homework Statement


$$
\begin{bmatrix}
a &b \\
c&d
\end{bmatrix}$$

I'm supposed to find the inverse

Homework Equations



Method of Gauss-Jordan

The Attempt at a Solution


So I tried putting zeros in this and I got the following :

$$
\begin{bmatrix}
ad-ac &0 &ad &ad-a \\
0&bc-ad &c &-a
\end{bmatrix}$$

Does my procedure seem to make any sense ?

I know that after that I must have two "1"
No, what you're doing doesn't make any sense.

the definition of the inverse of matrix A is

A * A-1 = I,

where A-1 is the inverse of A and I is the identity matrix, which contains 1 on the main diagonal and is zero everywhere else.

You can calculate the inverse by solving the system of equations above for multiple RHS as indicated:

Code:
| a   c | | x1  x2 |   | 1   0 |
| b   d | | x3  x4 | = | 0   1 |

The system above can be solved by Gaussian elimination. The values x1 ... x4 will be the elements of the inverse matrix.

http://mathworld.wolfram.com/MatrixInverse.html
 
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SteamKing said:
No, what you're doing doesn't make any sense.

the definition of the inverse of matrix A is

A * A-1 = I,

where A-1 is the inverse of A and I is the identity matrix, which contains 1 on the main diagonal and is zero everywhere else.

You can calculate the inverse by solving the system of equations above for multiple RHS as indicated:

Code:
| a   c | | x1  x2 |   | 1   0 |
| b   d | | x3  x4 | = | 0   1 |

The system above can be solved by Gaussian elimination. The values x1 ... x4 will be the elements of the inverse matrix.

http://mathworld.wolfram.com/MatrixInverse.html
I was finally able to solve the problem. Thank you anyway for your answer. I am still stuck on the Gaussian elimination of the other problem though :/
 
No, what you have shown there does NOT make any sense! Where did you get it?
It is true that one method of finding the inverse to a given matrix is to write the given matrix with the identity matrix beside it:
\begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}
then use "row operations" to reduce the given matrix to the identity matrix while at the same time applying those same row operations to the identity matrix.
A good way to do the row operations is to work one row and column at a time. For example, The first few step should be to divide the entire first row by a, in order to get "1" there, then subtract that new first row, times c, from the second row, to get "0" there:
\begin{bmatrix}1 & \frac{b}{a} \\ 0 & d- \frac{bc}{a}\end{matrix}\begin{bmatrix}\frac{1}{a} & 0 \\ -\frac{c}{a} & 1 \end{bmatrix}

Now, divide the second row, of both matrices, by d- \frac{ac}{b}= \frac{ad- bc}{b} to get a "1" in the last place (you may recognize that "db- ac" as the determinant of the original matrix) then subtract that times the b/a from the first row- again for both matrices. That will give
\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix}\frac{d}{ad- bc} & \frac{-b}{ad- bc} \\ \frac{c}{ad- bc} & \frac{a}{ad- bc} \end{bmatrix}
-
 
SteamKing said:
No, what you're doing doesn't make any sense.
I think what the OP was doing was setting up the augmented matrix
$$\left[\begin{array}{cc|cc} a & b & 1 & 0 \\ c & d & 0 & 1\end{array}\right]$$ and then using Gaussian elimination to turn the lefthand side into the identity matrix. The righthand side would then be the inverse. The method is fine, but the fact that the OP's matrix has no ##b## on the righthand side indicates something went wrong.
 
vela said:
I think what the OP was doing was setting up the augmented matrix
$$\left[\begin{array}{cc|cc} a & b & 1 & 0 \\ c & d & 0 & 1\end{array}\right]$$ and then using Gaussian elimination to turn the lefthand side into the identity matrix. The righthand side would then be the inverse. The method is fine, but the fact that the OP's matrix has no ##b## on the righthand side indicates something went wrong.
I solved the problem, there's no need now.
 
Physicaa said:
I solved the problem, there's no need now.
It's not all about you. ;) I was clarifying your unclear and confusing original post for the benefit of others, like SteamKing.
 
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