Linear Algebra Midterm 2
Leontif Model x - production vector d- final demand
(I-C)x=d or (I-C)^-1 d=x
The statement is true because it is the Multiplicative Property of determinants.
(det A)(det B)=det AB. Select the correct choice below and, if necessary, fill in the answer box to complete your choice
Basis of Col A {[5] [7]} {[5] , [3]} {[3] [3]} Find a Basis for Nul A {[4] [-5]} {[-7] [4]} {[1] , [0]} {[0] [1]}
A matrix A and an echelon form of A are shown below. Find a basis for Col A and a basis for Nul A. A= [5 7 29 -3] [1 2 10 -3] [5 3 1 13] ~ [0 1 7 -4] [3 3 9 3] [0 0 0 0]
True. If 0 is an eigenvalue of A, then there are nontrivial solutions to the equation Ax=0x. The equation Ax=0x is equivalent to the equation Ax=0, and Ax=0 has nontrivial solutions if and only if A is not invertible.
A matrix A is not invertible if and only if 0 is an eigenvalue of A. Choose the correct answer below
True. A number c is an eigenvalue of A if and only if the equation Ax=cx has nontrivial solutions, and Ax=cx and (A-cI)x=0 are equivalent equations.
A number c is an eigenvalue of A if and only if the equation (A-cI)x=0 has a nontrivial solution. Choose the correct answer below.
False; if A and B are invertible matrices, then left parenthesis AB right parenthesis Superscript negative 1equalsUpper B Superscript negative 1Upper A Superscript negative 1.
A product of invertible n x n matrices is invertible, and the inverse of the product is the product of their inverses in the same order.
The statement is false. Conditions (ii) and (iii) must be satisfied for each u and v in H, which is not specified in the given statement.
A subspace of |R^n is any set H such that (i) the zero vector is in H, (ii) u, v, and u+v are in H, and (iii) c is a scalar and Cu is in H.
True. An eigenspace of A corresponding to the eigenvalue lambda is the null space of the matrix (A-λI).
An eigenspace of A is a null space of a certain matrix. Choose the correct answer below.
The statement is false because scaling a row also scales the determinant by the same scalar factor.
An elementary row operation on A does not change the determinant. Choose the correct answer below.
A square lower triangular matrix is invertible when all entries on its main diagonal are nonzero. If all of the entries on its main diagonal are nonzero, then the nxn matrix has n pivot positions.
An mxn lower triangular matrix is one whose entries above the main diagonal are zeros. When is a square lower triangular matrix invertible? Justify your answer.
True. A steady-state vector for a stochastic matrix is actually an eigenvector because it satisfies the equation Ax=x.
A steady-state vector for a stochastic matrix is actually an eigenvector. Choose the correct answer below.
The matrix is not invertible. If a matrix has two identical columns then its columns are linearly dependent. According to the Invertible Matrix Theorem this makes the matrix not invertible.
Can a square matrix with two identical columns be invertible? Why or why not?
An example of a 2x2 matrix with only one distinct eigenvalue is [1 0] [0 1] .
Construct an example of a 2x2 matrix with only one distinct eigenvalue.
True; since each elementary matrix corresponds to a row operation, and every row operation is reversible, every elementary matrix has an inverse matrix.
Each elementary matrix is invertible.
False, because any subspace of |R^n must contain the zero-vector. Therefore, a line can only be a one-dimensional subspace of |R^n if it passes through the origin.
Each line in |R^n is a one-dimensional subspace of |R^n.
[ .7 0 -3.5] [0 1.4 8.4] [0 0 1 ]
Find the 3x3 matrix that produces the described composite 2D transformation below, using homogeneous coordinates. Translate by (- 5,6), and then scale the x-coordinate by 0.7 and the y-coordinate by 1.4.
The characteristic polynomial is λ^2 - 2λ - 5. The real eigenvalues are 1 + √6 , 1 - √6
Find the characteristic polynomial and the eigenvalues of the matrix. A= [ 3 2 ] [ 1 -1]
The characteristic polynomial is -λ^3 + 21λ + 16.
