🎯 Key Takeaways for quick navigation: 00:57 🔄 *A formula for the inverse of a 2x2 matrix is introduced: A inverse equals 1 over the determinant times the transpose of the matrix of cofactors.* 03:45 🤔 *The general formula for the inverse of an n x n matrix is established: A inverse equals 1 over the determinant of A times the transpose of the cofactor matrix of A.* 07:55 🔄 *The correctness of the inverse formula is checked by verifying that A times A inverse equals the identity matrix.* 19:17 🔄 *The solution to the system Ax=b is expressed as x equals A inverse times b, with A inverse given by the established formula.* 20:40 🧮 *Cramer's Rule is introduced, expressing each component of the solution vector x in terms of determinants of matrices obtained by replacing columns of A with the vector b.* 25:39 ⚠️ *Cramer's Rule is acknowledged as theoretically interesting but impractical for computation due to its reliance on computing multiple determinants.* 28:24 📐 *The determinant is revealed to be related to volume, setting the stage for discussing how determinants represent volume in specific cases and then generalizing to a broader understanding.* 28:51 📏 *Determinant of a matrix in three dimensions equals the volume of a parallelepiped (box), with each row forming an edge.* 31:37 🔄 *The volume of the box, given by the determinant, may be negative, indicating a change in handedness (right-handed to left-handed).* 33:04 📐 *For the identity matrix, the determinant equals the volume, representing a unit cube.* 35:24 🔄 *Orthogonal matrices, when used as transformation matrices, also represent cubes with a volume of 1 but may be rotated in space.* 38:38 🔄 *Determinant of an orthogonal matrix is either 1 or -1, ensuring that the volume formula remains valid.* 40:29 📐 *Volume (determinant) behaves linearly, doubling when an edge is doubled (property 3a).* 43:20 📐 *The determinant formula for area applies to parallelograms, simplifying the calculation to ad-bc.* 45:45 📏 *The determinant formula for area is ad-bc, providing a straightforward and general formula without square roots or complex calculations.* 47:38 📐 *The area of a triangle is half the determinant of the matrix formed by its vertices, a simple extension of the parallelogram formula.* 49:29 🔄 *Shifting the triangle's vertices does not change the determinant-based area formula, emphasizing its versatility and simplicity.* Made with HARPA AI
Really understanding inv(A) = 1/det(A) * CT because of prof. Strangs explanation is an amazing thing; it just hit me, epiphany style. Just marvellous teaching.
The author of Mathematics for the Million was Lancelot Hogben in 1936. Eric Temple Bell wrote Men of Mathematics in 1937. For decades, both were very popular books on math for the general audience back when computational cost meant paying people to crank mechanical devices and write down the answers. Now a smartphone can invert a giant matrix in the blink of an eye.
(21:29) "So, anytime I multiply cofactors by some numbers, I'm getting the determinant of *_something_* " - Important point in this lecture!! And (14:16) also the same idea("...determinant of *_some_* matrix...").
The readings are assigned in: Strang, Gilbert. Introduction to Linear Algebra. 4th ed. Wellesley-Cambridge Press, 2009. ISBN: 9780980232714. See the course on MIT OpenCourseWare for more info and materials at: ocw.mit.edu/18-06S05. Best wishes on your studies!
It’s a great book! I have it (might be a different edition), and it’s the best intro to linear algebra text I’ve ever seen. It’s just like these lectures: clear and intuitive and dispels for the learner all myths that linear algebra is abstruse and unapproachable.
...... i'm having math's exam tomorrow and i Optimistic that i can get a good mark after watching this ... this guy is knows how explain and justify !!!!!!!
It took me some time to understand the A screwed up thing, but as soon as I got the idea, it turned out to be a good explanation! I'm always impressed by his teaching style.
@@mohdfaisalquraishi8675 Think of this as a new matrix whose cofactor includes the first row. And the first row also act as factor of the cofactor above which contain itself. Hope it helps!
@@mohdfaisalquraishi8675 bro think about more dimensions instead of 2x2 and its gonna be more clear. when dont get the proper cofactors for that guy u always get singular
Cramer Rule is good for solving 2 by2 and 3 by 3 linear equations. Dr. Strang it seem me to that you are turn off by Cramer rule, however your lecture on the subject was another masterpiece. The volume of a box is also explained very well using determinants.
I drew a little sketch of a parallelogram and verified the area does equal the determinant of the matrix by finding the area of the big rectangle and subtracting off areas of neighboring triangles and rectangles, but I still don't have a good intuitive sense of WHY the determinate should give you this area.
Professor mentioned a book named 'Mathematics for the Million' which he had read in childhood However, the book is written by 'Lancelot Hogben', not by the guy named 'Bell'. =D (If you are interested, check 'THE ALGEBRA OF THE CHESSBOARD' chapter of book archive.org/details/HogbenMathematicsForTheMillion/page/n527/mode/2up )
I hope that some future use and simplification of Cramer's formula can be made. I would hate to think that Cramer wasted his time and that the knowledge was useless.
