linear algebra midterm study
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Multiple choice
Let T : R^n -> R^n be an isomorphism. Which of the following is always true?
B) [T] is invertible
Let {u1, u2, u3} be a basis for vector space V. Which of the following is also a basis for V?
C) {u1 −u2, u3 −u2, u1 +u3}
Let n ≥ 1 be an integer, and consider the mapping T_n : P2 -> R defined as T_n(a
+ bx + cx^2) = a + b^n − c^n. The mapping T_n is linear
A) only for n = 1
The eigenspace of the matrix [-4, 15; -2, 7] for eigenvalue λ = 2 is:
B) Span{[5; 2]}
The eigenvalues of the matrix [2, -1; -1, 2] are:
D) λ = 1 and λ = 3
The eigenvalues of the matrix [1, 2; 2, 1] are:
A) λ = −1 and λ = 3
The eigenvalues of the matrix [-2, -1; -1, -2] are:
D) λ = −1 and λ = −3
The eigenvalues of the matrix [3, 3; 5, 1] are:
D) λ = −2 and λ = 6
If A is a 4 x 4 matrix with det A = 10, then det(A + I) is:
D) Cannot be determined
If A is a 4×4 matrix with det A = 2, then det(cof A) is:
D)8
If A is a 4 × 4 matrix with det A = −2, then det(cof A) is:
D) −8
The determinant of the matrix [0, 0, -3; -2, -1, 3; 3, -4, -4] is:
C) −33
The determinant of the matrix [2, 0, 0; -1, 3, -2; 3, 3, 1] is:
B) 18
The determinant of the matrix [3, 2, 0, 0; -3, 1, 0, 0; 0, 0, -4, 2; 0, 0, 4, -1] is:
A)-36
The determinant of the matrix [3, 0, 1; 0, 2, -1; 0, 1, 1] is:
B) 9
Question - Gram-Schmidt
Use the Gram-Schmidt procedure to produce an orthonormal basis for the sub-
space spanned by
B = {[1; -1; 1], [1; 0; 1], [1; 1; 2]}
Do not change the order of the vectors.
Denote the given basis vectors by w1, w2, and w3 respectively. Step 1: Let v1 = w1.
Step 2: perp w2 = [1; 0; 1] - 2/3[1; -1; 1] = -1/3[1; 2; 1] so take v2 = [1; 2; 1]
Step 3: perp w3 = [1; 1; 2] - 2/3[1; -1; 1] - 5/6[1; 2; 1] = 1/2[-1; 0; 1] so take v3 = [-1; 0; 1]
Normalizing we get that an orthonormal basis for this subspace is
{1/sqrt(3) [1; -1; 1], 1/sqrt(6) [1; 2; 1], 1/sqrt(2) [-1; 0; 1]}
(b) Find the coordinates of the vector x = 2 with respect to the orthonormal basis obtained in part (a).
[x]_B = [x1 * u1; x2 * u2; x3 * u3] = [0; sqrt(6); 0]
Question - Gram schmid again
Use the Gram-Schmidt procedure to produce an orthonormal basis for the sub- space spanned by
{[1; -2; 1], [0; 1; 0], [0; 0; 1]}
Do not change the order of the vectors.
v1 = [1; −2; 1], S1 = Span{v1}.
perp_S1 = [0; 1; 0] = [0; 1; 0] - (-2/6) [1; -2; 1] = [1/3; 1/3; 1/3], choose v2 = [1; 1; 1], S2 = Span{v1 ,v2 }
perp_S2 [0; 0; 1] = [0; 0; 1] - 1/6[1; -2; 1] - 1/3[1; 1; 1] = [-1/2; 0; 1/2], choose v3 = [-1; 0; 1]
Then an orthonormal basis for the subspace is {1/sqrt(6)[1; -2; 1], 1/sqrt(3)[1; 1; 1], 1/sqrt(2)
[-1; 0; 1]}
(b) Find the coordinates of the vector x = [-2; 3; 0] with respect to the orthonormal basis obtained in part (a).
X = [-2; 3; 0] = (-8/sqrt(6)) 1/sqrt(6)[1; -2; 1] + (1/sqrt(3)) 1/sqrt(3)[1; 1; 1] + (2/sqrt(2)) 1/sqrt(2)[-1; 0; 1]
So [X]_B = [-8/sqrt(6); 1/sqrt(3); 2/sqrt(2)]
Question - linear mapping, bases
4. Let T : R^2 -> R^2 be the linear mapping defined by T[x; y] = [2y; x+y] and consider the bases B1 = {[0; 1], [1; -1]) and B2 = {[-1; 1], [1; 1]}
(a) Find the matrix M_B2B1 of T with respect to the given bases.
