It's not really trick it's more of just taking advantage of the row or column with the most zeros and understanding how the cofactor technique works for finding the determinant of a matrix.
@Golollisa_barii The reason to start with 2 is because it's the row that contains the most zeros and since the other entries in that same row are zero then we know automatically that the corresponding terms will be zero.
I should look at math lectures from UA-cam more often. I'm not sure why but they tend to be a lot more effective than the slideshows, textbooks, and lectures my course offers.
I believe videos can be helpful, but reading and understanding material from a book forces the mind to engage more deeply, resulting in better retention. When I was in college, we didn't have videos, so we had to thoroughly read the material. If we didn't understand something, we would re-read it or look for alternative explanations in other textbooks, often spending a lot of time in the library. Otherwise, we had to visit the professor during office hours, hoping they could provide a hint, as they definitely wouldn't solve it for us. Struggling with a problem or concept is a crucial part of learning math. That's why I always tell my students to try to solve the problem before looking at the detailed solution. It's fine if they don't get it right on the first attempt; they just need to review their work, think about it, and try again.
@@DoctrinaMathVideos Yep, looking at online lectures is a last resort. I said that I should do it more often because sometimes I am stubborn about that to a fault and don't use anything other than what the course gives to me for far too long.
The sign of the cofactor is (-1)^(i +j), so since 2 is located in the second row (i = 2) and third column (j = 3) then the sign will be (-1)^(2 + 3) = (-1)^5 = -1. Here is the sign chart (same one that is used in the video): tinyurl.com/cofactorsigns
If we perform coloum transformation and make all zero in row 1 expect one non zero values and again do the same for 2 row and find the determinant for 3x3 matrix and so then can we get the answer ?
Yes, but keep in mind that certain elementary row operations will change the determinant of the original matrix. This is discussed in my notes: blogs.nvcc.edu/mwesterhoff/files/2019/09/Section3.2-Properties-of-Determinants.pdf
It's negative because we have -(0 + 12 + 30) + (16 + 0 + 25) = -12 - 30 + 16 + 25 = -42 + 41 = -1. For the last part I am using a technique for finding the determinant of the 3 x 3 matrix which is widely used in engineering. The shortcut method is discussed here: www.algebrapracticeproblems.com/how-to-find-the-determinant-of-a-3x3-matrix/
When finding the determinant using cofactor expansion we can choose any row or column so a natural choice is to choose the row or column that contains the most zeros so we can avoid having to find other determinants. I like to think of it as choosing the path with least resistance.
Hello Steven, since 3 is in the second row and second column the sign will be positive. Here is the sign chart (same one that is used in the video): tinyurl.com/cofactorsigns
Hello, the negative sign is there because you subtract the set of numbers (diagonal lines that go from the top right to bottom left) from the other set of numbers (diagonal lines that go from top left to bottom right). This is part of the formula for applying the "shortcut" for finding the determinant of a 3x3 matrix.
Hello, that is a good question. What you would do in this case is to apply the cofactor expansion to that particular row or column, then apply within that sub-matrix and so on (iterative process). It's not too difficult but you have to keep track of the algebra. It's much easier to demonstrate this with an example. I received several requests to do an example of this situation and will eventually will post one on my channel. Another way to find the determinant (perhaps easier) is to put the matrix into echelon form then use the row-operations to find the determinant of the original matrix. It turns out that the most common row operation (R_i + R_j = R_j) does not change the determinant of the original matrix. Here is a specific example of this ua-cam.com/video/UR-MvWVfpQw/v-deo.html In real applications you would use a computational scientific program to calculate the determinant of the matrix. Image trying to find the determinant of a 100 x 100 matrix by hand and even on a computer this will take some time.
The general technique of finding the determinant of a matrix using co-factors will still apply however there will be more steps involved. One of the main points for this example, is that you should choose the row or column containing the most zeros. If there are no zero entries, then just choose any column or row to start the process. Keep in mind that for any matrix strictly larger than 3 x 3 it will eventually end up solving the determinant of a 3 x 3 sub-matrix (which implies you can still use the shortcut on the sub-matrix to find its determinant). The specific entries in the matrix has nothing to do with whether the technique will work or not.
This example was specifically chosen (by request) to illustrate that it is important to look at the rows and columns carefully to illustrate that the number of steps can be reduced by choosing the row or column with the most zeros.
Hello Ramamoorthy, since 2 is in the second row and second column the sign will be positive. Here is the sign chart (same one that is used in the video): mathonline.wdfiles.com/local--files/minor-and-cofactor-entries/Screen%20Shot%202014-06-08%20at%2010.02.07%20PM.png
There is no limitation for the cofactor expansion technique as long as the matrix is square. The last step where you see the diagonal lines can only be applied for 3 x 3 matrices. Instead of the diagonal lines approach you could have continued using the cofactor approach and end up with the same solution.
Once you learn the basic technique it's just an iterative process. If an application requires the use of the determinant of a 4 x 4 matrix or higher then I would highly recommend using some type of computational tool such as Matlab or Octave.
