Exam2MA262fall2011

# Then 2 t 2 y z y z t y 2 1 0 z 1 1 0 t 2 1 and 2 1 0

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Then (2 t - 2 y - z, y, z, t ) = y ( - 2 , 1 , 0 , 0) + z ( - 1 , 0 , 1 , 0) + t (2 , 0 , 0 , 1) , and { ( - 2 , 1 , 0 , 0) , ( - 1 , 0 , 1 , 0) , (2 , 0 , 0 , 1) } is a basis for S . Similarly solving x +2 y + z - 2 t = 0 for y, z and t , give the basis { (2 , - 1 , 0 , 0) , (0 , - 1 , 2 , 0) , (1 , 0 , 0 , 1) } , { (1 , 0 , - 1 , 0) , (0 , 1 , - 2 , 0) , (0 , 0 , 2 , 1) } and { (2 , 0 , 0 , 1) , (0 , 1 , 0 , 1) , (0 , 0 , 2 , 1) } , respectively. 7. Let T : R 3 R 3 be the linear transformation defined by (10 points) T ( x, y, z ) = ( x + 2 y - z, 2 x + 5 y - 4 z, x + 3 y - 3 z ) . Find a basis and the dimension of the kernel and the range of T . Solution. We have that the matrix of the given linear transformation is the one whose COLUMNS are T (1 , 0 , 0), T (0 , 1 , 0) and T (0 , 0 , 1), so it is A = 1 2 - 1 2 5 - 4 1 3 - 3 . To find the kernel we need to solve T ( x, y, z ) = (0 , 0 , 0), so applying Gaus- sian elimination on the matrix above, we get 1 2 - 1 2 5 - 4 1 3 - 3 1 2 - 1 0 1 - 2 0 1 - 2 1 2 - 1 0 1 - 2 0 0 0 . Hence T ( x, y, z ) = (0 , 0 , 0) if and only if y - 2 z = 0 y = 2 z x = - 2 y + z = - 3 z. Thus, Ker ( T ) = { ( - 3 z, 2 z, z ) } has dimension 1 with a basis given by { ( - 3 , 2 , 1) } . To find a basis and the dimension of Rgn ( T ) all we need is to apply Gaussian elimination on A T . So 1 2 1 2 5 3 - 1 - 4 - 3 1 2 1 0 1 1 0 - 2 - 2 1 2 1 0 1 1 0 0 0 . Thus Rgn ( T ) has dimension 2 with a basis given by { (1 , 2 , 1) , (0 , 1 , 1) } . 4

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8. If T : P 2 P 2 is a linear transformation such that T ( x 2 - 1) = x 2 + x, (10 points) T (2 x + 3) = 2 and T ( x 2 - 2 x ) = x + 1, then find T ( x 2 + x + 1) .
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