I am learning math to Machine Learning, and your video sealed the deal on the subject. I have been reading so much and your video simplified it, and is perfect complement for a challenging day. Thank you!
putting sign before matrix is okay in the form u were using i.e [a][x] = [b] ~ ax = b because product of matrix [a] & [x] is equal to matrix [b] but if ur using augmented form [ x y | b] then seperation is needed to r emember which is the constant however it is not compulsory to use it and can be written [x y b]
I have problem for you Prove or disprove that trace of A^m is equal to the sum(λ_{k}^{m},k=1..n) where λ is a scalar which satisfies following equation Ax = λx assuming A is square matrix and x is nonzero vector Trace of A is the sum of elements on main diagonal Here m is nonnegative integer I need it for justifing one thing in my video In my opinion this problem should appear in the video series but later
@@PrimeNewtons I suspect that this claim is true and I'm looking for simple, easy to understand proof without any knowlege from advanced algebra course
Yes, you can choose either x₁ or x₂ as the free variable. Here's why: x₃ is bound: Since x₃ is bound to another variable, it cannot be a free variable. x₄ is free This variable is already designated as free, meaning its value can be chosen independently. Equation's constraint: The equation itself imposes one constraint on the variables. Since x₄ is already free, we have one remaining degree of freedom to choose another free variable. Therefore, either x₁ or x₂ can be chosen as the free variable. Once you choose a value for one of them, the value of the other will be determined by the equation to satisfy the constraint. Example: If you choose x₁ as free: You can assign it any value you want. Let's say x₁ = 3. Then, the equation becomes: 3 + x₂ - 1 + x₄ = 0 Solving for x₂, you get: x₂ = -2 - x₄ Now, x₂ is determined based on your choice of x₁ and the value of x₄. If you choose x₂ as free: The same logic applies. You choose a value for x₂, and then x₁ is determined by the equation.
Sir i think its not the final answer..i think we let the free variables s and t... and then became equation...nd then put any constant in the place of s amd t then final answer we get...
![Parametric Vector Form and Free Variables [Passing Linear Algebra]](http://i.ytimg.com/vi/OnCLWE-ohSE/mqdefault.jpg)
You're an incredible instructor. Thank you for uploading these. I'm taking linear algebra at university and this is much clearer than my textbook
This video is awesome. your good mood is contagious
"FREE. Free like a bird." I'm always smiling at the aha moment when it all makes sense. THANK YOU
Thank you, I had to see lot of vid to understand free variables and at last i found ur vid . it was such a nice exp.❤
your mood is so contagious and your explanation couldn't be easier!!
I am learning math to Machine Learning, and your video sealed the deal on the subject. I have been reading so much and your video simplified it, and is perfect complement for a challenging day.
Thank you!
this was explained much better than the way my professor explained it.
Easier to understand than lectures in my language!!
You are such a nice teacher
Thank you! Great explanation
Such an Amazing Video !!!!!
It was really good video and helpful.
Exellent explanation
putting sign before matrix is okay in the form u were using
i.e [a][x] = [b] ~ ax = b because product of matrix [a] & [x] is equal to matrix [b]
but if ur using augmented form [ x y | b] then seperation is needed to r emember which is the constant however it is not compulsory to use it and can be written [x y b]
Thank you sir you helped so much
I have problem for you
Prove or disprove that
trace of A^m is equal to the sum(λ_{k}^{m},k=1..n)
where λ is a scalar which satisfies following equation Ax = λx
assuming A is square matrix and x is nonzero vector
Trace of A is the sum of elements on main diagonal
Here m is nonnegative integer
I need it for justifing one thing in my video
In my opinion this problem should appear in the video series but later
I'll look into it.
@@PrimeNewtons I suspect that this claim is true and I'm looking for simple, easy to understand proof without any knowlege from advanced algebra course
Tnx Love you sir❤❤.
Linear algebra is so much harder than calculus! 😂
🤣
GREAT!
Oga sir ! Well done sir , so is this the finally answer?
Yes sir!
why is putting = sign before the matrix incorrect?
Could I decide to make x_1 free instead of x_2? right?
Yes, you can choose either x₁ or x₂ as the free variable. Here's why:
x₃ is bound: Since x₃ is bound to another variable, it cannot be a free variable.
x₄ is free This variable is already designated as free, meaning its value can be chosen independently.
Equation's constraint: The equation itself imposes one constraint on the variables. Since x₄ is already free, we have one remaining degree of freedom to choose another free variable.
Therefore, either x₁ or x₂ can be chosen as the free variable. Once you choose a value for one of them, the value of the other will be determined by the equation to satisfy the constraint.
Example:
If you choose x₁ as free: You can assign it any value you want. Let's say x₁ = 3. Then, the equation becomes:
3 + x₂ - 1 + x₄ = 0
Solving for x₂, you get:
x₂ = -2 - x₄
Now, x₂ is determined based on your choice of x₁ and the value of x₄.
If you choose x₂ as free: The same logic applies. You choose a value for x₂, and then x₁ is determined by the equation.
@@user-uy3qf5sd6l we could also choose x3 or x4 to be the free variable right ?
superiyk bro
Sir i think its not the final answer..i think we let the free variables s and t... and then became equation...nd then put any constant in the place of s amd t then final answer we get...
love from india sir🫡🫡🫡
hay please check this out .😎😎💖💖
didnt know black people were cracked at teaching math