This example was specifically chosen (by request) to illustrate that it is important to look at the rows and columns carefully to illustrate that the number of steps can be reduced by choosing the row or column with the most zeros.
@@FreshDolby I see where you went wrong. You were looking at the original matrix when you should have been using the 4 x 4 sub-matrix. The number 3 is in the second row, second column, so the sign of 3 is positive.
Since there is a 2 is in the second row and third column the sign will be negative (-1)^(3+2) = (-1)^5 = -1. Here is the sign chart (same one that is used in the video): tinyurl.com/cofactorsigns
This example was specifically chosen (by request) to illustrate that it is important to look at the rows and columns carefully to illustrate that the number of steps can be reduced by choosing the row or column with the most zeros. Once you learn the basic technique of the cofactor expansion concept it becomes an iterative process. If an application requires the use of the determinant of a 4 x 4 matrix or higher then I would highly recommend using some type of computational tool such as Matlab or Octave.
The answer is 6, I would suggest reviewing your work and checking your co-factor signs. matrixcalc.org/det.html#%7B%7B4,0,-7,3,-5%7D,%7B0,0,2,0,0%7D,%7B7,3,-6,4,-8%7D,%7B5,0,5,2,-3%7D,%7B0,0,9,-1,2%7D%7Dexpand-along-column1
This 7 minute video was more helpful than 3 hours worth of physical lectures. Thank you!
Glad this was helpful.
it's true, thank you@@DoctrinaMathVideos
this trick is EXTREMELY helpful
It's not really trick it's more of just taking advantage of the row or column with the most zeros and understanding how the cofactor technique works for finding the determinant of a matrix.
Thank you so much.
This is really helpful.
Glad it was helpful!
@@DoctrinaMathVideos
Why start from 2
@Golollisa_barii The reason to start with 2 is because it's the row that contains the most zeros and since the other entries in that same row are zero then we know automatically that the corresponding terms will be zero.
Thank you man
Really helpful
Glad to hear that!
This lecture was so helpful
I am glad that you found this lecture to be helpful.
What if the second original row and coloumn was with two vaules different from zero and 3 zeros what should we do in this case ?
Just apply the co-factor expansion to each of the other columns that do not contain zero.
Thank you very much, now I understand the issue that I did not understand since morning.
Glad it helped!
I should look at math lectures from UA-cam more often. I'm not sure why but they tend to be a lot more effective than the slideshows, textbooks, and lectures my course offers.
I believe videos can be helpful, but reading and understanding material from a book forces the mind to engage more deeply, resulting in better retention. When I was in college, we didn't have videos, so we had to thoroughly read the material. If we didn't understand something, we would re-read it or look for alternative explanations in other textbooks, often spending a lot of time in the library. Otherwise, we had to visit the professor during office hours, hoping they could provide a hint, as they definitely wouldn't solve it for us. Struggling with a problem or concept is a crucial part of learning math. That's why I always tell my students to try to solve the problem before looking at the detailed solution. It's fine if they don't get it right on the first attempt; they just need to review their work, think about it, and try again.
@@DoctrinaMathVideos Yep, looking at online lectures is a last resort. I said that I should do it more often because sometimes I am stubborn about that to a fault and don't use anything other than what the course gives to me for far too long.
Thank you so much for this video
You are welcome.
Why is there a subtraction sign just before you start the multiplication?
The sign of the cofactor is (-1)^(i +j), so since 2 is located in the second row (i = 2) and third column (j = 3) then the sign will be (-1)^(2 + 3) = (-1)^5 = -1. Here is the sign chart (same one that is used in the video): tinyurl.com/cofactorsigns
If we perform coloum transformation and make all zero in row 1 expect one non zero values and again do the same for 2 row and find the determinant for 3x3 matrix and so then can we get the answer ?
Yes, but keep in mind that certain elementary row operations will change the determinant of the original matrix. This is discussed in my notes: blogs.nvcc.edu/mwesterhoff/files/2019/09/Section3.2-Properties-of-Determinants.pdf
thanks for making this video 😃😃
You are welcome.
why in the last part is - ? I did not really get the point of the last part
It's negative because we have -(0 + 12 + 30) + (16 + 0 + 25) = -12 - 30 + 16 + 25 = -42 + 41 = -1. For the last part I am using a technique for finding the determinant of the 3 x 3 matrix which is widely used in engineering. The shortcut method is discussed here: www.algebrapracticeproblems.com/how-to-find-the-determinant-of-a-3x3-matrix/
Why we mark 2 in starting....???????
When finding the determinant using cofactor expansion we can choose any row or column so a natural choice is to choose the row or column that contains the most zeros so we can avoid having to find other determinants. I like to think of it as choosing the path with least resistance.
Why 2nd coefficient value of determination is 3 not -3,in case of 4*4 matrix a22 value is negative
Hello Steven, since 3 is in the second row and second column the sign will be positive. Here is the sign chart (same one that is used in the video): tinyurl.com/cofactorsigns
@@DoctrinaMathVideos thanks for replying 💙🤗
You are welcome. :)
Hi, why is there negative sign in outside the parenthesis before zero?