I have a trivial doubt, 23:38 professor says x1 = (b1.c11)/detA But won't x1 = b1.(c11 + c21 + c31 +.....cn1)/detA by definition transpose(C) b = x.detA
can someone explain why the determinant of the last matrix he drawn, the one with 111 on the rightmost works? I know it works by computing, but does it has to do with the volume with the new three columns?
No, since there is already a C transpose, we replace columns of A, or rows of A transpose, to obtain the same expression as the sum. Work it out and see.
I connected the properties of a determinant (1-7) to the volume of a box. Now, can anyone please explain the connection of properties 8,9,10 to the volume of the box?
I think that if you have the base 3 rules applicable, then they are in the same category( or group, I am not quite good with abstract algebra). It is the same issue as it was in the previous sections where a vector space was created from functions and 3x3 matrices. Things just fall into place. Now visualizing it may be a hassle but it is not something that doesn't folllow from the rules.
Think about what the matrices do to a unit box. Then the determinant becomes the "change in area". Det(AB) = Det(A)Det(B) means if A scales by a, and B scales by b, then AB scales by ab.
8) det(A) = 0 -> A is singular When det(A) is 0, the transformation A collapses some dimension of the space, meaning there is no volume left. This collapse causes the transformation to lose information (it is no longer one-to-one) and so it is singular and non-invertible 9) det(AB) = det(A) * det(B) If you start with a box of unit volume, applying the transformation AB scales the volume of the box by the same amount as applying B first and then A. 10) det(A^T) = det(A) This one is a bit harder to grasp intuitively. I can only do it using a certain visualization of covectors. If someone else has an easier intuition, I'd love to hear it.
Thought there would be some Bj jokes 24:42 , but then again it's a MIT math class taught by Gilber Strang, maybe most are too absobed by his awsome lecture for that haha.
I recommend drawing a picture. Put the points (a,b), (c,d), and (a+a',b+b'), and let A=(volume of parallelogram given by (a+a',b+b') and (c,d)). Let A_1=(volume of parallelogram given by (a,b) and (c,d)), A_2=(volume of parallelogram given by (a',b') and (c,d)), so A=(A_1)+(A_2). Property 3b follows immediately because the LHS is just A, while the RHS is just A_1 plus A_2. It's harder to draw and visualize in higher dimensions but it's the exact same concept.
It took me some time to convince myself why As has two identical rows. The example only shows an example of 2x2 matrix, but the question of why is still there. Someone helps explain this? I have a thought on this. We know that the reduced row echelon form of A has no more than 1 of "1" at each row and cols. If we multiply a row of A (let's say the 1st row) with a col of C^T (let's say 2nd col), we get 0 because each co-factor matrix in 2nd col of C^T (e.g. C21) always consist a zero vector. (C21 consists of zero vector of col 1 of A). Thus, their product is 0. Is it the right thought?
Row 2 of A times the cofactor of row 1. Then Row 2 is the factor of the cofactor of row 1. So Row 2 is copied into row 1. And the cofactor of row 1 has every row expect the first. So row 2 is in it. So you got 2 identical rows.
Can anyone explain further why appending the 1s to the triangle matrix at the end shifts the triangle back to the origin? Also, if possible a little elaboration on why the cofactors columns equal 0 when multiplied by a different row number fro it's column number?
ranvideogamer Appending a column of 1s (let's call it M), technically, does not really shift the triangle back to the origin. Rather, by appending a column of 1s, you get a determinant whose value is the volume. Refer to Figure 5.1 in [Strang 4th ed, 272] for a geometric rendering of a *general triangle* (no corner at origin) made up of 3 *special triangles* (each with a corner at the origin). The area of a *special triangle* is clearly (ad-bc) for each of those triangles. det(M) turns out to be the sum of the areas of those 3 *special triangles*. The easiest way to find det(M) is to take cofactors of column 3 (the 1's column). Seen geometrically (as a sum of 3 *special triangles*), the constant in the third column has to be 1. The textbook also clearly explains that the area of a *special triangle* can be calculated as det(M) where the third corner/vector is the origin (0,0). In this case, you can take cofactors of row 3 (with 2 zeros and a 1), for the easy calculation to give you (ad-bc).
You could just shift all the coordinates back to the origin by subtracting (x₁, y₁) from the three coord, then take the determinant and you will get +/- the same number as his determinant with some simple/tedious algebraic multiplying out. Or alternatively with a bit of geometric imagination, imagine the parallelepiped that the determinant (or rather its transpose) with 1s determines the volume of, and seeing that one end is in the x-y plane and the other is parallel to it but 1 unit away, the volume of that 'box' is the same as the area of the parallelogram!
