Inverse matrix by row reduction

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SUMMARY

The discussion focuses on the method of inverting a matrix using row reduction, specifically for the matrix [[4, 5], [-2, 6]]. Participants explain the process of setting up an augmented matrix with the identity matrix and performing row operations to achieve the inverse. A key calculation discussed involves transforming the second row using the operation R2' = R2 + 2R1', which leads to the result of 17/2. The importance of verifying the solution by multiplying the original matrix by its inverse to obtain the identity matrix is emphasized.

PREREQUISITES
  • Understanding of matrix operations
  • Familiarity with augmented matrices
  • Knowledge of row operations in linear algebra
  • Ability to perform basic arithmetic with fractions
NEXT STEPS
  • Study the process of setting up augmented matrices for different types of matrices
  • Learn about the Gauss-Jordan elimination method for matrix inversion
  • Explore the properties of matrix multiplication and the identity matrix
  • Practice solving linear equations using row reduction techniques
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Students and educators in mathematics, particularly those studying linear algebra, as well as anyone seeking to understand matrix inversion techniques through row reduction.

zuby
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Hi,
Can anyone help me to inverse the below matrix by row reduction method.
I know determinant method but I don't know row reduction method please help me.

4 5
-2 6

thanks.
 
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zuby said:
Hi,
Can anyone help me to inverse the below matrix by row reduction method.
I know determinant method but I don't know row reduction method please help me.

4 5
-2 6

thanks.

You need to set up an augmented matrix with the identity matrix next to it, then go through a series of row operations until the matrix on the left becomes the identity matrix...
 
zuby said:
Hi,
Can anyone help me to inverse the below matrix by row reduction method.
I know determinant method but I don't know row reduction method please help me.

4 5
-2 6

thanks.
Hello,
here you got some exemple as you can see you can do it in Two way! (Look at second exemple) Mathwords: Inverse of a Matrix
Remember to always check your soultion! When you find your inverse and multiply by the non inverse Then you should get unit matrix! ( the one with 1 on diagonal and zero at rest!) With other words

$$AA^{-1}=I$$ some use I or E for unit matrix but that I is unit matrix!

Regards,
$$|\pi\rangle$$
 
Petrus said:
Hello,
here you got some exemple as you can see you can do it in Two way! (Look at second exemple) Mathwords: Inverse of a Matrix
Remember to always check your soultion! When you find your inverse and multiply by the non inverse Then you should get unit matrix! ( the one with 1 on diagonal and zero at rest!) With other words

$$AA^{-1}=I$$ some use I or E for unit matrix but that I is unit matrix!

Regards,
$$|\pi\rangle$$

Thanks for quick response.
I checked the site that you sent and understood row reduction method but that was easy. When I calculate the matrix values that I posted in previous. it will not give the same result that I have in my book here is my above matrix solution from my book. there is confusion below the image at red circled area how (17/2) is calculated by the book author.

View attachment 1577

Please help me thanks
 

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Hi zuby! Welcome to MHB! :)

Where is the confusion?

The 17/2 is calculated from:
$$R_2'=R_2+2R_1'$$
$$6 + 2 \cdot \frac 5 4 = \frac{12}{2} + \frac 5 2 = \frac{17}{2}$$
 
zuby said:
Thanks for quick response.
I checked the site that you sent and understood row reduction method but that was easy. When I calculate the matrix values that I posted in previous. it will not give the same result that I have in my book here is my above matrix solution from my book. there is confusion below the image at red circled area how (17/2) is calculated by the book author.

https://www.physicsforums.com/attachments/1577

Please help me thanks
what they did is that they did multiply 2 to R1 and add to R2. It is exactly what they mean with $$R_2'=R_2+2R_1'$$
$$\frac{2*5}{4}+6=\frac{17}{2}$$
does that make it clear for you?

Edit: I like Serena was faster
Regards,
 
Petrus said:
what they did is that they did multiply 2 to R1 and add to R2. It is exactly what they mean with $$R_2'=R_2+2R_1'$$
$$\frac{2*5}{4}+6=\frac{17}{2}$$
does that make it clear for you?

Edit: I like Serena was faster
Regards,

When I calculate it says 16 over 4. Where am I making mistake?

$$\frac{2*5}{4}+6$$

$$\frac{10}{4}+\frac{6}{1}=\frac{10}{4}+\frac{6}{4}=\frac{16}{4}$$
 
zuby said:
When I calculate it says 16 over 4. Where am I making mistake?

$$\frac{2*5}{4}+6$$

$$\frac{10}{4}+\frac{6}{1}=\frac{10}{4}+\frac{6}{4}=\frac{16}{4}$$

$$\frac{6}{1} \ne \frac{6}{4}$$
This should be:
$$\frac{6}{1} = \frac{24}{4}$$
 
I like Serena said:
$$\frac{6}{1} \ne \frac{6}{4}$$
This should be:
$$\frac{6}{1} = \frac{24}{4}$$

Oh I was not multiplying 4 by numerator 6.

Under of thanks.
 

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