Linear Algebra Exam 1 Concept Problems

Suppose A is a 3x3 matrix and b is in R3. If the solution set to Ax=b is span{(1 0 0)} then b is the zero vector.

span{91 0 0)} includes 0(1 0 0)=0 so A0=b thus b = 0

Suppose we are given a consistent system of 3 linear equations in 4 variables. Which of the following are possible for the solution set of the system? Circle all that apply. (i) The system has a unique solution (ii) The solution set is a line in R4 (iii) The solution set is a plane in R3

(ii) The solution set is a line in R4

Say A is an mxn matrix. Which of the following statements are equivalent to the statement that the columns of A span Rm? Select all that apply. (a) A has a pivot in each row (b) A has a pivot in each column (c) Ax=b is consistent for every b in Rm (d) Ax=0 is consistent

(a) A has a pivot in each row (c) Ax=b is consistent for every b in Rm

Which of the following statements are equivalent to the statement that the matrix equation Ax=b is consistent? Select all that apply. (a) the augmented matrix (A|b) has no pivot in the last column (b) the augmented matrix (A|b) has a pivot in every row (c) b is a linear combination of the columns of A (d) b lies in the span of the columns of A

(a) the augmented matrix (A|b) has no pivot in the last column (c) b is a linear combination of the columns of A (d) b lies in the span of the columns of A

Suppose I have two linear equations in three variables. Which of the following are possible solutions to the system? Select all that apply. (a) the empty set (b) a single point (c) a line (d) a plane (e) R3

(a) the empty set (c) a line (d) a plane

Suppose A is a 2x2 matrix and the set of solutions to Ax=0 is the line y=x. Also suppose that b is a nonzero vector in R2. Which of the following can possibly be the set of solutions to Ax=b? Select all that apply. (a) the line y=x (b) the line y=x+1 (c) the line x=0 (d) the origin

(b) the line y=x+1

What best describes the geometric relationship between the solutions to Ax=0 and the solutions to Ax=b? (a) they are both lines through the origin (b) they are parallel lines (c) they are both planes through the origin (d) they are parallel planes

(b) they are parallel lines

Suppose that the set of solutions to Ax=b is a plane z=1 in R3. Which of the following is the set of solutions to Ax=0? (a) the x-axis in R3 (b) the z-axis in R3 (c) the xy-plane in R3 (d) (0,0,0)

(c) the xy-plane in R3

Suppose we are given a consistent system of 4 linear equations in 3 variables. Which of the following are possible for the solution set of the system? Circle all that apply. (i) The system has a unique solution. (ii) The solution set forms a line in R4. (iii) The solution set forms a plane in R3.

(i) The system has a unique solution. (iii) The solution set forms a plane in R3.

Suppose v1 and v2 are vectors in R3. Which of the following statements must be true? Circle all that apply. (i) If the set {v1,v2} of vectors in R3 is linearly independent then span{v1,v2}=R2. (ii) There must be at least one vector b in R3 which is not a linear combination of v1 and v2.

(ii) There must be at least one vector b in R3 which is not a linear combination of v1 and v2.

Which of the following conditions guarantee that a system of 3 linear equations in 4 variables has infinitely many solutions? (circle all that apply) (1) The reduced row echelon form of the augmented matrix corresponding to the system of equations has a row of zeros. (2) The rightmost column of the augmented matrix is not a pivot column.

2. The rightmost column of the augmented matrix is not a pivot column. (system consistent, and at most three pivots so one or more free variables)

Give an example of a matrix A so the solutions to Ax=0 form a line in R4.

A = (1 0 0 0, 0 1 0 0, 0 0 1 0)

Suppose A is an mxn matrix and the equation Ax=0 has only the trivial solution. Then A must have exactly m pivots.

A must have n pivots. A pivot in every column.

Suppose we have three equations in three variables. Which of the following can be the set of solutions? Select all that apply. (a) a point (b) a plane (c) a line (d) no solution

Can be all options.

If the augmented matrix (A|b) has more columns than pivots then the matrix equation Ax=b has infinitely many solutions.

False.

If A is an mxn matrix and n > m, then the equation Ax=b must be inconsistent for some b in Rm.