At 38:30, Strang says that det(Q)^2 = 1 means that det(Q) is 1 or -1...but that's not true if det(Q) is allowed to be complex. Twenty lectures into this course and I have to mentally question everything so far to see what does and doesn't apply to the complex numbers.
8 ปีที่แล้ว +2
+Lobezno Meneses as long as every element in Q is real, there's no way det(Q) would be complex. If the pivots are real, det(Q) is also real because det(Q) is the product of the pivots.
98% of what he says applies to complex matrices as well. the only things to be modified are orthogonality definitions as well as transposition definitions.
@MIT OpenCourseWare , at 18:20 when we take Det(A) out from every row shouldn't it result into Det(A)^n like it happened in last chapter for 2^n volume increase case?
We would need to take out det(A) from each row only if we were taking the determinant of the matrix with det(A) in the diagonals. But in this case we are aren't taking the determinant of the whole matrix itself, we're just multiplying I by det(A) and equating it to A times C^T.
I've been watching in 1.75 and feel it is just the right pace. I usually watch vids in x2 but he is going too fast for me to keep it at x2 at all time and I don't want to constantly switch speed. But I agree it would be nice to have a dimer instead of fixed speed. In any case, watching it in higher speed definately makes him more dynamic which is quite refreshing given his awsome personality haha.
@@neurolife77 I always watch his lectures at 1x. I'm doing the course and I have to stop frequently to take good notes and such. I sometimes watch other things at higher speeds, but not often. You're right, a speed slider isn't such a bad idea
@@arjundevendrarajan Really is. I'm starting lecture 30 in a few minutes, almost done with the course, and its amazing how much 1x and note taking has helped me understand. I studied electrical engineering (before I dropped out lol) and linear algebra was never in the syllabus. Its scary to think an engineering student can graduate without appreciating the beauty that is LINEAR ALGEBRA. Best wishes to you, friend
@@ozzyfromspace JUST found out. There is one! You just have to go in settings, speed and there is a "personnalized" button on the top right of the window that is opened. It allows you to adjust the speed with 0.05 of precision. :D
Could someone help me understand "where does the 2 by 2 matrix inverse formula come from?" 1:33 I don't think this formula was taught previously. In my memory, only the Gauss-jordan method were taught to get inverse of a matrix.Thank youuu!
You're right, it hasn't been taught from the previous lectures, only Gauss-Jordan. Prof Strang laid out the formula of the inverse for a 2x2 matrix then generalized it then gave an explanation of why that is
@@danielnarcisozuglianello8281 I actually didn't understand your point here. From the cofactor formula we actually get det(A) in each diagonal element which makes det(A) come out from each row. I know this is wrong but just cant understand where I am going wrong.
det(A)*I (a determinant of a matrix) IS equal to (det(A))^n*det(I) = (det(A))^n*1 (a scalar), NOT (det A)^n*I (a matrix)... det(A)*I = [ [detA,0,0...], [0,detA,0,0,0]...[0,0,...detA])], which is a diagonal matrix and hence det(A)*I = det(A)^n, a scalar, not a matrix!
@@prso8594 Not sure if I am too late, but Daniel was right. The concept will become easier if you substitute det(A) with actual numbers to the diagonal matrix in 12:31. Say det(A) = 5, then a 2 * 2 diagonal matrix is [5 0 0 5], It is the same as "the identity matrix I = [1 0 0 1] times the scalar 5 (i.e., det(A))" rather than I * 5^2 (i.e., det(A)^2). The action "get det(A) in each diagonal element which makes det(A) come out from each row" is right only if you want to calculate det(det(A)*I). (Note that det(A)*I is a matrix, not a scalar, since det(A) is a scalar and I is a matrix.) Take the matrix mentioned above again, its determinant is 25, which equals to det(A)^2 * det(I) = 5^2 * 1.
@@jinzhonggu8276 Determinants always compute the volume of the "parallelogram" in n-dimensions. In this case, the column of 1s just allows you to translate the "parallelogram" by an arbitrary vector, so you end up with the volume in (n-1)-dimensions. He showed the 2D case, where the 3x3 matrix with a column of 1s computes the area of a parallelogram (except he cut it in half and called it a triangle). So the 4x4 case computes the volume of a parallelopiped in 3D. The 5x5 case computes the hyper-volume in 4D, and so on. If you want a specific geometric figure for the 4x4 case: you can divide the determinant by 1/6th to get the volume of the tetrahedron which has the 4 points used to construct the matrix as it's vertices.
