Problem 3. Find the matrix of the reflection across the...

Problem 3. Find the matrix of the reflection across the plane x + 2y + 3z = 0 in standard coordinates. Some steps: (1) The plane is the kernel of the 1 × 3 matrix (1 2 3). Find a basis for this kernel. You can also do this by simply solving x + 2y + 3z = 0 as a system of one equation with 3 unknowns. (2) Let v1, v2 be the basis you found above, and add v3 = egin{pmatrix} 1 \ 2 \ 3 end{pmatrix}. This vector is perpendicular (in 3D space) to v1 and to v2 (in fact it is in the normal direction of the plane x + 2y + 3z = 0). Let B = (v1, v2, v3). This is a basis of R^3. Write the matrix B of the reflection with respect to the coordinates B (or B, B is you prefer). (3) Let T be the matrix with columns v1, v2, v3. Is the answer to the question T^-1BT, or is it TBT^-1?

Transcript

00:01 In this question, we need to find out what is the matrix of reflection across the plane which is given as x plus 2y plus 3z equals to 0 in the standard coordinate that is in the xy plane we have to find out.

00:20 Correct so first in the first step our aim is to find out what is the basis of the kernel for this matrix okay so corresponding to this we can write this in the form of 1 2 3 x y z equals to 0 0 or simply 0 we can write in the matrix matrix form.

00:50 So this is of the form ax equals to 0.

00:53 We try to find out what is the kernel here.

00:56 We will get here x plus 2y plus 3z equals to 0.

01:01 X comes out to be minus 2y minus 3z.

01:05 Hence the solution first of all we can write down of ax equals to 0 is going to be xyz but x is minus 2y minus 3z and this is y comma z.

01:21 This we can write it as minus 2y y and 0 plus minus 3z 0 and here we have z.

01:36 Y if we take common this is going to be minus 2 1 0 and z if we take common this is minus 3 0 1 this is what we get so we write down that the basis of the kernel is going to be the set minus 3 0 1 and minus 2 1 0 this is the first step now let us see the next step in step 2 we have been given a vector v3 which is 1, 2, 3, correct.

02:16 And we can observe that v3 is perpendicular to both the vectors v1 and v2.

02:27 Now what we can do here? we are taking b equals to v1, v2, v3.

02:41 This is going to be a basis of r3.

02:45 Now we have to find the matrix with respect to this basis so what we are going to do let l be the matrix of reflection with the respect to the basis be here also we know that l of v 1 will be v 1 l of v 2 is going to be v 2 but l of v3 will be minus of v3 since v3 is perpendicular to both v1 and v2 and v1 v2 they all lie in the same plane so from this what we can do hence we get the matrix b as 1 0 0 0 1 0 0 0 minus 1 since v3 is minus v3 so we write this and now comes the third step we take t value as the vectors which we already wrote minus 2 1 0 minus 3 0 1 basis of the kernel and we have added one more which is 1 2 3 we are taking this as a matrix t from this we get t inverse it it comes out to be equal to that we can easily find.

04:20 So from this, hence tbt inverse, that is what we need to find out, which will be the required matrix...

Question

Please give Ace some feedback