You're very welcome! I'm so glad to hear that the video was helpful to you. It means a lot to me to know that my content is making a difference. If you have any more questions, please don't hesitate to let me know. Thanks for watching and for your kind words 😇
You're very welcome! I’m glad the video was helpful. If you’re interested in more practice, feel free to check out my other video on a similar topic: "Find the values of k which give one Solution, no Solution, or infinitely Many Solutions", and here is the video link: ua-cam.com/video/61vt1nDHOKo/v-deo.htmlsi=3Dt2q1Q91RAzlJeq I'm planning to create more videos like this in the future, so stay tuned! 😇
hi sir, your playlist in linear algebra is the BEST in youtube , but i just want to ask about step in this video 5:23 I think we should also do the R1-2R2 too to make the 2 above the leading 1 in the second column 0 too right?
Glad to hear that 😇 We only care about the value of k in the last row, so you don't have to do it because it will not affect the answer. Also, you can't make the above (k^2)-4 zeros (can't make the rest of the third column zeros) unless k is given.
If there were no k's in the 3x1 right hand matrix, and so you end up with k + or - some constant in the left hand matrix in the reduced echelon form is just equal to another non-zero constant, does that mean it can only have infinitely many solutions or no solutions, and there is no way it could have a unique solution?
If there is no k on the right-hand side ( it means that the last row will be [-4 -5 (k^2)-9 | 1] ), then there is no way to have infinitely many solutions since you cannot make a zero row for any value of k ( e.g. [0 0 0|0] ).
Usually, the variable k is placed in the 3rd row. However, placing k in the 1st column doesn't change the solution set. This is similar to how the equation x+y+z=1 is equivalent to z+y+x=1. To illustrate this, let's take the linear system: x + 2y + z = 1 x + 3y + 2z = 1 ((k^2)-9)x - 5y - 4z = k + 1. You can rearrange this system by swapping the 1st and 3rd columns, which leads to another system of equations with the same solutions. Note that in this rearranged system, the solution vector would be x=[z;y;x] instead of x=[x,y,z]: z + 2y - x = 1 2z + 3y + x = 1 -4z - 5y + ((k^2)-9)x = k + 1. This gives the same solutions as discussed in this video. If you have further questions or would like additional illustrations, don't hesitate to ask & Good luck!
When you start solving a problem using the Reduced Row Echolen Form (RREF), you cannot swap columns. As you can see from the name of the form, the Reduced 'Row' Echolen Form is designed to work only between rows. As an example, consider the following equations: x - z + 2y = 1 2x + z + 3y = 1 -4x + ((k^2)-9)z - 5y = k + 1 You can then rearrange the equations as follows: x + 2y - z = 1 2x + 3y + z = 1 -4x - 5y + ((k^2)-9)z = k + 1 All we have changed is how we have arranged them from the beginning, then the steps will still be the same. In the end, you just care about k values.
For this problem, you can use row echelon form to determine the type of solution. Reduced row echelon form is also an option, but it’s not necessary unless you want further simplification. Good luck, and feel free to ask if you have any more questions 👍
My brother you are the man!!... Thanks you so much for the video brother. A similar problem to that one has been giving me a headache since morning. Thanks man !
Thank you so much for your kind words! I'm really glad the video was able to help you out with that problem 😇 If you have any more questions or need further explanations, feel free to ask. Thanks for watching, and keep up the great work 👍
Of course, I'd be happy to help! Please share your question here, and I'll do my best to provide a clear explanation. Additionally, I'll consider addressing these types of questions in more detail in future videos. Looking forward to assisting you!
Thank you so much.... You don't know how helpful this video is.... Highly recommend to friend circle.. Already shared and subscribed!!!
Glad it helped! thanks, and good luck 😇
absolute life saver! i have a test in linear algebra after 2 days and this made me understand the concept easily and in a clear way.
Glad it helped & Good luck!
It's great seeing young Arabs entering the English side of educational UA-cam videos.
thank you for explaining it. you don't know how much help you have done
You're very welcome! I'm so glad to hear that the video was helpful to you. It means a lot to me to know that my content is making a difference.
If you have any more questions, please don't hesitate to let me know. Thanks for watching and for your kind words 😇
You are a real one for making this man! Thank you so much!
I appreciate that & you are most welcome
Wherever you’re man God bless you....This is a great video...
I appreciate your comment and I am delighted to hear that, as well as wishing you all the best 🙏
Thank you very much. You explained it very well and clearly, so it helps me to understand it very well. Thank you again.
Glad to hear that & Good luck!
Tysm. The book decided to not explain this but still put excersises about this topic. I also have a test tomorrow so you're an absolute life saver👍
You're very welcome! I’m glad the video was helpful.
If you’re interested in more practice, feel free to check out my other video on a similar topic:
"Find the values of k which give one Solution, no Solution, or infinitely Many Solutions", and here is the video link:
ua-cam.com/video/61vt1nDHOKo/v-deo.htmlsi=3Dt2q1Q91RAzlJeq
I'm planning to create more videos like this in the future, so stay tuned! 😇
Actually understood the concept. Thanks!
Great! & Thanks for watching 😇
Thank you 😊 I'm writing on this concept this afternoon
All the best & thanks for watching 😇
hi sir, your playlist in linear algebra is the BEST in youtube , but i just want to ask about step in this video 5:23 I think we should also do the R1-2R2 too to make the 2 above the leading 1 in the second column 0 too right?