M_B2B1 = [C_B2(T[0; 1]) C_B2(T[1; -1)]
= [C_B2([2; 1]) C_B2([-2; 0)]
= [C_B2(-1/2[-1; 1] + 3/2[1; 1]) C_B2(1[-1; 1] - 1[1; 1])
= [[-1/2; 3/2], [1; 1]] = [-½, 1; 3/2, -1]
(b) Find C_B2 (T (u)) where u = [1; 1] using the results in part (a).
u = [1; 1] = 2[0; 1] + [1; -1] so C_B1(u) = [2; 1]
So C_B2(T(u)) = M_B2B1 C_B1 (u) = [-1/2, 1; 3/2, -1] [2; 1] = [0, 2]
Question - linearly dependent or independent
5. (a) Is the set {x^2 + x + 1, x^2 − 1, −x + 1} linearly dependent or independent in P^2? Justify your answer.
Let a(x^2+x+1) + b(x^2 - 1) + c(-x + 1) = 0x^2+0x+0. Then
[1, 1, 0; 1, 0, -1; 1, -1, 1] [a; b; c] = [0; 0; 0]
[1, 1, 0; 1, 0, -1; 1, -1, 1] row reduces to [1, 0, 0; 0, 1, 0; 0, 0, 1] therefore a = b = c = 0.
So yes, the set is linearly independent.
(b) Consider the vector space M_2,2. Is the matrix A = [2, 1; 1, 4] in the set
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Span{[1, -3; 0, -2], [-4, 2; 2, 1], [1, -4; -2, -1]}? Justify.
Let a[1, -3; 0, -2] + b[-4, 2; 2, 1] + c[1, -4; -2, -1] = [2, 1; 1, 4]
Then [1, -4, 1; -3, 2, -4; 0, 2, -2; -2, 1, -1] [a; b; c] = [2; 1; 1; 4]
[1, -4, 1; -3, 2, -4; 0, 2, -2; -2, 1, -1 | 2; 1; 1; 4] row reduces to [1, -4, 1; 0, -1, -5; 0, 0, -1; 0, 0, 0 | 2; 11; 11; -109] so this system has no solutions.
So no, A is not in the set
Question - Transformation
6. Let T : R^3 -> R^3 be the transformation induced by the matrix A and let S : R^2 -> R^3 be defined by:
S[a;b] = [a-b; 2a+2b; -a+2b] where A = [1, -3, -3; 1, -2, 6; -2, 6, 1]
(a) Is S one-to-one? Explain.
[S] = [1, -1; 2, 2; -1, 2] which row reduces to [1, -1; 0, 1; 0, 0] so rank([S]) = 2 = dim(R^2), therefore yes, S is one-to-one.
(b) Is S onto? Explain.
Since rank([S]) = 2 does not equal 3 = dim(R^3) no, S is not onto.
(c) Is S an isomorphism? Explain.
S is not onto, therefore no it is not an isomorphism.
(d) Is T one-to-one? Explain.
[T] = A = [1, -3, -3; 1, -2, 6; -2, 6, 1] which row reduces to [1, -3, -3; 0, 1, 9; 0, 0, 1] so rank(A)
= 3 = dim(R^3), therefore yes, T is one-to-one.
(e) Is T onto? Explain.
Since rank(A) = 3 = dim(R^3) yes, T is onto.
(f) Is T an isomorphism? Explain.
T is both one-to-one and onto, therefore yes it is an isomorphism.
Question - Transformation rotation through an angle
Let T : R2 -> R3 be defined by T[x; y] = [x-y; 2x; x+y] and let S : R2 -> R2 be rotation through
an angle θ = 2pi/3
(a) Show that T is a linear transformation.
Let u
= [x; y], v= [a; b], and p, q ∈ R. Then we have:
T(pu+qv) = T [px+qa; py+qb]
= [(px + qa) − (py + qb); 2(px + qa); (px + qa) + (py + qb)]
= [px−py; 2px; px+py] + [qa−qb; 2qa; qa+qb]
= p[x−y; 2x; x+y] + q[a−b; 2a; a+b]
= pT (u) + qT (v)
Therefore T is a linear transformation.
(b) Determine the standard matrix of T * S.
[T] = [1, -1; 2, 0; 1, 1] and [S] = [cos(2pi/3) - sin(2pi/3); sin(2pi/3) - cos(2pi/3)] = [-1/2, -
sqrt(3)/2; sqrt(3)/2, -1/2]
So [T] * [S] = [T][S] = [1, -1; 2, 0; 1, 1] [-1/2, -sqrt(3)/2; sqrt(3)/2, -1/2] = [(-1-sqrt(3))/2, (1-
sqrt(3))/2; -1, -sqrt(3); (-1+sqrt(3))/2, (1-sqrt(3))/2]
Question - find all eigenvalues
Let A = [-4, 3; 8, -2]
(a) Find all eigenvalues of A and their corresponding eigenvectors.
det(λI −A) = 0 -> (λ+4)(λ+2)−24 = 0 -> λ2 +6λ−16 = 0 -> λ = −8 or λ = 2
For λ = −8, −8I − A = [-4, -3; -8, -6] RREF-> [1, 3/4; 0, 0] so eigenvectors are t [3; -4]
For λ = 2, 2I − A = [6, -3; -8, 4] RREF-> [1, -1/2; 0, 0] so eigenvectors are t [1; 2]
(b) Is A diagonalizable? If so, find a matrix P and diagonal matrix D such that P^(−1) AP = D.