Find the characteristic polynomial of the matrix, using either a cofactor expansion or the special formula for 3x3 determinants. [ 0 2 1 ] [ 2 0 4 ] [ 1 4 0]
True. Checking whether a given vector u is in fact an eigenvector is easy because it only requires checking that u is a nonzero vector and finding if Au is a scalar multiple of u.
Finding an eigenvector of A may be difficult, but checking whether a given vector u is in fact an eigenvector is easy. Choose the correct answer below.
One eigenvalue of A is λ=0. This is because the columns of A are linearly dependent, so the matrix is not invertible.
For A= [-1 2 5] [-1 2 5] [-1 2 5], find one eigenvalue, with no calculation. Justify your answer.
Nonzero vector in Nul A [ 3 ] [-3] [ 1 ] [ 0 ] Nonzero vector Col A [ 4 ] [ -16 ] [ 12 ]
For the matrix A below, find a nonzero vector in Nul A and a nonzero vector in Col A. A = [4 3 -3 4] [-16 -9 21 4] [12 3 -27 12]
Nul A is a subspace of |R^p for p=3 and Col A is a subspace of|R^q for q=5
Give integers p and q such that Nul A is a subspace of set of |R^p and Col A is a subspace of |R^q. A= [-5 4 3] [-2 1 0] [3 -4 0] [8 9 0] [6 -6 1]
Find det(A-λI), where lambda is a scalar.
How can the characteristic polynomial of a matrix A be determined?
−x⋅cos(θ)+y⋅sin(θ)+x −x⋅sin(θ)−y⋅cos(θ)+y
How do you find the last coordinates of the last column when rotating around a point other than the origin.
False; B Superscript negative 1 Baseline Upper A Superscript negative 1 is the inverse of AB.
If A and B are n x n and invertible, then A Superscript negative 1 Baseline Upper B Superscript negative 1 is the inverse of AB.
True; since A can be row reduced to the identity matrix, A is row equivalent to the identity matrix. Since every matrix that is row equivalent to the identity is invertible, A is invertible.
If A can be row reduced to the identity matrix, then A must be invertible.
True; since A is invertible, A Superscript negative 1 b exists for all b in set of real numbers R Superscript n. Define x= A Superscript negative 1 b. Then Ax=b.
If A is an invertible n x n matrix, then the equation Ax=b is consistent for each b in set of real numbers R Superscript n.
False; if A is invertible, then the row operations required to reduce A to the identity correspond to some product of elementary matrices E 1 E2 E3 . . . Ep. Then the row operations required to reduce A Superscript negative 1 to the identity would correspond to the product Ep Superscript negative 1 . . . E3 Superscript negative 1 E2 Superscript negative 1 E1 Superscript negative 1.
If A is invertible, then elementary row operations that reduce A to the identity Upper I Subscript n also reduce Upper A Superscript negative 1 to Upper I Subscript n.
It is a known theorem that if A is invertible then Upper A Superscript negative 1 must also be invertible. According to the Invertible Matrix Theorem, if a matrix is invertible its columns form a linearly independent set. Therefore, the columns of Upper A Superscript negative 1 are linearly independent.
If A is invertible, then the columns of Upper A Superscript negative 1 are linearly independent. Explain why.
True; since Upper A Superscript negative 1 is the inverse of A, Upper A Superscript negative 1AequalsIequalsAUpper A Superscript negative 1. Since Upper A Superscript negative 1AequalsIequalsAUpper A Superscript negative 1, A is the inverse of Upper A Superscript negative 1.
If A is invertible, then the inverse of Upper A Superscript negative 1 is A itself.
False; if ad-bc=/0, then A is invertible.
If A=[a b, c d] and ab-cd=/0, then A is invertible.
True; if ad=bc then ad-bc=0, and 1/ad-bc, A=[a b, c d] is undefined.
If A=[a b, c d] and ad=bc, then A is not invertible.