Hello, the negative sign is there because you subtract the set of numbers (diagonal lines that go from the top right to bottom left) from the other set of numbers (diagonal lines that go from top left to bottom right). This is part of the formula for applying the "shortcut" for finding the determinant of a 3x3 matrix.
You are welcome.
hi what do we do if the only choice left is a row or column that has less than 4 zeros to start with?
Hello, that is a good question. What you would do in this case is to apply the cofactor expansion to that particular row or column, then apply within that sub-matrix and so on (iterative process). It's not too difficult but you have to keep track of the algebra. It's much easier to demonstrate this with an example. I received several requests to do an example of this situation and will eventually will post one on my channel. Another way to find the determinant (perhaps easier) is to put the matrix into echelon form then use the row-operations to find the determinant of the original matrix. It turns out that the most common row operation (R_i + R_j = R_j) does not change the determinant of the original matrix. Here is a specific example of this ua-cam.com/video/UR-MvWVfpQw/v-deo.html In real applications you would use a computational scientific program to calculate the determinant of the matrix. Image trying to find the determinant of a 100 x 100 matrix by hand and even on a computer this will take some time.
If we have a matrice with no zero on it I don’t think it’s gonna work
The general technique of finding the determinant of a matrix using co-factors will still apply however there will be more steps involved. One of the main points for this example, is that you should choose the row or column containing the most zeros. If there are no zero entries, then just choose any column or row to start the process. Keep in mind that for any matrix strictly larger than 3 x 3 it will eventually end up solving the determinant of a 3 x 3 sub-matrix (which implies you can still use the shortcut on the sub-matrix to find its determinant). The specific entries in the matrix has nothing to do with whether the technique will work or not.
This example was specifically chosen (by request) to illustrate that it is important to look at the rows and columns carefully to illustrate that the number of steps can be reduced by choosing the row or column with the most zeros.
Finally I'm student and i see 😅
Perfect method . Thanks
Thank you.
sir how 3 gets plus sign.....
Hello Ramamoorthy, since 2 is in the second row and second column the sign will be positive. Here is the sign chart (same one that is used in the video): mathonline.wdfiles.com/local--files/minor-and-cofactor-entries/Screen%20Shot%202014-06-08%20at%2010.02.07%20PM.png
is there a limitation to this method? i find it works sometimes and sometimes it doesn't
There is no limitation for the cofactor expansion technique as long as the matrix is square. The last step where you see the diagonal lines can only be applied for 3 x 3 matrices. Instead of the diagonal lines approach you could have continued using the cofactor approach and end up with the same solution.
Thank you so much 👍
You are welcome.
thanks so much
You're welcome!
Try calculating determinant of 5×5 matrix manually when none of its rows and columns have zero in it 😂
Once you learn the basic technique it's just an iterative process. If an application requires the use of the determinant of a 4 x 4 matrix or higher then I would highly recommend using some type of computational tool such as Matlab or Octave.
This example was specifically chosen (by request) to illustrate that it is important to look at the rows and columns carefully to illustrate that the number of steps can be reduced by choosing the row or column with the most zeros.
The sign on 3 should be negative
@@FreshDolby I see where you went wrong. You were looking at the original matrix when you should have been using the 4 x 4 sub-matrix. The number 3 is in the second row, second column, so the sign of 3 is positive.
Best méthode !
Merci!
Thank you so much sir
You are welcome.
Thank you so much❤
You're welcome 😊
why -2 not 2 ?
Since there is a 2 is in the second row and third column the sign will be negative (-1)^(3+2) = (-1)^5 = -1. Here is the sign chart (same one that is used in the video): tinyurl.com/cofactorsigns
Thank you sir
You are welcome.
Life saver
Glad you found it useful.
thanks a lot
You are welcome.
thanks man
You are welcome.
This only works for matrices with 1 row or column value and the rest is zero
I think you will have to add additional terms and include non zero row or column values
This example was specifically chosen (by request) to illustrate that it is important to look at the rows and columns carefully to illustrate that the number of steps can be reduced by choosing the row or column with the most zeros. Once you learn the basic technique of the cofactor expansion concept it becomes an iterative process. If an application requires the use of the determinant of a 4 x 4 matrix or higher then I would highly recommend using some type of computational tool such as Matlab or Octave.
@@DoctrinaMathVideos Tell that to my linear algebra professor who expects us to do the determinant of a 5x5 by hand without a calculator! 😭😭
Been there done that, but I do find it kind of ridiculous. Better hope that there is a column or row with some zeros in it. :)
Thnx
You are welcome.
Thankyou sir
You are welcome.
Sir final answer will be -6
The answer is 6, I would suggest reviewing your work and checking your co-factor signs. matrixcalc.org/det.html#%7B%7B4,0,-7,3,-5%7D,%7B0,0,2,0,0%7D,%7B7,3,-6,4,-8%7D,%7B5,0,5,2,-3%7D,%7B0,0,9,-1,2%7D%7Dexpand-along-column1
final is in 2 days heh thanks!
Good luck!