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TEXT 2.1 # 12 We set x1 = t and solve for x2. Setting x1 = t yields 2t + 3x2 = 5. Solving for x2 yields 3x2 = 5 – 2t x2 = So, a complete set of solution written in parametric form is [t, 5 2 t. 3 3 5 2 t ]. 3 3 Note: We could have set x2 = t and solved for x1 to get the parametric [ 5 3 t , t]. 2 2 # 16 The lines intersect at (3, –2), so the unique solution is [3, –2]. To solve, subtract 3×1st from 2nd 7y = –14 y = –2, so substitution x = 3. # 22 We find the solution [x1, x2, x3] = [0, 0, 0] using back substitution. Note: This follows immediately from the fact that all three equations are equal to zero. TEXT 2.1 #8 This matrix is in row echelon form, but not reduced row echelon form. Why not? The leading entry in row 4 is not a 1. Could we have given another reason? # 18 R1+R2 2 0 1 A = 1 1 0 1 1 1 R2+2R3 3 1 1 1 1 0 1 1 1 R3+R1 3 1 1 2 1 3 1 1 1 3 1 1 2 1 3 1 2 0 R2+ 12 R3 R2+R1 3 1 1 3 1 1 2 4 1 3 5 1 = B 2 2 2 2 0 0 Therefore, the matrices A and B are row equivalent. # 26 We form the augmented matrix and row reduce it as follows: R3 – 5R1 –2 R2 0 1 1 1 0 1 1 1 1 3 1 5 3 1 5 1 0 3 0 0 8 1 7 2 The third row is equivalent to the equation 0 = –8 which clearly has no solution. Therefore, the system in inconsistent. Does R3 = 5R1 + 2 R2 (excluding constants) cause the system to be inconsistent. # 34 When there are 4 equations and 4 variables, if the solution exists it is unique. Why? Because the Rank Theorem (2.2) tells us there are 4 – 4 = 0 free variables. 1 1 1 1 4 1 0 2 3 4 10 0 1 → 0 0 3 6 10 20 0 0 4 10 20 35 1 1 1 0 0 1 0 0 1 The solution is 1 0 1 0 1 1 a 1 b 1 = . c 1 d 1 # 40 First row reduce the system [A│x] and then answers parts (a), (b) and (c). R1 ↔ R2 R2 – 12 kR1 2 3 k 2 4 6 2 4 6 2 3 k 4 6 2 0 2 2k 3 3k (a) There are no values of k for which this system has no solution. Why? The system has no solution when A has a zero row with corresponding constant ≠ 0. 2+ 2k = 0 k = –1 is the only value of k that creates a row of all zeros. But k = –1 the constant 3 + 3k = 3 – 3 = 0. What does this imply? (b) The system has a unique solution for k ≠ –1. Why? From (a), we see when k ≠ –1 then rank A = 2. So, there are 2 – 2 =0 free variables. (c) The only value of k for which system infinitely many solutions is k = –1. The system has infinitely many solutions when A has a zero row with constant = 0. From (a), we see this is exactly the case when k = –1. # 42 First row reduce the system [A│x] and then answers parts (a), (b) and (c). R3 – R1 – R2 1 2 3 2 1 2 3 2 1 1 k 1 1 k 1 1 2 1 4 k 2 2 1 4 k 2 (a) When k ≠ –1, 2, this system has no solution. Why? The system has no solution when A has a zero row with corresponding constant ≠ 0. k2 – k – 2 = 0 k = –1, 2 are the only values of k that that make the constant 0. (b) This system never has a unique solution because rank A = 2. Why? Since rank A = 2, there is always at least 3 – 2 = 1 free variable. (c) When k = –1, 2, this system infinitely many solutions. The system has infinitely many solutions when A has a zero row with constant = 0. From (a), we see this is exactly the case when k = –1, 2. # 42 (a) The following system of n equation has infinitely many solutions: x1 + x2 + … + xn = 0 2x1 + 2x2 + … + 2xn = 0 nx1 + nx2 + … + nxn = 0 Likewise, the following system of n +1 equation has infinitely many solutions: x1 + x2 + … + xn = 0 2x1 + 2x2 + … + 2xn = 0 nx1 + nx2 + … + nxn = 0 (n+1) x1 + (n+2) x2 + … + (n+1) xn = 0 (b) The system of n equation x1 = 0, x2 = 0, … , xn = 0 has the unique solution xi = 0. As does the system of 2n equations x1 = 0, x2 = 0, … , xn = 0.