Inverse matrix by row reduction

[zuby]

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.

 

[Prove It]

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...

 

[Petrus]

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

\(\displaystyle AA^{-1}=I\) some use I or E for unit matrix but that I is unit matrix!

Regards,
\(\displaystyle |\pi\rangle\)

 

[zuby]

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

\(\displaystyle AA^{-1}=I\) some use I or E for unit matrix but that I is unit matrix!

Regards,
\(\displaystyle |\pi\rangle\)

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

 

[I like Serena]

Hi zuby! Welcome to MHB! :)

Where is the confusion?

The 17/2 is calculated from:
$$R_2'=R_2+2R_1'$$
$$6 + 2 \cdot \frac 5 4 = \frac{12}{2} + \frac 5 2 = \frac{17}{2}$$

 

[Petrus]

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 \(\displaystyle R_2'=R_2+2R_1'\)
\(\displaystyle \frac{2*5}{4}+6=\frac{17}{2}\)
does that make it clear for you?

Edit: I like Serena was faster
Regards,

 

[zuby]

what they did is that they did multiply 2 to R1 and add to R2. It is exactly what they mean with \(\displaystyle R_2'=R_2+2R_1'\)
\(\displaystyle \frac{2*5}{4}+6=\frac{17}{2}\)
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?

\(\displaystyle \frac{2*5}{4}+6\)

\(\displaystyle \frac{10}{4}+\frac{6}{1}=\frac{10}{4}+\frac{6}{4}=\frac{16}{4}\)

 

[I like Serena]

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

\(\displaystyle \frac{2*5}{4}+6\)

\(\displaystyle \frac{10}{4}+\frac{6}{1}=\frac{10}{4}+\frac{6}{4}=\frac{16}{4}\)

$$\frac{6}{1} \ne \frac{6}{4}$$
This should be:
$$\frac{6}{1} = \frac{24}{4}$$

 

[zuby]

$$\frac{6}{1} \ne \frac{6}{4}$$
This should be:
$$\frac{6}{1} = \frac{24}{4}$$

Oh I was not multiplying 4 by numerator 6.

Under of thanks.