@@canned_heat1444 this was not explicitly derived in last lecture so I disagree. However, this resource seems to provide the answer based on the adjoint matrix www.sosmath.com/matrix/inverse/inverse.html
+Sizhan Shi: He does it in a sneaky way: the properties that Strang labelled 1,2, 3a and 3b define the determinant, so what he does is prove that they also define the volume (up to a minus sign from property 2). If volume and the determinant follow exactly the same rules, then the determinant and the volume have to be the same. So he doesn't need to directly address the problem of non-right angles as long as he shows that, however it works out in the nitty gritty, it does the same as the determinant.
The paralleloopiped squeezing and emission dynamics producing electric signals in a way elongated diagonally and the volume is getting reduced to zero but the piezoelectric current getting increased every time a pressure is applied as you say screwing by one degree producing topological effect of producing superconductivity in special materials may be elaborated for understanding superconductivity please. Sankaravelayudhab Nandakumar on behalf of Hubble Telescope Research Unit for discussion on Superconductivity by screwing for Strangler explanation.
By increasing the row but decreasing the column my volume getting decreased the square becomes a rectangle parallellogram diagonalised volume decreased piezo electricity flows parallelgram getting squeezed for reduction in volume but critically screwed at one degree topological screwing to produce superconductivity in certain materials requiring some information from the Hon.Prof Strangler please. Sankaravelayudhan Nandakumar on behalf of Hubble Telescope Research Committee.
"our little matrix" - a Gilbert Strang production
15:5 "Let me call that matrix A....s, A Screwed up" Great professor :D
He's a legend !
this is hilarious!!!!
lol
FUCK YOU
27:02 I didn't like Cramer's rule since my school days, others did like it. It's so satisfying that he thinks the same about it.
SHUBHAM AGRAWAL same here, that’s tedious
Now i can say i have a "strang" understanding of linear algebra
hahah partially true, he definitely know more than ya!
At 10:30 I actually went:"Ohhhh hohohohohoho". Fantastic work as always! Greetings from Munich
@31:00"The proper word, of course, is parallelepiped. But for obvious reasons, uh..., I wrote box." rofl!
🎯 Key Takeaways for quick navigation:
00:57 🔄 *A formula for the inverse of a 2x2 matrix is introduced: A inverse equals 1 over the determinant times the transpose of the matrix of cofactors.*
03:45 🤔 *The general formula for the inverse of an n x n matrix is established: A inverse equals 1 over the determinant of A times the transpose of the cofactor matrix of A.*
07:55 🔄 *The correctness of the inverse formula is checked by verifying that A times A inverse equals the identity matrix.*
19:17 🔄 *The solution to the system Ax=b is expressed as x equals A inverse times b, with A inverse given by the established formula.*
20:40 🧮 *Cramer's Rule is introduced, expressing each component of the solution vector x in terms of determinants of matrices obtained by replacing columns of A with the vector b.*
25:39 ⚠️ *Cramer's Rule is acknowledged as theoretically interesting but impractical for computation due to its reliance on computing multiple determinants.*
28:24 📐 *The determinant is revealed to be related to volume, setting the stage for discussing how determinants represent volume in specific cases and then generalizing to a broader understanding.*
28:51 📏 *Determinant of a matrix in three dimensions equals the volume of a parallelepiped (box), with each row forming an edge.*
31:37 🔄 *The volume of the box, given by the determinant, may be negative, indicating a change in handedness (right-handed to left-handed).*
33:04 📐 *For the identity matrix, the determinant equals the volume, representing a unit cube.*
35:24 🔄 *Orthogonal matrices, when used as transformation matrices, also represent cubes with a volume of 1 but may be rotated in space.*
38:38 🔄 *Determinant of an orthogonal matrix is either 1 or -1, ensuring that the volume formula remains valid.*
40:29 📐 *Volume (determinant) behaves linearly, doubling when an edge is doubled (property 3a).*
43:20 📐 *The determinant formula for area applies to parallelograms, simplifying the calculation to ad-bc.*
45:45 📏 *The determinant formula for area is ad-bc, providing a straightforward and general formula without square roots or complex calculations.*
47:38 📐 *The area of a triangle is half the determinant of the matrix formed by its vertices, a simple extension of the parallelogram formula.*
49:29 🔄 *Shifting the triangle's vertices does not change the determinant-based area formula, emphasizing its versatility and simplicity.*
Made with HARPA AI
"Well it takes approximately for ever to compute these determinants"!! Wow, such great sense of humour!!
Really understanding inv(A) = 1/det(A) * CT because of prof. Strangs explanation is an amazing thing; it just hit me, epiphany style. Just marvellous teaching.
The author of Mathematics for the Million was Lancelot Hogben in 1936. Eric Temple Bell wrote Men of Mathematics in 1937. For decades, both were very popular books on math for the general audience back when computational cost meant paying people to crank mechanical devices and write down the answers. Now a smartphone can invert a giant matrix in the blink of an eye.
the whole lecture was incredible.
godlike.
gausslike.
stranglike..!!!!!!!!!!!!
