practiceFinalSolutions

# practiceFinalSolutions - MATH2940 Solutions to Practice...

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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 row-redeuced 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 row-reduced form of the matrix. Taking the last matrix from the row-reduction 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 row-reduced 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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practiceFinalSolutions - MATH2940 Solutions to Practice...

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