Linear Algebra Math 221 Open Book Exam 1 Open Notes 3 Sept, 24 Calculators Permitted Show all work (except #4) 1 2 3 4 2 1. (25 pts) Given A 1 2 1, b 2 and c 4. 1 a) (7 pts) Bring matrix A to echelon form. 1 2 3 1 2 3 1 2 3 1 2 3 1 2 1 2 2 2 8 b) (2 pts) Which columns of A are pivot columns? Columns 1 and 3 are pivot columns. (Column 2 is non-pivot.) c) (16 pts) Find all solutions to: i. (8 pts) Axb ii. (8 pts) Axc It is simplest to do both of these computations at the same time (in fact, as we will see later in the course with LU decomposition, it is easiest to do some of the computations without reference to b and c). We perform Gaussian reduction of the augmented matrix (A bc), thus killing two problems with one stone. 1 2 3 4 2 1 2 3 4 2 1 2 3 4 2 1 2 1 2 4 2 2 2 2 2 2 2 4 2 1 2 4 2 1 8 8 3 1 2 3 4 2 2 2 2 11 At this point it is plain that the equation Axc has no solutions as the last row of (A c) (ignoring the column corresponding to b) has form ( -11), giving the obviously inconsistent equation -11. Thus we will not waste any further time with the equation Axc and will drop the column corresponding to c before reducing (A b) to reduced echelon form. 1 2 3 4 1 2 3 4 1 2 1 2 2 1 1 1 1
As the second column has no pivot we see that y is a free variable. The second row tells us that z1, and the first row tells us that (1)x+(2)y1, so x1-2y. Solutions: x 1 2y i) y is free z 1 ii) Axc is inconsistent and has no solutions Check: i) Arbitrarily choosing y we get the solution vector (x,y,z)(1,,1). A*(1,,1)(4,2,) as desired. ii) (No way to check a result of inconsistent short of re-doing the work) 2. (25 pts) Consider the set of vectors 1 r 1, r 2, r 2 6? 2 5 1 a) (1 pts) Is the set { } linearly independent? If not, find a non-trivial linear relation between the members of the set. We are asking if the homogeneous vector equation x 1 + x 2 + x 3 has any non-trivial solutions. We check this by setting up a matrix A, whose columns are these vectors, and asking if the equation Ax has any non-trivial solutions, which is the same as asking if the system of equations has any free variables, or equivalently if A has any non-pivot columns. Note that we can answer this question by taking the matrix to echelon form. 1 2 2 1 2 2 1 2 2 1 2 2 1 6 2 4 2 4 2 4 2 5 1 2 5 1 1 3 1 Solution: YES, the set is linearly independent. As all of the columns have pivots the set of three vectors is linearly independent. b) (15 pts) Can b r 2 8 be expressed as a linear combination of {v, v, 2 3 }? If 1 so, write b as a linear combination of { } in all possible ways. We are asking for solutions to the vector equation x 1 + x 2 + x 3 b. This can be solved by generating an augmented matrix and taking it to echelon form (to see if the equation can be solved) and then, if there are solutions, on to reduced echelon form (to read off any possible solutions). Note that we can re-use some of the work from part (a) (in fact, if we do part (b) first we will answer part (a) in the process).
1 2 2 2 1 2 2 2 1 2 2 2 1 2 2 2 1 6 8 2 5 1 1 2 5 1 1 1 3 5 1 2 We see from this echelon form that the equation can be solved (and hence b can be written as a linear combination of { }). 1 2 2 2 1 2 2 1 2 2 1 4 2 2 1 1 1 1 1 2 1 2 1 2 1 2 We see that the unique (no free variables or non-pivot columns) solution to the linear system is x 1-4, x 2 1, x 3 2. Solution: b (-4) +(1) +(2) Check: We confirm by plugging back in: ( 4) v r 1 + (1) v r 2 + (2) v r 1 2 2 2 3 ( 4) 1 + ( 1) + (2) 6 8 b r 2 5 1 1 3. (25 pts) Given the system of equations x + 2y + 2z + 6w 1 2x 4 y 2z 9w 3 x + 2y + 4z +12w 3 a) (5 pts) Express this system as an augmented matrix. 2 4 2 9 3 1 2 4 12 3 b) (1 pts) Bring this matrix to reduced echelon form. 2 4 2 9 3 2 3 1 2 3 1 1 2 4 12 3 2 6 2 3 3 1 2 2 5 1 2 1 1 2 1 2 4 2 4 1 2 3 3 3 3 1 1 c) (1 pts) Is the system consistent? If so, write down all solutions of the original system.
YES The system is consistent as the reduced echelon form (the echelon form would be adequate) has no rows of the form ( 1), in other words no equations of the form 1. We note that the second column, the y column, has no pivot, so y is a free variable. We now read off the rest of the solution from the rows of the reduced echelon form. The first row reads (1)x+(2)y+()z+()w1, so x1-2y, the second row reads ()x+()y+(1)z+()w2, so z2, and the third row reads ()x+()y+()z+(1)w-1, so w -1. x 1 2y y y (free) z 2 w 1 Check: We check by plugging back into the original equations: x+2y+2z+6w 1(1-2y)+2(y)+2(2)+6(-1) -1-2x-4y-2z-9w -2(1-2y)-4(y)-2(2)-9(-1) 3 x+2y+4z+12w 1(1-2y)+2(y)+4(2)+12(-1) -3 4. (25 pts) ALWAYS True, NEVER True or SOMETIMES True (5 pts per question) a) The matrix corresponding to the linear transform x x + y 1 1 x y y is 1 1. y 1 NEVER True A matrix taking a 2-vector to a 3-vector must have 2 columns, not 3. b) A linear transformation from R 5 to R 5 is 1-to-1. SOMETIMES True Recall that such a linear transform is represented by a 5x5 matrix (5 rows and 5 columns). When you reduce such a matrix you may have 5 or fewer pivots, thus you may have one pivot per column or you may not. Recall that being 1-to-1 is equivalent to having no non-pivot columns, as this means there are no free variables. Also note the two examples of the identity transformation, which is obviously 1-to-1, and the zero transformation (everything goes to ) which is obviously not 1-to-1. c) The span of three vectors { } is equal to the span of the three vectors { + + + }. SOMETIMES True If all three are zero vectors this is true. For most choices of { } this is false, as the linear transformation (equivalently, the matrix 1 1 1 1 1 ) is not invertible. For example, the choice, 1, {e 1, e 2, e 3 } 1 1 1
spans all of R 3, but {e 1 +e 2, e 2 +e 3, e 3 +e 1 } only spans a plane as e 1 +e 2 -(e 2 +e 3 ) + (e 3 +e 1 ). d) A set of 1 vectors in R 5 is linearly dependent. ALWAYS True A set of at most 5 vectors in R 5 can be linearly independent any larger set must be linearly dependent. We determine if a set of vectors is independent by reducing a matrix made up with these vectors as columns. In this case the matrix will have 1 columns and 5 rows. There can be at most 5 pivots, so some columns will be non-pivot, there will be free variables in the system, the system will have non-trivial solutions, there will be non-trivial linear dependencies between the vectors, and the set will be linearly dependent. e) A 1-to-1 linear transformation T:R 3 R 5 is onto. NEVER True A linear transformation from R 3 R 5 is never onto. Recall that a linear transformation from R n R m being onto is the same as it having a matrix with a pivot in each row. As the matrix corresponding to T has only 3 columns it can have at most 3 pivots thus it cannot have 5 pivots, or one per row.