(21:29) "So, anytime I multiply cofactors by some numbers, I'm getting the determinant of *_something_* " - Important point in this lecture!!
And (14:16) also the same idea("...determinant of *_some_* matrix...").
Man, watching this lecture after learning about the wedge product is a trip.
"soo...cramer found a rule" sounded like we're talking about the one from seinfeld stumbling upon some deeper mathematical truth
Can't wait for next Marvel's "Doctor Strang in the Multi-Dimension of Madness"
With these lectures i am falling in love with maths all over again
45:53- BEAUTIFUL that's why I love math !
"They're nice formulas, but I just don't want you to use them." - Prof. Strang on Cramer's Rule.
Best maths teacher ever in my life , respect sir. Love from India ♥️
Some of these topics might be common knowledge, but the way he explains them - Mind = Blown 🙏🏻
Took me some time but the whole Ascrewedup example is really neat for seeing why the off diagonals are 0
Cool. Waterloo University has very similar lectures. This one is very clear though and I can understand the professor.
I'm curious as to what book Professor Strang is referring to. These lectures, by the way, are absolute masterworks.
The readings are assigned in: Strang, Gilbert. Introduction to Linear Algebra. 4th ed. Wellesley-Cambridge Press, 2009. ISBN: 9780980232714. See the course on MIT OpenCourseWare for more info and materials at: ocw.mit.edu/18-06S05. Best wishes on your studies!
It’s a great book! I have it (might be a different edition), and it’s the best intro to linear algebra text I’ve ever seen. It’s just like these lectures: clear and intuitive and dispels for the learner all myths that linear algebra is abstruse and unapproachable.
...... i'm having math's exam tomorrow and i Optimistic that i can get a good mark after watching this ... this guy is knows how explain and justify !!!!!!!
I think I could watch this like TV
@31:00 " The proper word is PARALLELOPIPED. But for obvious reasons, I wrote box. " Hahahahaha
15:58 Gilbert, it is less confusing if you take a and b on the second row while doing the co-factor formula. lol
Yeah, it's technically the co-factors of the row [c d] (the 2nd row) in the original matrix haha
44:32 "I'm pausing on that proof for a minute..." doesn't complete it in the end :(
50:23 he mentions about it
35:20 Being an orthogonal matrix does not mean that it is square.
The point of this lecture... -> 26:59
it is so beautiful the way he explains...
It took me some time to understand the A screwed up thing, but as soon as I got the idea, it turned out to be a good explanation! I'm always impressed by his teaching style.
i didn't get it :(
@@mohdfaisalquraishi8675 Think of this as a new matrix whose cofactor includes the first row. And the first row also act as factor of the cofactor above which contain itself. Hope it helps!
@@mohdfaisalquraishi8675 bro think about more dimensions instead of 2x2 and its gonna be more clear. when dont get the proper cofactors for that guy u always get singular
cuz u rule out the row and colmn of the one u are tryin to find cofactors for when u find cofactors but in that case u dont rule it out. leads to a 0
24:42 "In general, what is BJ ?" hahahahah Love this guy
Cramer Rule is good for solving 2 by2 and 3 by 3 linear equations. Dr. Strang it seem me to that you are turn off by Cramer rule, however your lecture on the subject was another masterpiece. The volume of a box is also explained very well using determinants.
It surprised me that there were only 200k views for this amazing lecture over 10 years.
45:34 the other revolting stuff.
oh, my motherfricking god. this guy is a genius.
I drew a little sketch of a parallelogram and verified the area does equal the determinant of the matrix by finding the area of the big rectangle and subtracting off areas of neighboring triangles and rectangles, but I still don't have a good intuitive sense of WHY the determinate should give you this area.
For the last triangular case you can also convert into two vectors (x2-x1,y2-y1) and (x3-x1,y3-y1) and then do det of this 2x2 matrix
Roddy Rich took the box mainstream recently. Professor Strang was clearly ahead of him time
Eigenvalues "The big stuff". Awesome!!!!
Professor mentioned a book named 'Mathematics for the Million' which he had read in childhood
However, the book is written by 'Lancelot Hogben', not by the guy named 'Bell'. =D
(If you are interested, check 'THE ALGEBRA OF THE CHESSBOARD' chapter of book archive.org/details/HogbenMathematicsForTheMillion/page/n527/mode/2up )
u are worldclass...u are better than any german docent..
slatz20
🇺🇸🏆MIT
I hope that some future use and simplification of Cramer's formula can be made. I would hate to think that Cramer wasted his time and that the knowledge was useless.
in all the course thats my best lecture its like magic
This guy is a fantastic teacher.
i love the humor of him
On 29:50 why the row is the vector? I thought that the vector is a column, not a row.
