MATH 240 Final: MATH 240 NIU Final99f Solution
Document Summary
12/10/1999: (20 pts) let a be the following matrix. A = (a) reduce the matrix a to row echelon form. 0 0 (b) find a basis for the column space of a. The leading 1"s occur in the 1st and 3rd columns, so these columns form a basis: (c) find a basis for the nullspace of a. After row-reducing, we have the equivalent system of equations x1 = x2 x4 x5, x3 = 2x4 3x5. Letting each of the independent variables x2, x4, x5 equal 1, in turn, we get the following basis. 1: (20 pts) let a be the following matrix. A = (a) find the characteristic polynomial of a. = ( 2)( 2 5 + 4) = ( 2)( 4)( 1) p( ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) 1 5 (cid:12)(cid:12)(cid:12)(cid:12) (b) find the eigenvalues of a. To diagonalize a we need to be able to nd 3 linearly independent eigenvectors, since a is a 3 3 matrix.