False. 2x3 matrix where A= (1 0 0, 0 1 0) columns of A span R2.

If Ax=b is inconsistent then Ax=0 must also be inconsistent.

False. Example Ax=b inconsistent augmented matrix is (1 0 1, 0 0 1). Ax=0 consistent augmented matrix (1 0 0, 0 0 0).

If A is an m × n matrix and A has a pivot in every column, then the equation Ax = b has a solution for each b in R m.

False. For example, if A = (1,0,0,1,0,0) , then Ax = (0,0,1) is inconsistent

If the RREF of an augmented matrix has a row of zeros, then the corresponding linear system of equations either has no solutions or infinitely many solutions

False. For example, the augmented system (1 0 1, 0 1 2, 0 0 0) has a unique solution.

A system of three linear equations in four variables can have exactly one solution.

False. If the system is consistent, it will have at least one free variable (4 columns but max of 3 pivots), so it will have infinitely many solutions.

If the reduced row echelon form of an augmented matrix has a row of zeros, then the system of linear equations corresponding to the augmented matrix has infinitely many solutions.

False. It could be inconsistent, or it could have a unique solution.

The vector equation x (1 −1) + y (0 1) = (1 −1) has infinitely many solution

False. One step row-reduction gives the unique solution x = 1, y = 0

If u, v, and w are nonzero vectors in R 2 , then Span{u, v, w} is R 2 .

False. Take v1 = v2 = v3 = (1,0)

The vector equation x1 (1 −1) + x2 (2 −2) = (2 2) is consistent

False. The associated augmented matrix has a pivot in the rightmost column.

A system of 4 linear equations in 5 variables can have exactly one solution

False. The augmented matrix can have at most 4 pivots, so there will be at least one free variable, thus any associated system will either be inconsistent or have infinitely many solutions

If a system of linear equations has more variables than equations, then the system must have infinitely many solutions

False. The system can be inconsistent. For example: x + y + z = 5, x + y + z = 2.

If b is a vector in R3, then b must be a linear combination of the vectors (1 0 -1), (1 1 0), and (0 1 1).

False. These vectors don't span R3.

The set of solutions to a vector equation is always equal to a span

False. This is only true when the solution set contains the origin.

Are there three nonzero vectors v1 , v2 , v3 in R 3 so that Span{v1 , v2 , v3 } is a plane but v3 is not in Span{v1 , v2 }? If your answer is yes, write such vectors v1 , v2 , v3 and label each vector clearly.

For example, v1 = (1 0 0), v2 = (−1 0 0) , v3 = (0 1 0).

Suppose we have a consistent system of four linear equations in three variables, and the corresponding augmented matrix has two pivots. The set of solutions to the system is a:

Line in R3

Is the augmented matrix (0 1 1 1, 0 0 0 1) in reduced row echelon form?

No.

Suppose A is a 2x2 matrix and A(1 1) = (19 7). Is it possible that the set of solutions to Ax=0 is the line x1=x2?

No.

Is there a 2 × 2 matrix A so that the solution set for the equation Ax = 0 is the line x1 = x2 + 1? If yes, write such an A. If no, justify why there is no such A

No. The solution set to Ax = 0 must include the origin x1 = x2 = 0, which is not on the line x1 = x2 + 1.

Is there a 2x2 matrix A so that the solution set to Ax=0 is the line x1-x2=3? If your answer is yes, write such a matrix A. If answer is no, justify why there is no such matrix A.

No. x=0=(0 0) must be a solution to Ax=0 but 0-0 does not equal 3.

The equation (1 −1 2 0 4 3) x = b is consistent for every b in R 2 .

True, since (1 −1 2 0 4 3) has a pivot in every row

The augmented matrix ( 0 1 0 2 0 0 1 −3 ) is in reduced row echelon form.

True.

There is a 5x7 matrix A so that Ax=b is consistent for every b in R5.

True. A can be a 5x7 matrix with a pivot in every row.

If A is a 3 × 3 matrix and Ax = 0 has infinitely many solutions, then Ax = b must be inconsistent for some b in R 3 .