Aha so close to finish course 😏 only twelve lectures remained 🤗
I think the explanation from Pro book is more clear for the cofactor formula
I have a trivial doubt, 23:38 professor says x1 = (b1.c11)/detA
But won't x1 = b1.(c11 + c21 + c31 +.....cn1)/detA by definition transpose(C) b = x.detA
I have same question too
"so anytime I'm multiplying cofactors by numbers I think I'm getting the determinant of something"
@hypnoticpoisons does not matter since det A is equal to det A transpose
can someone explain why the determinant of the last matrix he drawn, the one with 111 on the rightmost works? I know it works by computing, but does it has to do with the volume with the new three columns?
6:00 couldnt get what he means by det(a) is product of n and cofactor is product of n-1 entry; can anybody explain?
We choose only one in each row and column in n by n matrix. The cofactor of one row has that row removed. So it's n -1
I think B_j should be A transpose replaced by b 24:16
No, since there is already a C transpose, we replace columns of A, or rows of A transpose, to obtain the same expression as the sum. Work it out and see.
love his sense of humour
I connected the properties of a determinant (1-7) to the volume of a box.
Now, can anyone please explain the connection of properties 8,9,10 to the volume of the box?
I think that if you have the base 3 rules applicable, then they are in the same category( or group, I am not quite good with abstract algebra). It is the same issue as it was in the previous sections where a vector space was created from functions and 3x3 matrices. Things just fall into place. Now visualizing it may be a hassle but it is not something that doesn't folllow from the rules.
Think about what the matrices do to a unit box. Then the determinant becomes the "change in area". Det(AB) = Det(A)Det(B) means if A scales by a, and B scales by b, then AB scales by ab.
8) det(A) = 0 -> A is singular
When det(A) is 0, the transformation A collapses some dimension of the space, meaning there is no volume left. This collapse causes the transformation to lose information (it is no longer one-to-one) and so it is singular and non-invertible
9) det(AB) = det(A) * det(B)
If you start with a box of unit volume, applying the transformation AB scales the volume of the box by the same amount as applying B first and then A.
10) det(A^T) = det(A)
This one is a bit harder to grasp intuitively. I can only do it using a certain visualization of covectors. If someone else has an easier intuition, I'd love to hear it.
Only determinant is the topic that’s does catch my interest.
Watching in 2022 ! #legends🗿
watching in 2024
at 5:42, wasn't it supposed to be b*d*i rather than b*f*g?
Thought there would be some Bj jokes 24:42 , but then again it's a MIT math class taught by Gilber Strang, maybe most are too absobed by his awsome lecture for that haha.
If you're not getting why those entries turned out to be zero. Here is a proof for it:
ltcconline.net/greenl/courses/203/MatricesApps/cofactors.htm
Can someone please explain why property 3b is also true for the volume?
I recommend drawing a picture. Put the points (a,b), (c,d), and (a+a',b+b'), and let A=(volume of parallelogram given by (a+a',b+b') and (c,d)). Let A_1=(volume of parallelogram given by (a,b) and (c,d)), A_2=(volume of parallelogram given by (a',b') and (c,d)), so A=(A_1)+(A_2). Property 3b follows immediately because the LHS is just A, while the RHS is just A_1 plus A_2. It's harder to draw and visualize in higher dimensions but it's the exact same concept.
It took me some time to convince myself why As has two identical rows. The example only shows an example of 2x2 matrix, but the question of why is still there. Someone helps explain this?
I have a thought on this. We know that the reduced row echelon form of A has no more than 1 of "1" at each row and cols. If we multiply a row of A (let's say the 1st row) with a col of C^T (let's say 2nd col), we get 0 because each co-factor matrix in 2nd col of C^T (e.g. C21) always consist a zero vector. (C21 consists of zero vector of col 1 of A). Thus, their product is 0.
Is it the right thought?
Row 2 of A times the cofactor of row 1. Then Row 2 is the factor of the cofactor of row 1. So Row 2 is copied into row 1. And the cofactor of row 1 has every row expect the first. So row 2 is in it. So you got 2 identical rows.
thank you very much sir. you made my day . I've been struggling a lot to understand inverse and adjoint relationship and you nailed it .
The determinand-by-box definition was simply brilliant!
Wow, this is just awesome!
I'm confused at around 17:25, detA*I shouldn't be detA^n*I? cause it is the matrix of detA,so each row can be divided over detA, so that is detA^n*I!
if u need calculate det(detA * I), it would be detA^n * I. But this is matrix multiple scalar.
