MHB Inverse matrix by row reduction

AI Thread Summary
To find the inverse of the matrix using row reduction, one must set up an augmented matrix with the identity matrix alongside it and perform row operations until the left side becomes the identity matrix. A user expressed confusion over a calculation involving the value 17/2, which was clarified through the operation R2' = R2 + 2R1'. The mistake was identified in not properly converting the number 6 into a compatible fraction for addition. The discussion emphasizes the importance of checking solutions by multiplying the original matrix by its inverse to ensure the result is the identity matrix. Understanding these row operations is crucial for accurately calculating the inverse.
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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