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04win-fsols - MATH 51 FINAL EXAM SOLUTIONS Professors...

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MATH 51 FINAL EXAM SOLUTIONS Professors Clingher, Munson, and White March 15, 2004 1 Consider the matrices A = 1 1 0 0 2 2 2 0 1 6 0 1 - 1 1 3 - 1 - 2 1 1 - 1 and R = 1 0 1 0 1 0 1 - 1 0 1 0 0 0 1 2 0 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 . Solution: R has pivots in columns 1, 2 and 4, so the corresponding columns of A form the basis: 1 2 0 - 1 , 1 2 1 - 2 , 0 1 1 1 1(b). Find a basis for the column space of R . Solution: R is its own rref, so again the pivot columns of R (that is, the first, second, and fourth columns) form the basis: e 1 , e 2 , and e 3 . 1(c). Find a basis for the nullspace of A . The nullspace of A consists of all solutions of A x = 0, or, equivalently, of R x = 0. From R x = 0, we see that the free variables are x 3 and x 5 and that x = - x 3 - x 5 x 3 - x 5 x 3 - 2 x 5 x 5 = x 3 - 1 1 1 0 0 + x 5 - 1 - 1 0 - 2 1 . Thus - 1 1 1 0 0 and - 1 - 1 0 - 2 1 form a basis for N ( A ). 2. Find all solutions of x 1 + 2 x 2 + x 3 + x 4 = 7 x 1 + 2 x 2 + 2 x 3 - x 4 = 12 2 x 1 + 4 x 2 + 6 x 4 = 4 . Solution: see example 6.4 on page 44 of the Levandosky text. The rref is 1 2 0 3 . . . 2 0 0 1 - 2 . . . 5 0 0 0 0 . . . 0 1
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from which we get the solution: x = 2 0 5 0 + x 2 - 2 1 0 0 + x 4 - 3 0 2 1 . 3(a). Find all eigenvalues of the matrix A = 5 0 0 1 2 1 1 1 2 . Solution: 0 = det( λI - A ) = λ - 5 0 0 - 1 λ - 2 - 1 - 1 - 1 λ - 2 = ( λ - 5) λ - 2 - 1 - 1 λ - 2 = ( λ - 5)(( λ - 2) 2 - ( - 1) 2 ) = ( λ - 5)( λ - 3)( λ - 1) so the eigenvalues are 5, 3, and 1. 3(b). The matrix M = 5 - 6 - 6 - 1 4 2 3 - 6 - 4 has λ = 2 as one of its eigenvalues. (You need not check this.) Let V be the eigenspace corresponding to this eigenvalue. (In other words, V consists of all eigenvectors with eigenvalue 2 together with the origin.) Find a basis for V . Solution: 2 I - M = - 3 6 6 1 - 2 - 2 - 3 6 6 . Note that V is the nullspace of 2 I - M . We find this nullspace by solving (2 I - M ) v = 0. Solving by Gaussian elimination, we get the single equation x - 2 y - 2 z = 0 so y and z are free and v = x y z = 2 y + 2 z y z = y 2 1 0 + z 2 0 1 . Thus 2 1 0 and 2 0 1 form a basis for V .
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04win-fsols - MATH 51 FINAL EXAM SOLUTIONS Professors...

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