Can anyone explain further why appending the 1s to the triangle matrix at the end shifts the triangle back to the origin? Also, if possible a little elaboration on why the cofactors columns equal 0 when multiplied by a different row number fro it's column number?
ranvideogamer Appending a column of 1s (let's call it M), technically, does not really shift the triangle back to the origin. Rather, by appending a column of 1s, you get a determinant whose value is the volume. Refer to Figure 5.1 in [Strang 4th ed, 272] for a geometric rendering of a *general triangle* (no corner at origin) made up of 3 *special triangles* (each with a corner at the origin). The area of a *special triangle* is clearly (ad-bc) for each of those triangles. det(M) turns out to be the sum of the areas of those 3 *special triangles*. The easiest way to find det(M) is to take cofactors of column 3 (the 1's column). Seen geometrically (as a sum of 3 *special triangles*), the constant in the third column has to be 1. The textbook also clearly explains that the area of a *special triangle* can be calculated as det(M) where the third corner/vector is the origin (0,0). In this case, you can take cofactors of row 3 (with 2 zeros and a 1), for the easy calculation to give you (ad-bc).
ah, thanks! I'l refer to the textbook!
You could just shift all the coordinates back to the origin by subtracting (x₁, y₁) from the three coord, then take the determinant and you will get +/- the same number as his determinant with some simple/tedious algebraic multiplying out.
Or alternatively with a bit of geometric imagination, imagine the parallelepiped that the determinant (or rather its transpose) with 1s determines the volume of, and seeing that one end is in the x-y plane and the other is parallel to it but 1 unit away, the volume of that 'box' is the same as the area of the parallelogram!
At 38:30, Strang says that det(Q)^2 = 1 means that det(Q) is 1 or -1...but that's not true if det(Q) is allowed to be complex. Twenty lectures into this course and I have to mentally question everything so far to see what does and doesn't apply to the complex numbers.
+Lobezno Meneses as long as every element in Q is real, there's no way det(Q) would be complex. If the pivots are real, det(Q) is also real because det(Q) is the product of the pivots.
98% of what he says applies to complex matrices as well. the only things to be modified are orthogonality definitions as well as transposition definitions.
What else could it be? One has exactly two square roots, both of which are real, i.e., 1 and -1.
where is the proof of 39b) for the paralellopipe?
Cramer's Rule starts at 20:00
Thanks!
@MIT OpenCourseWare , at 18:20 when we take Det(A) out from every row shouldn't it result into Det(A)^n like it happened in last chapter for 2^n volume increase case?
We would need to take out det(A) from each row only if we were taking the determinant of the matrix with det(A) in the diagonals.
But in this case we are aren't taking the determinant of the whole matrix itself, we're just multiplying I by det(A) and equating it to A times C^T.
@@cactuslover2548 ohh got it. Thanks a lot.
Dayum, stuff makes sense
Strang is great, TH-cam needs improvement... we need a speed between 1.5 & 1.75
I've been watching in 1.75 and feel it is just the right pace. I usually watch vids in x2 but he is going too fast for me to keep it at x2 at all time and I don't want to constantly switch speed. But I agree it would be nice to have a dimer instead of fixed speed. In any case, watching it in higher speed definately makes him more dynamic which is quite refreshing given his awsome personality haha.
@@neurolife77 I always watch his lectures at 1x. I'm doing the course and I have to stop frequently to take good notes and such. I sometimes watch other things at higher speeds, but not often. You're right, a speed slider isn't such a bad idea
@@ozzyfromspace Same the normal speed is for taking notes
@@arjundevendrarajan Really is. I'm starting lecture 30 in a few minutes, almost done with the course, and its amazing how much 1x and note taking has helped me understand. I studied electrical engineering (before I dropped out lol) and linear algebra was never in the syllabus. Its scary to think an engineering student can graduate without appreciating the beauty that is LINEAR ALGEBRA. Best wishes to you, friend
@@ozzyfromspace JUST found out. There is one! You just have to go in settings, speed and there is a "personnalized" button on the top right of the window that is opened. It allows you to adjust the speed with 0.05 of precision. :D
And my teacher just said "this is the formula"(explained a sum) and that's it for the topic
Could someone help me understand "where does the 2 by 2 matrix inverse formula come from?" 1:33 I don't think this formula was taught previously. In my memory, only the Gauss-jordan method were taught to get inverse of a matrix.Thank youuu!
You're right, it hasn't been taught from the previous lectures, only Gauss-Jordan. Prof Strang laid out the formula of the inverse for a 2x2 matrix then generalized it then gave an explanation of why that is
can anyone explain why A*C^t=det A*I
i though it equals to (det A )^n * I
because we will factor det A from each row by property 3a
ali hassan You're not calculating determinents here! What you're doing is matrix multiplication...
@@danielnarcisozuglianello8281 I actually didn't understand your point here. From the cofactor formula we actually get det(A) in each diagonal element which makes det(A) come out from each row. I know this is wrong but just cant understand where I am going wrong.
det(A)*I (a determinant of a matrix) IS equal to (det(A))^n*det(I) = (det(A))^n*1 (a scalar), NOT (det A)^n*I (a matrix)... det(A)*I = [ [detA,0,0...], [0,detA,0,0,0]...[0,0,...detA])], which is a diagonal matrix and hence det(A)*I = det(A)^n, a scalar, not a matrix!
