Verify (AB)^t = B^t A^t Using Example | Math Dot Com

A step by step verification of the expression (AB)t = Bt At is explained in this video.Take L.H.S of the expression and solve step by step:
Step 1: Find the product of two matrices A and B.
Step 2: Take the transpose of the product of two matrices.
Now take R.H.S and solve step by step:
Step 1: Find transpose of Matrix B.
Step 2: Find transpose of Matrix A.
Step 3: Find the product of step 1 and step 2 matrices.
Compare the L.H.S with R.H.S and check if they are equal.
For any query please comment or email at [e-mail]

[@paphinvitchrishartono964]

Thank you Math Dot Com! Super clear and easy to understand! 👌👌

[@MathDotCom]

Glad it was helpful!

[@shanesonn]

Very useful thank you

[@rohanhatkar6679]

Thank you soo much mam 🥰 for this video

[@anahim.mezaarroyo9544]

Muchas gracias, super claro 👍

[@Roy66802]

Thanks my mans

[@ka19jocker64]

THANK YOU MAM❤️❤️❤️

[@memeswift8379]

Thank you so much 🫶🫶🫶✨

[@MathDotCom]

You're welcome 😊

[@conniekolesar727]

Please can you try with a 3 by 3 matrix

[@TanvirClashOn]

thanks

[@AqilDaiyan]

Thank you

[@MathDotCom]

Welcome!

[@szymonsmagowski5062]

Well the verification is made correct, although it is ultra useless. It just shows that one example of pair A and B respect the equation. In order for it to be at least useful it should be proved for all cases using letter notation or disproved by counter example. I know that in the title there is verify not prove, but i hardly daubt that anywhere there is an exercise like this. I just lost some time comming here when i was looking for a proof of (AT)^t = A^t *B^t. Neverless film is correct.

[@MathDotCom]

It is to verify not to proof may b it is useless for u but useful for the others.

[@sham._.writes]

But why not (AB)^t =A^t B^t

[@MathDotCom]

Its same doesn’t matter