False. The condition that Ax=λx for some scalar λ is not sufficient to determine if x is an eigenvector of A. The vector x must be nonzero.
If Ax=λx for some scalar λ, then x is an eigenvector of A. Choose the correct answer below.
False. The condition that Ax=λx for some vector x is not sufficient to determine if λ is an eigenvalue. The equation Ax=λx must have a nontrivial solution.
If Ax=λx for some vector x, then λ is an eigenvalue of A. Choose the correct answer below.
True, because any coordinate in a subspace H, with basis B, can only be written in one way as a linear combination of basis vectors. The linear combination gives a unique coordinate vector [x]B that is composed of the coordinates of x relative to B.
If B={v 1, . . . vp} is a basis for a subspace H and if x=c 1v 1 + ... + cp vp, then c 1, . . . , cp are the coordinates of x relative to the basis B.
It is not possible. Since Cx=v is consistent for every v in set of real numbers R^6, according to the Invertible Matrix Theorem that makes the 6 x 6 matrix invertible. Since it is invertible, Cx=v has a unique solution.
If C is 6 x 6 and the equation Cx=v is consistent for every v in set of real numbers R Superscript 6, is it possible that for some v, the equation Cx=v has more than one solution? Why or why not?
True, because if a set of p vectors spans a p-dimensional subspace H of |R^n, then these vectors must be linearly independent. Any linearly independent spanning set of p vectors forms a basis in p dimensions.
If a set of p vectors spans a p-dimensional subspace H of |R^n, then these vectors form a basis of H
Equation Dx=b has a solution for each b in set of real numbers R Superscript 7. According to the Invertible Matrix Theorem, a matrix is invertible if the columns of the matrix form a linearly independent set; this would mean that the equation Dx=b has at least one solution for each b in set of real numbers R^n.
If the columns of a 7 x 7 matrix D are linearly independent, what can you say about the solutions of Dx=b? Why?
Dimension of solution space=Nul A= 4
If the rank of a 4x6 matrix A is 2, what is the dimension of the solution space Ax=0?
Rank A=Col Space of A=4
If the subspace of all solutions of Ax=0 has a basis consisting of two vectors and if A is a 4x6 matrix, what is the rank of A?
False. There may be linearly independent eigenvectors that both correspond to the same eigenvalue.
If v1 and v2 are linearly independent eigenvectors, then they correspond to distinct eigenvalues. Choose the correct answer below.
The statement is true. The column space of A and S are both the set of all linear combinations of v1, . . . vp
If v1, . . . vp are in |R^n, then S= Span{v 1, . . . vp} is the same as the column space of the matrix A=[v1 . . . vp].
The statement is false because in order for 5 to be an eigenvalue of A, the characteristic polynomial would need to have a factor of λ - 5.
If λ+5 is a factor of the characteristic polynomial of A, then 5 is an eigenvalue of A. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
True, by definition of invertible.
In order for a matrix B to be the inverse of A, both equations AB=I and BA=I must be true.
It is not possible; according to the Invertible Matrix Theorem an n x n matrix cannot be invertible when its columns do not span set of real numbers R Superscript n.
Is it possible for a 5 x 5 matrix to be invertible when its columns do not span set of real numbers R Superscript 5? Why or why not?
The value of h for which the eigenspace for λ=7 is two-dimensional is h=6.
It can be shown that the algebraic multiplicity of an eigenvalue lambda is always greater than or equal to the dimension of the eigenspace corresponding to λ. Find h in the matrix A below such that the eigenspace for λ=7 is two-dimensional. [ 7 -2 6 -2 ] [ 0 5 h 0 ] [ 0 0 7 8 ] [ 0 0 0 4]
Since Col A is p-dimensional, then the p columns of A span Col A and therefore, the spanning set of p columns is automatically a basis for Col A. This implies that the p columns of A are linearly independent.
Let A be an nxp matrix whose column space, denoted Col A, is p-dimensional. Explain why the columns of A must be linearly independent.