Glad to hear that 😇
We only care about the value of k in the last row, so you don't have to do it because it will not affect the answer.
Also, you can't make the above (k^2)-4 zeros (can't make the rest of the third column zeros) unless k is given.
Thanks sir...for discussing such example..it cleared remained doubts
Glad to hear that 😇
Man thank you so much, you've explained it so well💪💪
Glad you liked it, thank you for watching & goodluck 😇
in just 13 mini l have understood the topic thank you
Glad to hear that & thank you for watching 😇
Thank you! Well explained!
Thanks for watching!
thank you very much....this was very helpful
Glad it helped & you're most welcome 👍
If there were no k's in the 3x1 right hand matrix, and so you end up with k + or - some constant in the left hand matrix in the reduced echelon form is just equal to another non-zero constant, does that mean it can only have infinitely many solutions or no solutions, and there is no way it could have a unique solution?
If there is no k on the right-hand side ( it means that the last row will be [-4 -5 (k^2)-9 | 1] ), then
there is no way to have infinitely many solutions since you cannot make a zero row for any value of k ( e.g. [0 0 0|0] ).
Thank you so much. Greatly appreciate this video
You are welcome, and glad it was helpful! Best wishes 😇
Thanks a lot, but please tell me how to solve the equation a^2-5. I couldn't find a solution.
Thank you for your comment! Just to clarify, did you mean k^2−4 instead of a^2-5?
Let me know, and I'd be happy to help 😇
Really good explanation!Hoping for more videos
Thank you so much for your kind words! I'm glad you found the explanation helpful. Stay tuned, more videos are on the way soon😊
Thanks for overcome my pressure
Any time & glad to heart that!
thank you so much ❤
You're welcome & thank for watching 😊
What do you do if the k variable is in the 1st column instead of the 3rd?
Usually, the variable k is placed in the 3rd row. However, placing k in the 1st column doesn't change the solution set. This is similar to how the equation x+y+z=1 is equivalent to z+y+x=1.
To illustrate this, let's take the linear system:
x + 2y + z = 1
x + 3y + 2z = 1
((k^2)-9)x - 5y - 4z = k + 1.
You can rearrange this system by swapping the 1st and 3rd columns, which leads to another system of equations with the same solutions. Note that in this rearranged system, the solution vector would be x=[z;y;x] instead of x=[x,y,z]:
z + 2y - x = 1
2z + 3y + x = 1
-4z - 5y + ((k^2)-9)x = k + 1.
This gives the same solutions as discussed in this video.
If you have further questions or would like additional illustrations, don't hesitate to ask & Good luck!
great problem and solution!
Glad you enjoyed it 😇
May I ask if the K is in column 2, can I just swap the entire column 2 with column 3?
Thanks alot !
When you start solving a problem using the Reduced Row Echolen Form (RREF), you cannot swap columns. As you can see from the name of the form, the Reduced 'Row' Echolen Form is designed to work only between rows.
As an example, consider the following equations:
x - z + 2y = 1
2x + z + 3y = 1
-4x + ((k^2)-9)z - 5y = k + 1
You can then rearrange the equations as follows:
x + 2y - z = 1
2x + 3y + z = 1
-4x - 5y + ((k^2)-9)z = k + 1
All we have changed is how we have arranged them from the beginning, then the steps will still be the same.
In the end, you just care about k values.
@@Mulkek this comment literally made me pass linear algebra thank you!!
Great video!
Glad you enjoyed it
Thanku ❤️
Glad to hear that & thanks for watching 😇
I know this is a kind of old video, but just want to say this really helped me. You're a great teacher. Liked and subscribed.
I appreciate that & Glad it helped
Amazing 🤩🤩
Thanks 🤗
finally understood this, thank you
I'm so glad to hear you've finally understood it! You're very welcome. Keep up the good work, and if you have any more questions, just let me know 😇
Thanks alot.
Most welcome!
Thanks a lot sir❤
Most welcome 😇
we need to do row echelon form or reduced?
For this problem, you can use row echelon form to determine the type of solution. Reduced row echelon form is also an option, but it’s not necessary unless you want further simplification. Good luck, and feel free to ask if you have any more questions 👍
thank you
You're so welcome & Good luck👍
You literally saved me😭😭
Thanks
You are welcome & thanks for watching 😇
its supposed to be k^2 - 8 not "-4" but thanks for the explanation
Could you explain more about what did you mean by k^2 - 8 not "-4" (e.g. you can tell the time)? & thanks for watching 🙂
My brother you are the man!!... Thanks you so much for the video brother. A similar problem to that one has been giving me a headache since morning. Thanks man !
Thank you so much for your kind words! I'm really glad the video was able to help you out with that problem 😇
If you have any more questions or need further explanations, feel free to ask. Thanks for watching, and keep up the great work 👍
sir can you help me with a certain question
Of course, I'd be happy to help! Please share your question here, and I'll do my best to provide a clear explanation. Additionally, I'll consider addressing these types of questions in more detail in future videos. Looking forward to assisting you!
Don Deezy
Thank you
Thank you for watching & Good luck 😇
Thanks
You're so welcome
Thank you
Thank you for watching 😇