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anonymous
 5 years ago
Let T: R3 > R4 be the linear transformation defined as follows:....?
T(x,y,z) = (xy+3z, 2x+2y5z, 3x3y+7z, x+y+2z)
1. Find a formula for all elements in ker(T).
2. Is there an (x,y,z) in R3 so that T(x,y,z) = (1,1,1,4)? (You do not need to find (x,y,z) if it exists, but explain how you know you answer is correct.)
3. Is there an (x,y,z) in R3 so that T(x,y,z) = (1,2,3,4)? (You do not need to find (x,y,z) if it exists, but explain how you know you answer is correct.)
anonymous
 5 years ago
Let T: R3 > R4 be the linear transformation defined as follows:....? T(x,y,z) = (xy+3z, 2x+2y5z, 3x3y+7z, x+y+2z) 1. Find a formula for all elements in ker(T). 2. Is there an (x,y,z) in R3 so that T(x,y,z) = (1,1,1,4)? (You do not need to find (x,y,z) if it exists, but explain how you know you answer is correct.) 3. Is there an (x,y,z) in R3 so that T(x,y,z) = (1,2,3,4)? (You do not need to find (x,y,z) if it exists, but explain how you know you answer is correct.)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01. The kernel of a transformation contains the vectors which T maps to (0,0,0,0). So to find ker(T) you should solve the following system of linear equations: \[\begin{array}{rl}xy+3z=0&(I)\\2x+2y5z=0&(II)\\3x3y+7z=0&(III)\\x+y+2z=0&(IV)\end{array}\] From equations (I) and (IV) you immediately get that z must be zero which implies that x is equal to y. So \[\ker(T)=\{(t,t,0)\in\mathbb{R}^3\mid t\in\mathbb{R}\}.\] Well if you'd wanted to find (x,y,z) such that \[(x,y,z)\mapsto(1,1,1,4)\] and \[(x,y,z)\mapsto(1,2,3,4)\] you'd have to solve \[\begin{array}{rl}xy+3z=1&(I)\\2x+2y5z=1&(II)\\3x3y+7z=1&(III)\\x+y+2z=4&(IV)\end{array}\] and \[\begin{array}{rl}xy+3z=1&(I)\\2x+2y5z=2&(II)\\3x3y+7z=3&(III)\\x+y+2z=4&(IV)\end{array}\] If you're just interested in existence of solutions you should use KroneckerCapelli theorem which claims that if the coefficient matrix \[A=\left[\begin{array}{ccc}1&1&3\\2&2&5\\3&3&7\\1&1&2\end{array}\right]\] and the matrix extended by an extra column of the righthand side (in our case (1,1,1,4) and (1,2,3,4)) have the same rank then there is a solution. Let Ab denote the extended matrix. In exercise 2: rank(A)=2 and rank(Ab)=2 so there is a vector v (namely v=(1,1,1)) such that T(v)=(1,1,1,4). In exercise 3: rank(A) is still 2 but rank(Ab)=3 so there isn't a vector v=(x,y,z) such that T(v)=(1,2,3,4).
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