Rowspace solution to Ax = b example | Linear Algebra | Khan Academy

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Visualizing the rowspace solution to Ax=b
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[@KatherineRogers]

Question for students: In the equation Ax=b, which is "the solution"? In algebra you are often used to creating some left hand side and manipulating that to find "the answer" on the right hand side. In Linear Algebra however, note that "the solution" for the set of linear equations is typically the column matrix x. Sure column matrix b is a linear combination of columns. ONLY vectors in that column space may make up some possible vector b. We use that to know which vectors b we may write as part of the question. The "solution" however is vector x. Try solving THIS m1 -1 m2 -1 multiplied by the column vector results in the column vector The equations are m1x - y = -b1 m2x - y = - b2 Rearranging y = m1x + b1 y = m2x + b2 Note the 1 and 2 following m and b should be taken as subscripts. In the form you are used to....the values b1 and b2 are part of the equation defining the location of the line on the graph. Instead it is x and y which are the unknowns. (Note that you can take ANY 2x2 matrix A and multiply each line through by some factor to where the second column consists only of -1s. Yeah ya are allowed to do that. This gets your line equations into the y=mx+b sort of format albeit some things are moved to other sides of the = sign. This can make it much easier to graph both equations by hand. You can then go on with standard methods to solve the equations from the original matrix A or from your current one with the y coefficient as -1. You can also multiply or divide rows in a 2x2 matrix through by a factor to get the b values to be 1. This is useful if you know the intercept intercept form of the line equation. Another form of the line equation that is useful is the normal form. You can take the square of the x coefficient and add this to the square of the y coefficient. Take the square root of that. Divide your row by that. Do that to each row and you have normalized line equations. This form allows you to read off things about direction cosines and distance from the origin of the line. In other words....dont forget all the things you know about lines from algebra and analytic geometry. If you never took analytic geometry, pick up a copy of a text book to use as a reference. Even an older text book from the 50s or so....will still work as a reference. ) "The solution" means...at what point ie some (x,y) DO these lines cross to where there is "a solution". Note that this solution is written as the vector (matrix) from the origin to that point. Now if you understand that each row represents an equation....and you are looking for "the point" where the lines cross, you should understand that if the lines NEVER cross ie are parallel, there is no solution. Example 2x -y -4 2x -y -3 Equivalent to y=2x+4 and y=2x+3 Subtract the second equation from the first and you get 0 0 -1 in the row ie zero of x and zero of y yields -1 which is impossible. If lines are parallel and you fail to notice this....you will get nonsense when you try to come up with a solution Look again and realize what went wrong ie the lines were parallel. A great possibility is when the lines cross at one point. That point....is "the solution". Again, you write that as the vector pointing from the origin to that point. This is written as a column vector and shows up in your equation Ax=b as x. There is yet another possibility when you hunt for "the solution" ie the point where the lines cross is where line in equations given are actually multiples of the same line or are identical equations. Then the two lines coincide at EVERY point, so there are infinitely many solutions. Where may some point (x, y) be found on one line and also on the other line? At every point on both lines. Part of the confusion is because the solution matrix x.....tells what linear combination of the columns of A which would lead to column vector b. Realize again that this only tells you what vectors b are possible. Of those, a vector b is chosen and THEN ....the solution asked for is still the column vector x in the equation Ax = b. Notice that the row vectors of A in the non augmented matrix have a certain number of elements. "The solution" matrix x MUST have this same number of elements. The vectors in the null space (not the null space of A transpose....but the null space of A)....have the same number of elements as the row vectors of the non augmented matrix, and of the solution vector X. The number of elements in the row of A (non augmented) and the solution matrix x must be the same or it would be impossible to get the product Ax. This is true no matter if the matrix b is the zero matrix or some other matrix. Text books make a HUGE deal about the number of elements in the row matrices of non augmented matrix A. If it is 2 elements, then the books say the row space is in R2. Usually the 2 is written as a superscript. Notice that the number of elements in the columns in A is the same as the number of elements in the matrix b. If you transposed matrix A and used this as before....the number of elements in the columns of A would become the number of elements in the rows, and of solution matrices of A transpose. If....you see a matrix A that has a different number of elements in its rows vs its columns and you believe any of the following, please ask your teacher. Myth ..."the solution is matrix b" No! Myth...."the solution should have the same numer of elements as the columns of A do. No If you can not look at a 2x2 matrix and draw the equations on the graph....and point on the graph to what the question asks for when it asks for "the solution" then please DO ask your teacher to clarify this again. If you can NOT explain why the solution MUST have the same number of elements as the rows of the non augmented matrix A, please ask your teacher. You should be able to explain what will happen if the lines are parallel, cross at one point, or coincide when you try to "solve the equations". You should be able to explain what is meant by vectors in the column space of A, what matrix x does to these....and how that causes matrix b. Make sure you can draw examples from 2x2 matrices of column vectors that then combine to form some matrix b. If not....please DO ask your teacher to explain this again. Know what is meant by vector b must be in the column space of A. Please DO be able to explain....why when we are looking for a point....how come "the solution" is actually a matrix with a single column only ie a vector. Be able to show on a graph exactly what that vector is telling you. Know that "the solution" is another way of saying....is there some (x,y) that will work for the first line equation....that ALSO works for the second line equation? Know this happens where the graph of the lines cross. Know how to express a vector pointing from the origin to exactly this point. I saw in looking over several Khan Academy videos....comments that showed marked confusion on these issues.