Yes, A is diagonalizable since the geometric multiplicity of each eigenvalue is equal to its algebraic multiplicity.
P = [3, 1; -4, 2] and D = [8, 0; 0, 2]
Question - cofactor
Let A = [2, -5, 4; −5, −4, −2; 4, 1, 2]
Cofactor matrix
cofA = [-6, 14, 26; 2, -12, -16; 11, -22, -33]
Cofactor method detA = 22 so A^(−1) = 1/det(A) * (cofA)^T = 1/22 * [-6, 14, 26; 2, -12, -16; 11, -
22, -33]
Question - cramers rule
6. Use Cramer’s Rule to solve the following system of equations:
2x1+ x2+ x3=1
x1− x2+4x3=0
x1 +2x2 −2x3 =3
D = |2, 1, 1; 1, -1, 4; 1, 2, -2| = -3
D1 = |1, 1, 1; 0, -1, 4; 3, 2, -2| = 9
D2 = |2, 1, 1; 1, 0, 3; 1, 2, -2| = -15
D3 = |2, 1, 1; 1, -1, 4; 1, 0, 3| = -6
So
X1 = D1/D = -3
X2 = D2/D = 5
X3 = D3/D = 2
Question - System of equation determinants
7. Consider the system of equations
kx1 − kx2 + 3x3 = 3
(k+1)x2+ x3 =0
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kx1 − 8x2 + (k−1)x3 = 1
(a) Using determinants, find all values of k for which the system has a unique solution.
Let A= [k, -k, 3; 0, (k+1), 1; k, -8, (k-1)] . Then detA = k(k−2)^2.
Therefore the system has a unique solution if and only if k does not = 0, 2.
(b) Find det A^(−1) if k=1.
When k = 1, detA = 1(−1)^2 = 1, so det(A^−1) = 1/detA = 1.
(c) Use Cramer’s rule to determine x2 in the above system when k = 1.
We already know from part (b) that det A = 1 in this case. Det N2 =det [1, 3, 3; 0, 0, 1; 1, 1, 0] = 2
Therefore x2 = 2/1 = 2.
Question - eigenvalues eigen vectors diagonalizable
Let A = [-1, 6, 3; 3, −4, −3; -6, 12, 8]
(a) Find all eigenvalues of A and their corresponding eigenvectors.
det(A−λI)= |-1-λ, 6, 3; 3, −4−λ, −3; −6, 12, 8−λ| =−λ^3 +3λ^2 −4
= −(λ − 2)^2(λ + 1)
So λ1 = 2 and λ2 = -1 are the eigenvalues of A
A - λ1I = [−3, 6, 3; 3, -6, -3; -6, 12, 6] ~ [1, -2, -1; 0, 0, 0; 0, 0, 0] so E_2 span {[2; 1; 0], [1; 0; 1]}
A - λ2I = [0, 6, 3; 3, -3, -3; -6, 12, 9] ~ [1, 0, -1/2; 0, 1, 1/2; 0, 0, 0] so E_-1 span {[1; -1; 2]}
Where E_λ is the eigenspace corresponding to eigenvalue λ.
(b) Is A diagonalizable? Explain why or why not.
Yes, A is diagonalizable since the geometric multiplicity of each eigenvalue (2 and 1 for λ1 = 2 and λ2 = −1, respectively) is equal to its algebraic multiplicity.
(c) If A is diagonalizable find matrix P and diagonal matrix D such that P^(−1)AP = D.
P = [2, 1, 1; 1, 0, -1; 0, 1, 2] and D = [2, 0, 0; 0, 2, 0; 0, 0, -1]
Question - a is not invert values
For the matrix A = [1, a, 2+a; a, 4, 4; a, 4, 6]
(a) Compute det(A) as a function of a.
|1, a, 2+a; a, 4, 4; a, 4, 6| = |4, 4; 4, 6| - a|a, 4; a, 6| + (2+a)|a, 4; a, 4|
= 8 − a(2a) + (2 + a)(0) = 8 − 2a^2.
(b) Find all values of a for which A is not invertible.
A is not invertible if det(A) = 0 if 8−2a^2 =2(2−a)(2+a)=0 if a = +-2.
(c) For a = 3, find det(AA^T ).
When a = 3, det(A) = −10 so det(AA^T) = det(A)det(A^T) = (det(A))^2 = (−10)^2 = 100.
(d) For a = 3, find A−1 using the cofactor method.
For a=3, det(A)= −10, and cof(A)= [8, -6, 0; 2, −9, 5; -8, 11, -5],
A^−1 = 1/det(A) (cof(A))^T = [−8/10, 6/10, 0; −2/10, 9/10, −5/10; 8/10, −11/10, 5/10]^T
= [−4/5, −1/5, 4/5; 3/5, 9/10, −11/10; 0, −1/2, 1/2]