Preparation problems for the discussion sections on September 2nd and 4th1. For the following systems determine(1) the augmented matrix,(2) an echelon form of the matrix,(3) the reduced echelon form of the matrix,(4) whether the system is consistent,(5) the parametric description of the set of solutions,(6) how many solution the system has,(7) the geometric interpretation of the set of solutions.System A:x2= 3x1+ 2x2= 4System B:x1+ x2= 32x1+ 2x2= 6System C:x1+ x2= 32x1+ 2x2= 7Solution: System A: For (1): The augmented matrix is0 1 31 2 4.For (2): We reduce it to echelon form as follows:0 1 31 2 4R1↔R2−−−−→1 2 40 1 3.Note that this is not the reduced echelon form of the augmented matrix, but it is enough toanswer (4). The system is consistent, because in echelon form there is no row of the form0 0 x ,where x is non-zero.For (3): To get to the reduced echelon form:0 1 31 2 4R1↔R2−−−−→1 2 40 1 3R1↔R1−2R2−−−−−−−→1 0 −20 1 3.For (5): By (3), we get that System A has the same solutions asx1= −2x2= 3.1Hence the parametric description of the set of solutions is simplyx1x2=−23.For (6): The system has exactly one solution.For (7): The set of solutions is a point in R2.System B: For (1): The augmented matrix is1 1 32 2 6.For (2): We reduce it to echelon form as follows:1 1 32 2 6R2↔R2−2R1−−−−−−−→1 1 30 0 0.Note that is also the reduced echelon form of the augmented matrix, answering (3).For (4): The system is consistent, because in echelon form there is no row of the form0 0 x ,where x is non-zero.For (5): By (3), we get that System B has the same solutions asx1+ x2= 3.Hence the parametric description of the set of solutions is simplyx1x2=3 − x2x2=30+ x2−11.For (6): The system has infinitely many solution, because it is consistent and has a free variable(x2).For (7): The set of solutions is a line in R2.System C: For (1): The augmented matrix is1 1 32 2 7.For (2): We reduce it to echelon form as follows:1 1 32 2 7R2↔R2−2R1−−−−−−−→1 1 30 0 1.For (3): To get to the reduced echelon form:1 1 30 0 1R1↔R1−3R2−−−−−−−→1 1 00 0 1.For (4): The system is inconsistent, because in echelon form there is a row of the form0 0 x ,2where x is non-zero.For (5): By (4), we get that System C is inconsistent and hence there is no solution.For (6): No solution.For (7): The set of solutions is empty.2. Some question to check your understanding:a) What is the largest possible number of pivots a 4 × 6 matrix can have? Why?b) What is the largest possible number of pivots a 6 × 4 matrix can have? Why?c) How many solutions does a consistent linear system of 3 equations and 4 unknownshave? Why?d) Suppose the coefficient matrix corresponding to a linear system is 4 × 6 and has 3 pivotcolumns. How many pivot columns does the augmented matrix have if the linear systemis inconsistent?Solution: For (a): 4. Each row can have at most one leading entry and hence at most onepivot.For (b): Again 4. Each column has at most one pivot.For (c): The augmented matrix of this system has 3 rows and 4 columns. Hence it has at most3 pivots. So one variable cannot correspond to a pivot column. Hence the system has one freevariable. Any consistent system with a free variable has infinitely many solutions.For (d): 4. In order to be inconsistent, the augmented matrix in echelon form has a row0 . . . 0 x , where x 6= 0. This row gives the augmented matrix a forth pivot.3. Find the parametric description of the set of solutions ofx1+ 3x2− 5x3= 4x1+ 4x2− 8x3= 7−3x1− 7x2+ 9x3= −6Solution: We bring the augmented matrix to reduced echelon form:1 3 −5 41 4 −8 7−3 −7 9 −6R2→R2−R1,R3→R3+3R1−−−−−−−−−−−−−−−→1 3 −5 40 1 −3 30 2 −6 6R3→R3−2R2−−−−−−−→1 3 −5 40 1 −3 30 0 0 0R1→R1−3R2−−−−−−−→1 0 4 −50 1 −3 30 0 0 0Hence the parametric description of the set of solutions isx1x2x3=−5 − 4x33 + 3x3x3=−530+ x3−431.34. For which values of h1and h2is the following system consistent?x1= h1x2= 5x1+ 2x2= h2Solution: We bring the augmented matrix to echelon form:1 0 h10 1 51 2 h2R3→R3−R1−−−−−−−→1 0 h10 1 50 2 h2− h1R3→R3−2R2−−−−−−−→1 0 h10 1 50 0 h2− h1− 10.Hence the augmented matrix contains a row of the form0 0 x , where x is non-zero, ifand only if h2− h1− 10 6= 0. Hence the system is consistent if and only if h2− h1= 10.5. Show that the interchange of two rows of a matrix can be accomplished by a finite sequenceof elementary row operations of the other two types.solution: We can replace Ri↔ Rjwith the following sequence of row operations: (i 6= j)Ri↔ Ri+ Rj, Rj↔ Rj− Ri, Rj↔ −Rj, Ri↔ Ri− Rj6. Let A = [aij]3×4, and let B = [bij]3×4be an echelon form of A.(1) Is it true that, if a11= 0, then b11= 0?(2) Is it true that, if A has a column of zeros, then B also has a column of zeros?(3) Suppose B has a row of zeros. What can you say about rows of A? (Explain.)(4) Suppose we form a new matrix using some columns of A, let’s say the first and the thirdcolumn. What is an echelon form corresponding to this new matrix?solution:(1) False, consider A =0 0 1 00 1 0 01 0 0 0.(2) True, since row operations do not change a column of zeros.(3) Each row of B is a linear combination of rows of A with at least one non-zero coefficient.Therefore, a linear combination of rows of A with at least one non-zero coefficient isequal to zero. (In language that we have not yet learned, this means that the rows ofA are “linearly dependent”.)(4) The matrix consisting of the first and the third column of B has the same reducedechelon form as the new matrix introduced in the problem. So, instead of transformingthe new matrix into echelon form you can transform the matrix consisting of the firstand the third column of B into echelon form.(The example A =1 1 1 10 1 1 10 0 1 1and B = A shows that just taking first and thirdcolumn does not necessarily result in an echelon
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