Inverse matrix by row reduction

zuby

New member
Hi,
Can anyone help me to inverse the below matrix by row reduction method.

4 5
-2 6

thanks.

Prove It

Well-known member
MHB Math Helper
Hi,
Can anyone help me to inverse the below matrix by row reduction method.

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

Well-known member
Hi,
Can anyone help me to inverse the below matrix by row reduction method.

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

New member
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.

Klaas van Aarsen

MHB Seeker
Staff member
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

Well-known member
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

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

New member
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?

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

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

Klaas van Aarsen

MHB Seeker
Staff member
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

New member
$$\frac{6}{1} \ne \frac{6}{4}$$
This should be:
$$\frac{6}{1} = \frac{24}{4}$$
Oh I was not multiplying 4 by numerator 6.

Under of thanks.