## The inverse of a Matrix-inverse of a 2×2 matrix.

### A brief description of the content of the lecture.

The objective of this lecture is to determine the inverse of a matrix by elementary row operation and use it for solving the simultaneous equation by lower/upper factorization or Doolittle’s factorization for the lower /upper factorization.

### How to find the inverse of a matrix?

First, let us discuss how to get the inverse of a matrix by using elementary operations, we are using the Augmented matrix on the left-hand side will be written a matrix denoted by A, separated by a line and this is the identity matrix I.

For example, it is 3 by 3 or 2 by 2 or 4 by 4 and it’s written in the capital letter L.

Through elementary operations, we are converting our original matrix A to the identity matrix I, and through the elementary operations the I which was in the augmented matrix on the right-hand side of the Augmented matrix, will be converted into A-1, where *A*^{-1 }is the inverse of A. Our original matrix, which is the one we are looking for, what are the elementary row operations.

## The possible operations to a matrix.

There are three operations one is the swap of rows changing from R1 to R2 changes location for example and multiplying or dividing each element in the row by a constant. The 3rd of operation is replacing rows by adding or subtracting multiple other rows. We can say R1 +2*R2 or R1-R3 where R1 is the first-row donation and R3 is the third row, etc.

### A solved example of how to get the inverse of a matrix of a 2×2 matrix?

For example, if we have a matrix that is 2×2, it’s required to find the inverse of a 2×2 matrix where is A (-3 -2, 7 4), respectively.

The solution that we are going to use is the augmented matrix, we write on the left-hand side our matrix A.

We divide the matrix by a vertical line and we are putting this identity matrix in our case it will be 2 by 2 the diagonal as we can see is (1 and 1) and the other elements are zeros.

We want to convert the left-hand side into (1 1 0 0), how this can be done?

First, to make -3 converted to +1, we multiply R1*1/3, and the result we are going to put it in the place of R1, so this (-3*-1/3) to 1,(-2*-1/3), it will give us (+2/3) and the right-hand-side will be (-1/3,0).

While the second row will be left unchanged because there is no operation, it will move the same elements which are a(7 4 0 1) here.

The next step will be making this 7 = 0 and this can be done by multiplying the first row by(-7) and adding to the second row R2, so (-7*1+7) will become 0, here 2/ 3 -7 +4=-2/3, -7*-1/3 will be 7/3 and the application of 0 will be 0 + R2. It will become here, now we want to make this one equal to one. The next step is to multiply the second row by (-3/2) and place it in our to zero, which is the same this is reciprocal would become 1 and 7/3 RI.(-3/2) it will become (-7/3) -3/2 or the last item which is equal 1.

The next target is to make the 2/3 =0, this can be done by multiplying the second row which is 1 by -2/3, and adding to 1.

We get (1 0 2 +1 ), and since the second row is unchanged so it will be maintained at ( 0 1 – /2 (-3/2), on the left-hand side we find that the diagonal is=1 and the other diagonal is 0.

This is the inverse we are looking for it is divided and lies in the right-hand portion of this line.

### Another way to get the inverse of a matrix is by the determinant.

There is a rule for the matrix A if it is (2×2), we replace this we swap these -3 will go down, while we are going to change the sign of 7, it will become -7, and the(-2) will become the +2.

But again we are going to divide on the determinant which is(-3*4-1*-2), which at the end will become +2 so the inverse of the matrix is called adjoint divided by the determinant, the adjoint is (4 -2, and (-7 -3/2). The details of the operations can be seen in the next slide image.

The result, we have obtained by using the determinant is identical to the result obtained by the augmented matrix.

This is the pdf file used for the illustration of this post.

The next post is Easy Illustration for LU decomposition method for 2×2 matrix.

HELM-Helping Engineers Learn Mathematics.

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