Linear Algebra - Inverse of Matrices

In summary, the conversation discusses a problem where a 2-by-2 matrix A needs to be written as a product of elementary matrices and its inverse also needs to be written in the same form. The steps used to convert A into reduced row echelon form are shared and the correct elementary matrices are determined. The conversation ends with the person figuring out the solution after some time.
  • #1
Lorelyn
7
0
I have a problem in which I am given a 2-by-2 matrix A (0,8 / 5,7) which I am asked to write as a product of elementary matrices. Then I am asked write the inverse of A as a product of elementary matrices. So I turned A into reduced row echlon form using the following steps:

-Switch R1 and R2
-1/5 * R1
-1/8 * R2
- (-7/5) * R2 + R1 into R1

So the elementary matrices are:
(0,1 / 1,0) * (1/5,0 / 0,1) * (1,0 / 1/8,0) and (1,-7/5 / 0,-7/5)

Right?

Then the inverse of A as a product of elementary matrices is just the inverse of each of (1,-7/5 / 0,-7/5) * (1,0 / 1/8,0) * (1/5,0 / 0,1) and (0,1 / 1,0).

Does this make sense? Because I keep getting the wrong answer and I don't know where I've gone wrong...
 
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  • #2
Lorelyn said:
I have a problem in which I am given a 2-by-2 matrix A (0,8 / 5,7) which I am asked to write as a product of elementary matrices. Then I am asked write the inverse of A as a product of elementary matrices. So I turned A into reduced row echlon form using the following steps:

-Switch R1 and R2
-1/5 * R1
-1/8 * R2
- (-7/5) * R2 + R1 into R1
So the elementary matrices are:
(0,1 / 1,0) * (1/5,0 / 0,1) * (1,0 / 1/8,0) and (1,-7/5 / 0,-7/5)

Right?
No. Since you did not change the second row in the last step, the bottom row of the last elementary matrix is "0 1" not"0 -7/5". The elementary matrix corresponding to a row-operation is the matrix derived by applying that row operation to the identity matrix. Sutracting 7/5 of the second row from the first row of the identity matrix gives (1, -7/5, 0, 1).

Then the inverse of A as a product of elementary matrices is just the inverse of each of (1,-7/5 / 0,-7/5) * (1,0 / 1/8,0) * (1/5,0 / 0,1) and (0,1 / 1,0).

Does this make sense? Because I keep getting the wrong answer and I don't know where I've gone wrong...
The last elementary matrix is (1 , -7/5, 0, 1) and its inverse is (1, +7/5, 0, 1).
 
  • #3
Thanks! I actually just figured it out by myself all it took was a couple hours staring at it blankly before something clicked in my brain! :)
 

1. What is the inverse of a matrix?

The inverse of a matrix is a matrix that, when multiplied by the original matrix, results in the identity matrix (a square matrix with 1s along the main diagonal and 0s everywhere else). In other words, the inverse "undoes" the original matrix.

2. How do you find the inverse of a matrix?

To find the inverse of a matrix, you can use the Gauss-Jordan elimination method or the adjugate method. The Gauss-Jordan method involves performing row operations on the original matrix until it is in reduced row echelon form. The resulting matrix will be the inverse of the original. The adjugate method involves finding the adjugate matrix (a matrix where each element is the cofactor of the corresponding element in the original matrix) and multiplying it by the reciprocal of the determinant of the original matrix.

3. Can every matrix be inverted?

No, not every matrix can be inverted. A matrix can only be inverted if it is a square matrix (same number of rows and columns) and if its determinant (a value calculated from the elements of the matrix) is not equal to 0. If the determinant is 0, the matrix is said to be singular and does not have an inverse.

4. What is the significance of the inverse of a matrix?

The inverse of a matrix is useful in solving systems of linear equations and in finding solutions to other mathematical problems. It is also used in the process of finding eigenvalues and eigenvectors of a matrix, which have applications in fields such as physics and engineering.

5. Can the inverse of a matrix be its own inverse?

Yes, the inverse of a matrix is always its own inverse. This means that if you multiply a matrix by its inverse, and then multiply it again by its inverse, you will get back the original matrix.

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