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MARTINA BALAGOVIC: Hi. Welcome back. Today's problem is about solving homogeneous linear systems, A*x equals 0, but it's also an introduction to the next lecture and next recitation section, which are going to be about solving non-homogeneous linear systems, A*x equals b. The problem is fill the blanks type. And it says the set S of all points with coordinates x, y, and z, such that x minus 5y plus 2z equals 9 is a blank in R^3. It is in a certain relation to the other blank S_0 of all the points with coordinates x, y, and z that satisfy the following linear equation, x minus 5y plus 2z equals 0. After we solve this, we have the second part of the problem, which says all points of x have a specific form, x, y, z equals blank, 0, 0, plus some parameter times blank, 1, 0 plus some other parameter times blank, 0, 1. And we need to fill out all six blanks. Now you should pause the video, fill in the blanks, and then come back and see some pretty pictures that I prepar...

MARTINA BALAGOVIC: Hi. Welcome back. Today's problem is about solving homogeneous linear systems, A*x equals 0, but it's also an introduction to the next lecture and next recitation section, which are going to be about solving non-homogeneous linear systems, A*x equals b. The problem is fill the blanks type. And it says the set S of all points with coordinates x, y, and z, such that x minus 5y plus 2z equals 9 is a blank in R^3. It is in a certain relation to the other blank S_0 of all the points with coordinates x, y, and z that satisfy the following linear equation, x minus 5y plus 2z equals 0. After we solve this, we have the second part of the problem, which says all points of x have a specific form, x, y, z equals blank, 0, 0, plus some parameter times blank, 1, 0 plus some other parameter times blank, 0, 1. And we need to fill out all six blanks. Now you should pause the video, fill in the blanks, and then come back and see some pretty pictures that I prepar...

MARTINA BALAGOVIC: Hi. Welcome to recitation. Today's problem is about change of basis. It says the vector space of polynomials in x of degree up to 2 has a basis 1, x, and x squared. That's the obvious basis that you would write for that vector space. But today we're going to consider another basis, w_1, w_2, and w_3. And we don't know what w_1, w_2, and w_3 are explicitly. What we know is that their values at x equals minus 1, 0, and 1 are given by this table here. So they are 1, 0, 0; 0, 1, 0; and 0, 0, 1. We're asked to do the following. We're asked to express this polynomial-- so y of x is minus x plus 5-- in this basis, w_1, w_2, w_3. We're asked to find the change of basis matrices between these two bases, 1, x, x squared, and w_1, w_2, w_3. And finally, we're asked to find the matrix of taking derivatives, which is a linear map on this space, in both of these basis. And let me give you an extra level of challenge, which is to try to...

MARTINA BALAGOVIC: Hi. Welcome to recitation. Today's problem is about change of basis. It says the vector space of polynomials in x of degree up to 2 has a basis 1, x, and x squared. That's the obvious basis that you would write for that vector space. But today we're going to consider another basis, w_1, w_2, and w_3. And we don't know what w_1, w_2, and w_3 are explicitly. What we know is that their values at x equals minus 1, 0, and 1 are given by this table here. So they are 1, 0, 0; 0, 1, 0; and 0, 0, 1. We're asked to do the following. We're asked to express this polynomial-- so y of x is minus x plus 5-- in this basis, w_1, w_2, w_3. We're asked to find the change of basis matrices between these two bases, 1, x, x squared, and w_1, w_2, w_3. And finally, we're asked to find the matrix of taking derivatives, which is a linear map on this space, in both of these basis. And let me give you an extra level of challenge, which is to try to...