True. If Ax = 0 has infinitely many solutions then A has at most two pivots. Since A has three rows, this means A cannot have a pivot in every row, so Ax = b must be inconsistent for some b in R 3

Suppose v1 , v2 , v3 , b are vectors in R 3 . If the vector equation x1 v1 + x2 v2 + x3 v3 = b is inconsistent, then the vector equation x1 v1 + x2 v2 = b must also be inconsistent.

True. If b is not in Span{v1 , v2 , v3 } then it cannot be in Span{v1 , v2 }, every vector in Span{v1 , v2 } is contained in Span{v1 , v2 , v3 }.

If A is a 3 × 4 matrix and b is a vector so that the set of solutions to Ax = b is a line through the origin, then b =0.

True. If the solution set contains the origin then x = 0 is a solution so b = A(0) = 0. Very similar to a 3.3-3.4 supplemental problem

If A is a 4×3 matrix and the solution set for Ax = 0 is a line, then A has 2 pivots.

True. If the solution set to Ax = 0 is a line, then we must have exactly one free variable in the homogeneous solution, which means that exactly 2 out of the 3 columns of A will have pivots.

If A is a 5x7 matrix then at least 2 vectors are required to span the solution set for Ax=0

True. Minimum number of free variables would be 2 because 7-5=2. Thus need two different vectors for each free variable in the solution set

Suppose w1=(1 0 0), w2=(0 1 2), and w3=(-2 3 5). Then span{w1,w2,w3} is R3.

True. Pivot in every row so vectors span R3.

If A is a 2 × 3 matrix and the solution set to Ax = 0 is a plane in R 3 , then the equation Ax = b must be inconsistent for some b in R 2 .

True. Since A is 2 × 3 and the solution set to Ax = 0 is a plane, we have two free variables and thus A only has one pivot, so it cannot have a pivot in every row and thus Ax = b will be inconsistent for some b in R 2 .

If A is an m×n matrix and Ax = b has a unique solution for some b in R m, then Ax = 0 has only the trivial solution.

True. Since Ax = b is consistent, its solution set is a translation of the solution set to Ax = 0. Since Ax = b has a unique solution, this means Ax = 0 has a unique solution (namely the trivial solution).

If A is an m × n matrix and m > n, then then there is at least one vector b in R m which is not in the span of the columns of A.

True. The matrix A has at most n pivots, but it has m rows and m > n so it cannot have a pivot in every row

If the set of solutions to Ax=b is a line, then the set of solutions to Ax=0 must also be a line

True. The solution set being a line means there is one free variable. This does not change between Ax=b and Ax=0. The line is just translated to pass through 0 instead of b.

The three vectors (1,0,0), (1,1,0), and (-1,0,1) span R 3 .

True. The three vectors form a 3 × 3 matrix with a pivot in every row

If the bottom row of an augmented matrix in RREF is (0 1 2 3) , then the corresponding system of equations must be consistent

True. With the bottom row in view, we see that the rightmost pivot in the system is to the left of the augment, so the right column cannot be a pivot. In fact, the system has infinitely many solutions!

Suppose A is a 2x3 matrix and b is a vector so that Ax=b is consistent. Then -4b must be a linear combination of the columns of A.

True. b is in the column span of A, thus -4b is also in the column span of A.

If A is a 2 × 2 matrix and the equation Ax = (−2 −1) is consistent, then the vector (2 1) must be in the span of the columns of A.

True: the column span of A contains all scalar multiples of (−2 −1) , so it includes (2 1)

Is the vector (99 97) a linear combination of the vectors (3 4) and (5 6)?

Yes.

Suppose A is a 4x3 matrix. Must it be true that the set of solutions to Ax=0 is a span?

Yes.

Suppose A is a 4x5 matrix. Is it possible that Ax-b is consistent for all b in R4?

Yes.

Suppose that A is an 3x4 matrix and that b is the vector obtained by adding together the first two columns of A. Must it be true that Ax=b is consistent?

Yes.

Suppose we have four variables x1,x2,x3,x4. Is the set of solutions to x1+x2=0 a 3D plane?

Yes.

Complete the following definition (be mathematically precise!): Let w,v1,v2,...,vp be vectors in Rn. We say w is a linear combination of v1,v2,...,vp if ...

w=x1v2+...xpvp for some scalars x1,...,xp.