@@prso8594 Not sure if I am too late, but Daniel was right.
The concept will become easier if you substitute det(A) with actual numbers to the diagonal matrix in 12:31.
Say det(A) = 5, then a 2 * 2 diagonal matrix is
[5 0
0 5],
It is the same as "the identity matrix I =
[1 0
0 1]
times the scalar 5 (i.e., det(A))" rather than I * 5^2 (i.e., det(A)^2).
The action "get det(A) in each diagonal element which makes det(A) come out from each row" is right only if you want to calculate det(det(A)*I).
(Note that det(A)*I is a matrix, not a scalar, since det(A) is a scalar and I is a matrix.)
Take the matrix mentioned above again, its determinant is 25, which equals to
det(A)^2 * det(I) = 5^2 * 1.
Excellent lecture
Cool. He offers another direction here different from the book he wrote.
Can the determinants that compute the triangle(not including origin) be generailized to n dimension? I look around internet but found nothing related.
I mean for example using a 4x4 matrix with rows (x1 y1 z1 1), (x2 y2 z2 1), (x3 y3 z3 1), (x4 y4 z4 1), does this determinant compute anything?
@@jinzhonggu8276
Determinants always compute the volume of the "parallelogram" in n-dimensions. In this case, the column of 1s just allows you to translate the "parallelogram" by an arbitrary vector, so you end up with the volume in (n-1)-dimensions. He showed the 2D case, where the 3x3 matrix with a column of 1s computes the area of a parallelogram (except he cut it in half and called it a triangle).
So the 4x4 case computes the volume of a parallelopiped in 3D. The 5x5 case computes the hyper-volume in 4D, and so on. If you want a specific geometric figure for the 4x4 case: you can divide the determinant by 1/6th to get the volume of the tetrahedron which has the 4 points used to construct the matrix as it's vertices.
amazing !!!session once again
the formula that he starts with, where do we know it from???
same question
kind of late but if it is the determinant formula with cofactors it is from the last lecture.
@@canned_heat1444 this was not explicitly derived in last lecture so I disagree. However, this resource seems to provide the answer based on the adjoint matrix www.sosmath.com/matrix/inverse/inverse.html
@@canned_heat1444 I apologize, he uses the cofactor formula from last lecture, but derives it in the first few minutes of this lecture.
Can someone help me understand why property 3b explains |det A| = volume could be extended beyond cubes and rectangles but to all angles?
+Sizhan Shi: He does it in a sneaky way: the properties that Strang labelled 1,2, 3a and 3b define the determinant, so what he does is prove that they also define the volume (up to a minus sign from property 2). If volume and the determinant follow exactly the same rules, then the determinant and the volume have to be the same. So he doesn't need to directly address the problem of non-right angles as long as he shows that, however it works out in the nitty gritty, it does the same as the determinant.
He is wonderful
Excellent as always and thx MIT.
my uni should just play this video in class tbh
How polit is this professor. I am watching this after 14yrs 😢
31:41 I was thinking you mean the volume between a frame of determine
In Romania we do this in high-school, 10th grade or something like that... and you're doing it at MIT? Great course(video) though
Arbaces420
How’s Romania??
Lol
in Kazakhstan we do this in 8th grade
why are the vectors/edges of the cube not columns but rows suddenly ?
To show " |det A| = volume" we can make columns or rows the edges, it's the same because det A = det (A transpose)
The paralleloopiped squeezing and emission dynamics producing electric signals in a way elongated diagonally and the volume is getting reduced to zero but the piezoelectric current getting increased every time a pressure is applied as you say screwing by one degree producing topological effect of producing superconductivity in special materials may be elaborated for understanding superconductivity please.
Sankaravelayudhab Nandakumar on behalf of Hubble Telescope Research Unit for discussion on Superconductivity by screwing for Strangler explanation.
Thank you.
I Really Like The Video Cramer's Rule, Inverse Matrix, and Volume From Your
15:05
"it takes approximately forever" lmfao
Cramer's rule? more like cram...ming for that test that I'm 100% not prepared for in the slightest amirightohgodpleasehelpme
selam beyler
Great... Its just a great lecture... :)
What a cute professor!
want to understand why A_s has two identical rows-.-
He purposely did it. He said that "s" stands for "screw up", so he made it singular.
By increasing the row but decreasing the column my volume getting decreased the square becomes a rectangle parallellogram diagonalised volume decreased piezo electricity flows parallelgram getting squeezed for reduction in volume but critically screwed at one degree topological screwing to produce superconductivity in certain materials requiring some information from the Hon.Prof Strangler please.
Sankaravelayudhan Nandakumar on behalf of Hubble Telescope Research Committee.