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Unformatted text preview: MATH2940: Solutions to Practice Final 1 – (a) Recall that a basis for the image of a matrix is the set of linearly independent columns of the matrix. Let ~ c i be the i th column of A . By inspection, ~ c 2 = 2 ~ c 1 and ~ c 4 = 3 ~ c 3 . To see if ~ c 5 is a linear combination of ~ c 1 and ~ c 3 , we row reduce 1 2 1 3 3 2 4 1 4 8 2 6 5 → 1 2 1 3 3 0 0 2 6 7 0 0 2 6 7 → 1 2 1 3 3 0 0 2 6 7 0 0 Columns ~ c 1 and ~ c 3 have pivots in the rowredeuced form. Hence, they are the only linearly independent columns, and so a basis for the image of A is 1 2 4 ,  1 2 . (b) The linear dependencies that give a basis for the kernel of a matrix can be read off from the rowreduced form of the matrix. Taking the last matrix from the rowreduction in (a) and further reducing it gives 1 2 1 3 3 0 0 2 6 7 0 0 → 1 2 0 1 / 2 0 0 2 6 7 0 0 0 → 1 2 0 1 / 2 0 0 1 3 7 / 2 0 0 0 Let x i be the i th component of a vector ~x in the domain. Then for ~x to be in the kernel of A , we need x 1 = 2 x 2 1 / 2 x 5 and x 3 = 3 x 4 7 / 2 x 5 . In other symbols, ~x =  2 x 2 1 / 2 x 5 x 2 3 x 4 7 / 2 x 5 x 4 x 5 = x 2  2 1 + x 4 3 1 + x 5  1 / 2 7 / 2 1 . Components x 2 ,x 4 ,x 5 correspond to free variables. So a basis for the kernel of A is  2 1 , 3 1 ,  1 / 2 7 / 2 1 . (c) The rank of a matrix, the number of pivots in a matrix’s rowreduced form and the dimension of the image of a matrix are all equal. Hence, rank( A ) = 2. 1 2 – (a) Recall that k f k 2 = < f,f > . Hence, k e t k 2 = Z 1 1 e 2 t dt = 1 2 e 2 t 1 1 = e 2 e 2 2 = sinh(2) k e t k 2 = Z 1 1 e 2 t dt = 1 2 e 2 t 1 1 = e 2 + e 2 2 = sinh(2) . (The hyperbolic sine part is obviously not important for the exam; it’s just an interesting fact that you may run into later in life.) Hence, k e t k = k e t k = sinh(2) 1 / 2 ....
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 Spring '06
 PANTANO
 Math, Linear Algebra, Determinant, Multivariable Calculus, Linear combination, f2 − f1

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