04aut-fsols

# 04aut-fsols - MATH 51 FINAL EXAM SOLUTIONS December 6, 2004...

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MATH 51 FINAL EXAM SOLUTIONS December 6, 2004 Professors De Silva, Ionel, Ng, Storm, and White 1. Consider the matrices A = 0 0 1 1 1 3 1 2 2 6 1 3 and R = 1 3 0 1 0 0 1 1 0 0 0 0 . The matrix R is the row reduced echelon form of A . (You do not need to check this.) 1(a). Find a basis for the column space of A . The pivots in R are in columns 1 and 3, so the ﬁrst and third columns of A , namely 0 1 2 and 1 1 1 , form a basis for C ( A ). 1(b). Find a basis for the null space of R . Solution: A vector x is in the nullspace of R if and only if R x = 0 , i.e., if and only if x 1 x 2 x 3 x 4 = - 3 x 2 - x 4 x 2 - x 4 x 4 = - 3 1 0 0 x 2 + - 1 0 - 1 1 x 4 so - 3 1 0 0 and - 1 0 - 1 1 form a basis for N ( R ) (which, incidentally, is the same as N ( A ).) 1(c). Note that A 1 1 1 1 = 2 7 12 . Find all solutions to A x = 2 7 12 . Solution: We get all solutions by taking any particular solution and then adding to it vectors in the nullspace of A : x = 1 1 1 1 + c - 3 1 0 0 + d - 1 0 - 1 1 . 2. Consider the following system of equations: x 2 + x 3 = a x 1 + x 2 + 2 x 3 = b x 1 + 2 x 2 + 3 x 3 = c 2 x 1 + 3 x 2 + 5 x 3 = d 1

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Find the condition(s) on a , b , c , and d , for the system to have a solution. (Your answer should be one or more equations of the form ? a +? b +? c +? d =?.) Solution: 0 1 1 | a 1 1 2 | b 1 2 3 | c 2 3 5 | d 1 1 2 | b 0 1 1 | a 1 2 3 | c 2 3 5 | d 1 1 2 | b 0 1 1 | a 0 1 1 | c - b 0 1 1 | d - 2 b 1 1 2 | b 0 1 1 | a 0 0 0 | c - b - a 0 0 0 | d - 2 b - a so the conditions are c - b - a = 0 and d - 2 b - a = 0, or equivalently, a + b - c = 0 a + 2 b - d = 0 3(a). Find all eigenvalues of the matrix A = 1 0 2 7 3 5 2 0 1 . Solution: The number λ is an eigenvalue provided 0 = det( λI - A ) = λ - 1 0 - 2 - 7 λ - 3 - 5 - 2 0 λ - 1 . Expanding by column 2 gives: 0 = ( λ - 3) λ - 1 - 2 - 2 λ - 1 = ( λ - 3)(( λ - 1) 2 - ( - 2) 2 ) = ( λ - 3)(( λ - 1) 2 - 2 2 ) = ( λ - 3)(( λ - 1) + 2)(( λ - 1) - 2) = ( λ - 3)( λ + 1)( λ - 3) so the eigenvalues are 3 and - 1. 3(b). Consider the matrix M = 3 1 1 0 4 1 0 0 3 . Note that v 1 = 1 1 0 is an eigenvec- tor with eigenvalue 4. Find eigenvectors v 2 and v 3 so that v 1 , v 2 , and v 3 form a basis for R 3 . Solution: The matrix is triangular, so the eigenvalues are the diagonal elements, namely 3 and 4. The eigenspace corresponding to λ = 3 is the nullspace of 3 I - A .
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04aut-fsols - MATH 51 FINAL EXAM SOLUTIONS December 6, 2004...

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