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Math 333 - Practice Final Exam with Some Solutions
(Note that the exam will NOT be this long.)
1 Definitions
- (0 points) Let A be an m × n matrix and b ∈ Rm. Let T : V −→ V be a linear transformation on a vector space V. (a) Define “Ax = b is consistent”.
(b) Define “Ax = b is homogeneous”.
(c) If m = n, then define “A is diagonalizable”.
(d) Define “T is diagonalizable”.
(e) If m = n, then define “v is an eigenvector of A with eigenvalue λ.”
(f) Define eigenvalue and eigenvector of T.
(g) Define the eigenspace of A associated to eigenvalue λ.
2 Inverses
- (0 points) Consider the matrix A =
^11 10
(a) Find the inverse of A (using the Gaussian elimination method).
Solution: A−^1 =
^11 −^01 − 01
(b) Show that your matrix is, in fact, the inverse of A.
Solution: AA−^1 =
^11 10
^11 −^01 − 01
^10 01
and
A−^1 A =
^11 −^01 − 01
^11 10
^10 01
- (0 points) Derive the formula for the inverse of a 2 × 2 matrix. That is, what is
A−^1 if A =
a b c d
? (Don’t just state the formula.)
Solution:( Hint: The definition of inverse says A−^1 is also a 2 × 2 matrix, say A−^1 = x y z w
, such that AA−^1 = I. Multiply A and A−^1 , set it equal to I, and solve for x, y,
z, and w in terms of a, b, c, and d.
- (0 points) Consider the matrix A =
(a) Find the inverse of A.
(b) Suppose A is the change of basis matrix from the standard ordered basis α = {e 1 , e 2 , e 3 } of R^3 to some other ordered basis β = {b 1 , b 2 , b 3 } of R^3. Then what is the ordered basis β?
Solution: (a) A−^1 =
−^11 //^33 − 74 // 33 − 41 // 33
(b) Suppose β = {b 1 , b 2 , b 3 } and α = {e 1 , e 2 , e 3 }. Since A is the change of basis matrix from α to β, then A[bi]α = [bi]β. Since [bi]β = ei and [bi]α = bi, we get Abi = ei, or equivalently, A−^1 ei = bi. So the ith^ vector in β is the ith^ column of A−^1. Therefore,
β =
−^11 //^33
^ − 74 // 33
^ − 41 // 33
^12
^21
^ − 11
form the basis of the column space. The nullspace of A has basis
^.
(d) rank(A) = dimension of column space = 3 and nullity(A) = dimension of null(A) = 1.
- (0 points) Consider the following inhomogeneous system of linear equations.
x + y + z + w = 1 x + y = 0 x + w = 0
(a) Write the coefficient matrix A for this system.
(b) Find the solution space to the associated homogeneous system.
(c) By simple inspection, find just one solution to the given inhomogeneous system.
(d) Find all solutions to the given inhomogeneous system.
- (0 points) Consider the matrix A =
− 2 h
and vector b =
k 1
. Find all
possible values of h and k so that the matrix equation Ax = b has: (a) no solution.
(b) exactly one solution.
(c) infinitely many solutions.
Solution: For each case we need to look at the augmented matrix (A|b) =
1 1 k − 2 h 1
Performing one row operation we arrive at the new augmented matrix
(B|c) =
1 1 k 0 h + 2 1 + 2k
and the system Ax = b is equivalent to Bx = c. (a) Bx = c has no solutions if and only if rank(B) < rank(B|c). This is when the second column does not have a pivot,but the augmented column does. This occurs if and only if h = −2 and k 6 = − 1 /2.
(b) Bx = c has exactly one solution if and only if B is invertible which occurs if and only if rank(B) = 2 if and only if the first and second columns have pivots if and only if h 6 = −2.
(c) Bx = c has infinitely many solutions in the remaining cases and so if and only if h = −2 and k = − 1 /2.
- (0 points) Suppose A is an m × n matrix with rank(A) = 0. Prove A = O.
Solution: The fact that rank(A) = 0 implies that there are NO pivot columns. So there are no pivots, which means NO row has a first non-zero entry. Thus each row consists entirely of zeros. Therefore A = O.
4 Determinants
- (0 points) Find the determinants of the following matrices in two ways:
- By using cofactor expansion along a row or column.
- By invoking some basic facts about determinants. Please state which general facts about determinants you are invoking. (In other words, you must explain what you are doing.)
(a) A =
^00 01
5 Diagonalization and Eigenspaces
- (0 points) Consider A ∈ Mn×n(F ). Prove that the vector v ∈ Rn^ (v 6 = ~0) is an eigenvector of A corresponding to eigenvalue λ if and only if v ∈ null (A − λI).
Solution: The vector v is an eigenvector of A corresponding to eigenvalue λ ⇐⇒ Av = λv ⇐⇒ Av − λv = ~ 0 ⇐⇒ (A − λI)v = ~ 0 ⇐⇒ v ∈ null (A − λI)
- (0 points) Find all eigenvalues and the corresponding eigenspaces for the following matrices. For each eigenvalue, give the algebraic multiplicity and geometric multiplicity.
(a)
(b)
Solution: (a) p(λ) = det(A−λI) = det
1 − λ 0 2 1 − λ
= (1−λ)(1−λ) = λ^2 − 2 λ+1.
So the eigenvalues are repeated with λ = 1. The eigenspace is
Eλ = null(A − I) = null
t
The eigenvalue λ = 1 has algebraic multiplicity 2 and geometric multiplicity 1.
(b) p(λ) = det(A − λI) = det
1 − λ 2 2 1 − λ
= (1 − λ)(1 − λ) − 4 = λ^2 − 2 λ − 3 =
(λ − 3)(λ + 1). So the eigenvalues are λ 1 = 3 and λ 2 = −1. The eigenspace for λ 1 is
Eλ 1 = null(A − 3 I) = null
= null
t
The eigenspace for λ 2 is
Eλ 2 = null(A + I) = null
= null
t
The eigenvalues λ 1 = 3 and λ 2 = −1 both have algebraic multiplicity 1 and geometric multiplicity 1.
- (0 points) Consider the matrix A =
(a) Find the characteristic polynomial of A.
(b) A is diagonalizable. That is, [LA]β is a diagonal matrix for some basis β of R^3. Find such a basis β.
(c) For the basis β you found above, what is [LA]β?
(d) We also know that A is similar to a diagonal matrix D. Find the matrices D and Q such that D = Q−^1 AQ.
Solution: The characteristic polynomial is
p(λ) = det
^1 −− 1 λ 1 −^0 λ 01 1 0 −λ
= −λ(1 − λ) (^2).
Thus the eigenvalues of A are 0 and 1. To get the basis β we must determine the corresponding eigenspaces:
Row reducing A − 0 I =
leads to
. Thus the corresponding
eigenspace is 1-dimensional with basis
Row reducing A − 1 I =
leads to
. Thus the correspond-
ing eigenspace is 2-dimensional with basis
