MA1S11: SOLUTIONS TO 2010 ANNUAL EXAM 1. (a) x + y = 1, x + y = 2. Graphically this system represents two parallel lines. (b) (i) Augmented matrix. 20 6 21 0 5 0 −10 −60 −4 −34 2 1 40 −2 15 2 −9 −20 −6 −19 (ii) Reduced row echelon form. 28 63 5 1 0 0 5 0 1 0 14 29 5 5 0 0 1 −2 −2 5 5 0 0 0 0 0 (iii) There are infinitely many solutions. x1 = 63 28 − x4 5 5 29 14 − x4 5 5 2 2 x3 = − + x4 5 5 x2 = x4 is a free variable. 2. (a) → (i) P =< 4, 0, −1 >, Q =< 4, 2, 7 >, QP =< 0, −2, −8 >. √ P (ii) kPk = 17. Unit vector kPk = √117 < 4, 0, −1 >. 1 2 MA1S11: SOLUTIONS TO 2010 ANNUAL EXAM (iii) Projection of P along Q. projQ P = P·Q 3 Q = < 4, 2, 7 > kQk2 23 (b) Parametric equations: x = 4 − 2t, y = 5t − 2, z = 3 − 2t Cartesian equations: 4−x y+2 3−z = = 2 5 2 (c) A normal vector is < −1, 6, 5 >. The point (−21, 0, 0) is on the plane. (d) cos θ = u·v 11 = √ √ ≈ 0.21 kukkvk 69 38 3. (a) 14 0 5 − 17 2 13 8 m,n t (b) If A = (aij )m,n i,j=1 then the transpose of A is A = (aji )i,j=1 . (i.e. swap rows for columns). A is symmetric if At = A. (c) Ax = b where 1 −4 5 −2 2 3 7 2 A= −1 −12 −11 −16 1 2 −1 −1 MA1S11: SOLUTIONS TO 2010 ANNUAL EXAM 3 −3 18 b= −37 −3 x1 x2 x= x 3 x4 If B is the inverse of A then x = Bb. (d) Your picture should have 3 vertices. A 1 in the (i, j)-entry of the matrix means there should be a edge joining vertex i to vertex j. (e) Take for example A= 0 1 1 0 1 0 Note that A2 = 0 1 Then (trace(A))2 = (0)2 and trace(A2 ) = 2. 4. (a) (69)10 converted to base 2 is (1000101)2 = (1.000101)2 × 26 . The exponent (6)10 converted to base 2 is 110. The number is stored as follows: Bit 1 is the sign bit, which is 0. The next 8 bits store the exponent, which we pad out with zeros, 00000110. The remaining bits store the mantissa, 000101000 · · · 0. (b) ( 40 ) = (101.1011011011 · · · )2 . 7 10 (c) Generate your own spreadsheet to see what happens. (d) Check this by inputting the commands into Mathematica.