[@AmarVashishth]

14:34 r is a member of RowSpace not column space; Should have said RowSpace instead of Column Space.

[@fashionvella730]

MIND = BLOWN 😯

[@mcwhitfield09]

I've been picturing a line and plane in R3 in my head to get me through the last 3 or 4 videos. Thanks for finally getting to this and confirming my intuition :)

[@fashionvella730]

STILL GOOD THEN 99% OF THE AFTER 13 YEARS

[@LaureanoLuna]

There are a couple of things I'm not clear about. How can the 'shortest' solution to Ax=b be nonzero and lie in the rowspace of A if the solution set of that equation is orthogonal to that rowspace (see 12.42)? Second,it seems to me that the notion of vector addition has been changed: [0 3] + c[2 3] is no longer the diagonal of the corresponding parallelogram but c[2 3] displaced by [0 3]. I'm confused about a line representing the vectors in it or the vectors that point to it.

[@legendarynoob6732]

Replying here just in case: Q. How can the 'shortest' solution to Ax=b be nonzero and lie in the row space of A if the solution set of that equation is orthogonal to that row space (see 12.42)? Ans. The solution set to *Ax* = *b* is a line containing infinitely many points. Notice that the row space and the solution set are orthogonal to each other but intersect at a point. That point can be specified by some vector in the row space and only one. That is the solution to Ax = b Q.Second,it seems to me that the notion of vector addition has been changed: [0 3] + c[2 3] is no longer the diagonal of the corresponding parallelogram but c[2 3] displaced by [0 3]. I'm confused about a line representing the vectors in it or the vectors that point to it. Ans. Adding two vectors a + b is by placing b's tail on a's head and the tail of a and head of b are joined. This can also be done in reverse order (a + b = b + a) where a , b are vectors in some subspace IR^n.

[@norwayte]

Good. Thank you.

[@alex1114_]

amazing

[@Waranle]

Thank you Sal :)

[@KatherineRogers]

You didnt turn the row space vector into a unit vector. The denominator 13 is only there in your answer C because that was not done. Notice 13 is exactly the magnitude of that vector. The 3, -2 vector is an arbitrary basis vector. You could have chosen any multiple of that as the basis vector. Once you have the unit vector in the row space, your vector is 9 times that.

[@ngndnd]

"And let's replace the second row with the second row, minus 2 times the first, so 6, minus 2 times 3 is 0. And minus 4, minus 2 times minus 1 is 0." i cant be the only one completely lost rn