FP4 hints  May 2008 

If you spot any mistakes, please let me know, so that I can correct this page. 

Question 1 We discussed in class 

Question 2 We discussed in class 

Question 3 We discussed in class 

Question 4 We discussed in class 

Question 5 We discussed in class 

Question 6 This question is best tackled by the simplex method. Write a simplex table that has a vertical line instead of the column of = signs; 6, 3 & 4 on the right of the line; and the coefficients of x, y and z in three rows on the left of the line (remembering to substitute for a) Take or add an appropriate multiple of the top row from / to rows 2 and 3 in such a way as to give you zeros in the x column for rows 2 and 3 (you don't alter the top row in this operation  you just use it to modify rows 2 and 3. Don't forget to include the values to the right of the line!). Now take or add an appropriate multiple of row 2 from / to row 3 in such a way as to give you a zero in the y column for row 3. You should now have two zeros in row 3, enabling you to calculate z. Substituting this into row 2 gives you y, and substituting for y and z into the top row will yield x. In (b)(i) if there is no single solution, then the determinant of the matrix of coefficients is zero. I hope that's enough of a clue. For part (b)(ii), write out the Simplex tableau again, but with b at the top of the values column and your value you obtained for a in (b)(i) as the bottomright z coefficient. Work through the first iteration to produce zeros in the lower two x spaces  naturally this will result in functions of b all the way down the values column. If you haven't made any mistakes, you will find that you now have a problem, in that the simplex process needed to give a zero in the bottom y slot wil simultaneously give you a zero in the bottom z slot, thereby preventing you from proceeding to a full solution. So, instead, find an appropriate multiplier to apply to one of the bottom two rows to make their lefthand sides look the same. There are now two cases to consider:


Question 7 We started this in class, and dealt with (a)(i). For (a)(ii), write the matrix as a determinant (i.e. straight lines down the sides), and writing λ after each of the leading diagonal entries. Write out an expression for the determinant, and set it equal to zero. You will end up with a cubic equation that has to be factorised. Any one of the three factors could be zero, and that gives you the three eigenvalues. For (b)(i), write r as the column vector (x, y, z) [sorry that I'm writing it in row form!] and then evaluate its dot product with (1, 1, 1) to obtain an expression to set equal to zero. For part (b)(ii), multiply the original matrix W by (a, b, c), which will give a you a new column vector in which all three elements might include a, b and c terms. This column vector, of course, is a trio of values for x, y and z; so take the dot product with (1, 1, 1), and if it comes to 0, then the new column vector lies in the plane. [Notice that this is to be expected from (a), where we found that the transformation collapses 3D space onto a plane.] 

Question 8 This is the kind of question where you combine columns (or rows) together to produce new columns (or rows) with which you replace one of the columns (or rows) you used in the combining operation. For example, you could add the first two rows together to produce the new row (x+z, y+x, z+y) which you could write instead of the existing top row. As soon as you achieve a situation in which the elements of a given row (or column) have a common factor, you can take that factor out as a factor of the determinant. Before you start, you need to work out the value of the determinant the hard way, and you should find that the answer corresponds to something in the question. 
