Math 245 - Linear Algebra - Exam 2 - Fall 2022
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Math 245 - Linear Algebra - Exam 2 - Fall 2022
1. Circle your choice. No explanation required. (1 pts each)
(1) True – False: The set {(x, y) : x2 + y2 = 0} is a subspace of R2 .
True: Notice that this set is consists only of {(0, 0)} which is a subspace.
(2) True – False: The non-zero columns of A form a basis for the column space of A.
False: It should read the pivot columns of A form a basis for the column space of A.
(3) True – False: If A is a 4 × 6 matrix with 2 pivot columns, then N(A) = R4 .
False: The rank theorem says that dim N(A) = 4. However, N(A) is a subspace of R6 .
(4) True – False: If A is a 3 ×3 matrix with det(A) = 0, then one row must be a scalar multiple of another row.
False: The det(A) = 0 condition just says that the three rows are linearly dependent. There are several ways for three rows to be linearly dependent without one being a mul- tiple of the other. For example, row three can be the sum of the first two rows.
(5) True – False: The matrix [0(1) 1(1)] is diagonalizable.
False: The eigenvalues are 1, 1 but the corresponding eigenspace is one dimensional.
(6) True – False: If A is diagonalizable, then A is invertible.
False: The zero matrix is diagonalizable (in fact diagonal) but is not invertible
(7) True – False: If A is a 3 × 3 matrix with det(A) = 1, then the rank of A is 3.
True: The condition det A = 1 says that A is invertible and thus, since A is 3 × 3, it must have rank equal to 3.
(8) True – False: If A100 is invertible, then A is invertible.
True: If A100 is invertible, then 0 det(A100) = det(A)100 and thus det A 0 and hence A is invertible.
(9) True – False: If zero is an eigenvalue of A, then A is not invertible.
True: If 0 is an eigenvalue, then there is a nonzero vector v such that Av = 0. This says that N(A) {0} and thus A is not invertible.
(10) True – False: A 3 × 3 matrix with eigenvalues 1, 2, 4 must be diagonalizable.
True: Eigenvectors from different eigenspaces must be linearly independent. Since there are three distinct eigenvectors, there must be three linearly independent eigenvec- tors – enough to form a basis in R3 .
(11) True – False: A 3 × 3 matrix with eigenvalues 1, 1, 4 can never be diagonalizable.
False: The diagonal matrix with diagonal entries 1, 1, 4 is diagonalizable.
(12) True – False: If A is an n × n diagonalizable matrix, then each vector in Rn can be written as a linear combination of eigenvectors of A.
True: If A is diagonalizable, then A has a basis of eigenvectors and thus every vector an be written as a linear combination of eigenvectors of A.
2. Compute a basis for the subspace {(x, y, z, w) ∈ R4 : x − y + z + w = 0, x + 2z − w = 0, x + y + 3z − 3w = 0}. (8 points)
The subspace above is the null space of the matrix
l 」
「1 1 3 −3 .
This matrix row reduces to l0(1) 「 |
0 1 0 |
2 1 0 |
2(1)」 0 . |
The says that the variables z and w are free variables and thus every vector in the null space of A
can be written as
l x 」 l −2」
「w(z) = z「 0(1)
Thus
l 1(2)」
「 0(1)
l1」
「1 .
2
form a basis for the null space – and hence the subspace.
3. Find a basis for the subspace spanned by the vectors
(1, −1, −2, 3), (2, −3, −1, 4), (0, −1, 3, −2), (−1, 4, −7, 7), (3, −7, 6, −9).
(8 pts)
Consider the span of these vectors as the column space of
l 1 「3(−)2 |
2 −3 −1 4 |
0 −1 3 −2 |
−1 4 −7 7 |
7」 9 . |
This matrix row reduces to l1 0 |
0 1 |
−2 1 |
0 0 |
」 − |
0 0 |
0 0 |
0 0 |
1 − . 0 0 |
From here one can see that the first, second, and fourth columns of the original matrix are pivot columns and thus the vectors (1, −1, −2, 3), (2, −3, −1, 4), (−1, 4, −7, 7) form a basis for the span.
4. Solve the system x\(t) = − x(t) + y(t), y\(t) = x(t) − y(t) with initial condition x(0) = 2, y(0) = 1. Be sure to show all work. (10 points)
This system can be written in matrix form as
X\ = AX, X(0) = [ ]1(2) ,
where
A = [ 1] .
This matrix is diagonalizable with
A = [ 1 ][1 ] −1 .
Thus
X(t) = [ 1 2(1)t] [ 1 ] −1 [ ]1(2) .
A calculation yields
X(t) = t2(1)t] .
Thus
x(t) = e −2t + e − t and y(t) = −e−2t + 2e − t .
5. Match each system with its trajectories. Provide brief explanations.
(1) x\ = −x + 4y, y\ = −2x + 5y
(2) x\ = x − 4y, y\ = 2x − 5y
(3) x\ = −5x + 8y, y\ = −4x + 7y
(4) x\ = 5x − 8y, y\ = 4x − 7y
The general solution to x\ = −x + 4y, y\ = −2x + 5y in vector form is
X(t) = c1e3t [ ]1(1) + c2et [ ]1(2) .
This matches figure 3.
The general solution to x\ = x − 4y, y\ = 2x − 5y in vector form is
X(t) = c1e−3t [ ]1(1) + c2e−t [ ]1(2) .
This matches figure 2.
The general solution to x\ = −5x + 8y, y\ = −4x + 7y in vector form is
X(t) = c1e3t [ ]1(1) + c2e−t [ ]1(2) .
This matches figure 4.
The general solution to x\ = 5x − 8y, y\ = 4x − 7y in vector form is
X(t) = c1e−3t [ ]1(1) + c2et [ ]1(2) .
This matches figure 1.
2022-12-01