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You're very welcome! Thanks for the support! 🙂

You referring to the matrix prior to RREF?

Any time! Thanks for the support, too! 🙂

You're very welcome! And thanks for your support, it means a lot! 🙂

@KevinTheSoftwareTester Thanks, Kevin, and salute to you! 🫡 I'll share with you one of my favorite of these, which would be the Mortal Kombat characters. I bring them in when I begin the topic of Functions. I like to connect the buttons of an Xbox control the fighting moves. For example, the input "X" has an output of "High Punch" 🤜☠️⚰️🪦🔔, lol! 😂 Thanks for commenting! #FAL 🤜🤛💪💯😎

One has to go so that we can analyze the effects of it missing I'm the world of math! It's also an indicator of what personality type you have...j/k 😂

@@larrychoraszewski1125 gosh Personality type scares me hahha

@@homayraadiba3594 relatable!

I'm glad you found it helpful, and thanks for the support! :-)

Thanks for the support, Kevin! 🙂

You're very welcome, and thanks for the support! 🙂

Glad it all worked out! Thanks for the support, too! 🙂

Happy to answer that! You'll want the number of columns to be one more than the total number of variables present in the system. This extra column will be populated with 0's, which represent the constant values of the homogeneous linear system. As an example, if you have 3 equations in 3 variables, you'll define the "size" of your matrix as 3 x 4. This 4th column should contain 0's only. Then you can cycle through the rref process using the graphing calculator. :-) By the way, thanks for your question as well as support of the channel, it's greatly appreciated!

No problem! These buttons represent the different "lists" that are available for you to edit after pressing [STAT], then option [1] "edit". If for some reason these list headings are missing, let me know and I'll send you some tips on how you can get them back. In the meantime, thanks for your questions as well as your support of the channel! :-)

Glad to hear the video helped out! And thank you for the support! 🙂

Thank you for the support, and I'm happy to hear it helped you out on the midterm! 🙂

You're very welcome! Thanks for the support! 🙂

Thanks for the positive feedback, and I'm so glad you like the video! Your support is greatly appreciated! Haters gonna hate, but, it's no sweat off my back. 🙂

Sorry it didn't meet your standards. Thanks for the feedback either way! 🙂

Happy to help! Thanks for the support, too! 🙂

Thanks for question, I'm happy to clarify! Note that the vector (-5,-1,1) would help establish the line's "direction". Imagine connecting the origin to the points (-5,-1,1), and then tracing a line through the two points. Now, since this vector would lie along this line, they're technically parallel to each other. 🙂 On that note, indeed, all possible scalar multiples of the vector (-5,-1,1) would generate the line as well. Thus, your description would be equivalent to mine; so, nice observation!

You're welcome! And thanks for your support of my channel! 🙂

Glad to hear it helped! 🙂

Glad to hear that! Thanks for the support! 🙂

No problem! Thanks for the support! 🙂

You're so very welcome! I hope your test went well! And thanks for your support! 🙂

Happy to hear it helped out! Thanks for your support! 🙂

Thank you so much for the kind remarks, I'm so happy to hear you find them helpful! 🙂Feel free to share them out as you see fit!

Thanks for the feedback, and happy to hear it helped out! 🙂

Wow--thanks so much for the kind remarks! I'm so happy to hear you're benefiting from them! 🙂

Thanks for the feedback--it's much appreciated! #slay 🙂

So glad the video helped out! Indeed, best of luck towards a successful Linear Algebra final! You got this! 🙂

Thanks for your comment! I hope to eventually put some Math 122 videos on my channel. 🙂

Glad to hear it helped out! 🙂

Awesome! Thanks for watching! 🙂

Thanks for sharing this! Keep up on that grind, though--it will definitely pay off! 🙂

Glad you asked! You would first find the general solution for the homogenous linear system Ax=0, as demonstrated in the video. Next, you'll find the specific solution to Ax=b and simply add this to the general solution from Ax=0 that you first found. :-) Let me know if this helps out. Thanks!

Got it, thank you!

@@samr3100 Happy to help! 🙂

Since the basis has only 1 basis vector, it has a dimension of 1; hence, the object would be a line instead of a plane. :-)

Happy to hear you liked my trick! :-)

If the remainder is zero, then you've found a real zero of the function. :-) Note that this video addresses the boundaries in which any real zeros will be between (on the x-axis). Let me know if this helps clarify, or if you have any other questions--I'll be happy to help out! Also, thanks for watching!

Many many thanks! :-)

So happy to hear this. Being helpful is what I strive for. 🙂Thank you for watching!

No problem!

stonks! :-)

No problem! Thanks for watching! :-)

No problem! :-)

Thanks for watching!

No problem! :-)

No problem! Thanks for watching!

So sorry... :-(