The number of pivots=number of vectors in basis=dimension of Col A. Also = to rank of A
Relationship between basis for Col A and dimension of Col A?
number of free variables=number of vectors in Nul A=dimension to sol set
Relationship between basis of Nul A and dimension of solution set to Ax=0?
The statement is true. If a series of row operations is performed on a matrix A to form B, then the equations Ax=0 and Bx=0 have the same set of solutions.
Row operations do not affect linear dependence relations among the columns of a matrix.
If the equation Ax=b has a solution for each b in set of real numbers R Superscript n, then A has a pivot position in each row. Since A is square, the pivots must be on the diagonal of A. It follows that A is row equivalent to Upper I Subscript n. Therefore, A is invertible.
Suppose A is nxn and the equation Ax=b has a solution for each b in set of real numbers R Superscript n. Explain why A must be invertible. [Hint: Is A row equivalent to I Subscript n?]
If Col =/ F |R^5, then the columns of F do not span |R^5. Since F is square, the Invertible Matrix Theorem shows that F is not invertible and the equation Fx=0 has a nontrivial solution. Therefore, Nul F contains a nonzero vector.
Suppose F is a 5 x 5 matrix whose column space is not equal to |R^5. What can you say about Nul F?
Since Col A is the set of all linear combinations of a1, . . . ap, the set {a1, . . . ap} spans Col A. Because {a1, . . . ap} is also linearly independent, it is a basis for Col A.
Suppose the columns of a matrix A=[a1, . . . ap] are linearly independent. Explain why {a1, . . . ap} is a basis for Col A.
The statement is true. The columns of an invertible nxn matrix are linearly independent and span |R^n, so they form a basis for |R^n.
The columns of an invertible nxn matrix form a basis for |R^n.
Since all the entries are positive, every sector must increase its output by some positive quantity. An increase in demand for any sector will increase the demand for every sector.
The consumption matrix C for the a certain country's economy for a particular year has the property that every entry in the matrix (I-C)^-1 is nonzero (and positive). What does that say about the effect of raising the demand for the output of just one sector of the economy? Since all the entries are positive, ________________ must ________________output by some positive quantity. An increase in demand for any sector will ________________ the demand for ________________.
The statement is false because the determinant of the 2times2 matrix A = [ 2 1 ] [ 1 2 ] is not equal to the product of the entries on the main diagonal of A.
The determinant of A is the product of the diagonal entries in A. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
True, because the pivot columns of A form a basis for Col A. Therefore, the number of pivot columns of A is the same as the dimension of Col A.
The dimension of Col A is the number of pivot columns in A.
True, because the Rank Theorem states that if matrix A has n columns, then rank A + dim Nul A = n. Since rank A is the same as dim ColA, the dimensions of ColA and NulA add up to the total number of columns in A.
The dimensions of Col A and Nul A add up to the total number of columns in A.
False. If the matrix is a triangular matrix, the values on the main diagonal are eigenvalues. Otherwise, the main diagonal may or may not contain eigenvalues.
The eigenvalues of a matrix are on its main diagonal. Choose the correct answer below.
The statement is false. The described set is the null space of an mxn matrix A. This set is a subspace of |R^n.
The set of all solutions of a system of m homogeneous equations in n unknowns is a subspace of |R^m.
False. An echelon form of a matrix A usually does not display the eigenvalues of A.
To find the eigenvalues of A, reduce A to echelon form. Choose the correct answer below.
The point in |R^3 is (8, 4, -2)
What vector in set of |R^3 has homogeneous coordinates (1/6, 1/12, -1/24, 1/48)?
Set up and augmented matrix with [x]B as the last column. Row reduce and the if you get two pivot columns/rows, the numbers in the last column is x.
When given B1 and B2 (both bases) and [x]B, how do you solve for x?
set up the vector equation x=c1b1 + c2b2. Turn that into an augmented matrix
When given B1 and B2 (both bases) and the x vector, how do you solve for [x]B?
On their diagonal
Where do you find eigenvalues on a triangular matrix?