CHEM-51 Midterm: CHEM 51 Dartmouth 122Exam 2SolutionsSpring 07
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Math 2210 fall 2013 prelim 2 - solutions: perform the row reductions: 1 (a) since the rst 3 columns are pivot columns, choose the rst three columns of the unreduced matrix as a basis: There a many possible answers for this, some of them are listed as follows: 1 (c) the row reduced matrix has 4th columns as non-pivot columns, hence x4 is free. Hence, the solution is: x1 = the general solution is given by. 2 x4 ; x3 = x4 ; x4 free, x1 x2 x3 x4. 1 is a basis for nul(a): we begin by calculating the characteristic polynomial: = (2 )(2 )(1 ). We calculated the determinant by cofactor expansion, beginning with the rst column and then the. Note that this gives the factorization into linear factors right away, so there is no need to expand the polynomial. We now see there are two eigenvalues 1 and 2.
