• anonymous
Let T: R3 -> R4 be the linear transformation defined as follows:....? T(x,y,z) = (x-y+3z, -2x+2y-5z, 3x-3y+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.)
  • Stacey Warren - Expert
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  • jamiebookeater
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  • anonymous
1. 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: \[\begin{array}{rl}x-y+3z=0&(I)\\-2x+2y-5z=0&(II)\\3x-3y+7z=0&(III)\\-x+y+2z=0&(IV)\end{array}\] 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 \[\begin{array}{rl}x-y+3z=1&(I)\\-2x+2y-5z=-1&(II)\\3x-3y+7z=1&(III)\\-x+y+2z=4&(IV)\end{array}\] and \[\begin{array}{rl}x-y+3z=1&(I)\\-2x+2y-5z=2&(II)\\3x-3y+7z=3&(III)\\-x+y+2z=4&(IV)\end{array}\] If you're just interested in existence of solutions you should use Kronecker-Capelli 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 right-hand side (in our case (1,-1,1,4) and (1,2,3,4)) have the same rank then there is a solution. Let A|b denote the extended matrix. In exercise 2: rank(A)=2 and rank(A|b)=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(A|b)=3 so there isn't a vector v=(x,y,z) such that T(v)=(1,2,3,4).

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