Linear Algebra Quiz: Characteristic Polynomial
15 questions · exam conditions
0:00
Characteristic PolynomialQuestion 1 of 15

Two 3×33 \times 3 matrices AA and BB are similar. If the characteristic polynomial of AA is pA(λ)=−λ3+4λ2−5λ+2p_A(\lambda) = -\lambda^3 + 4\lambda^2 - 5\lambda + 2, what is the characteristic polynomial of BB, denoted pB(λ)p_B(\lambda)?

pB(λ)=−λ3+4λ2−5λ+2p_B(\lambda) = -\lambda^3 + 4\lambda^2 - 5\lambda + 2
pB(λ)=λ3−4λ2+5λ−2p_B(\lambda) = \lambda^3 - 4\lambda^2 + 5\lambda - 2
pB(λ)p_B(\lambda) has the same roots as pA(λ)p_A(\lambda), but the polynomials may be different.
The constant terms of pA(λ)p_A(\lambda) and pB(λ)p_B(\lambda) are the same, but other coefficients may differ.
← Back to quizzes

Linear Algebra Quiz

Linear Algebra Quiz: Characteristic Polynomial

Practice Characteristic Polynomial in Linear Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Characteristic Polynomial, giving you a quick way to practice the rules, question types, and explanations that matter most for Linear Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Two 3×33 \times 3 matrices AA and BB are similar. If the characteristic polynomial of AA is pA(λ)=−λ3+4λ2−5λ+2p_A(\lambda) = -\lambda^3 + 4\lambda^2 - 5\lambda + 2, what is the characteristic polynomial of BB, denoted pB(λ)p_B(\lambda)?

  1. pB(λ)=−λ3+4λ2−5λ+2p_B(\lambda) = -\lambda^3 + 4\lambda^2 - 5\lambda + 2 (correct answer)
  2. pB(λ)=λ3−4λ2+5λ−2p_B(\lambda) = \lambda^3 - 4\lambda^2 + 5\lambda - 2
  3. pB(λ)p_B(\lambda) has the same roots as pA(λ)p_A(\lambda), but the polynomials may be different.
  4. The constant terms of pA(λ)p_A(\lambda) and pB(λ)p_B(\lambda) are the same, but other coefficients may differ.
Explanation: A fundamental property of similar matrices is that they share the same characteristic polynomial. If AA and BB are similar, there exists an invertible matrix PP such that B=P−1APB = P^{-1}AP. The characteristic polynomial of BB is: pB(λ)=det⁡(B−λI)=det⁡(P−1AP−λP−1P)=det⁡(P−1(A−λI)P)p_B(\lambda) = \det(B - \lambda I) = \det(P^{-1}AP - \lambda P^{-1}P) = \det(P^{-1}(A - \lambda I)P). Using the property det⁡(XYZ)=det⁡(X)det⁡(Y)det⁡(Z)\det(XYZ) = \det(X)\det(Y)\det(Z), we get: pB(λ)=det⁡(P−1)det⁡(A−λI)det⁡(P)=det⁡(P)−1pA(λ)det⁡(P)=pA(λ)p_B(\lambda) = \det(P^{-1})\det(A - \lambda I)\det(P) = \det(P)^{-1} p_A(\lambda) \det(P) = p_A(\lambda). Therefore, the characteristic polynomials are identical. Distractor B is the negative of the correct polynomial, which is incorrect. Distractor C is subtly wrong; because the characteristic polynomials are identical, they cannot be different. Distractor D is an incomplete statement; while the constant terms (the determinants) are the same, all other corresponding coefficients (like those related to the trace) are also the same.

Question 2

Let MM be the matrix M=(41−20−13004)M = \begin{pmatrix} 4 & 1 & -2 \\ 0 & -1 & 3 \\ 0 & 0 & 4 \end{pmatrix}. Which of the following is the characteristic polynomial of MM?

  1. p(λ)=(λ−4)2(λ+1)p(\lambda) = (\lambda-4)^2(\lambda+1)
  2. p(λ)=−(λ−4)2(λ+1)p(\lambda) = -(\lambda-4)^2(\lambda+1) (correct answer)
  3. p(λ)=−λ3+7λ2+8λ−16p(\lambda) = -\lambda^3 + 7\lambda^2 + 8\lambda - 16
  4. p(λ)=(λ−4)(λ+1)(λ−4)(λ−1)(λ+2)(λ−3)p(\lambda) = (\lambda-4)(\lambda+1)(\lambda-4)(\lambda-1)(\lambda+2)(\lambda-3)
Explanation: For a triangular matrix (either upper or lower), the eigenvalues are its diagonal entries. The diagonal entries of MM are $4, -1,$ and 44. The eigenvalues are λ1=4\lambda_1 = 4, λ2=−1\lambda_2 = -1, and λ3=4\lambda_3 = 4. The characteristic polynomial, p(λ)=det⁡(M−λI)p(\lambda) = \det(M - \lambda I), can be written in factored form using the eigenvalues: p(λ)=(−1)3(λ−λ1)(λ−λ2)(λ−λ3)p(\lambda) = (-1)^3 (\lambda - \lambda_1)(\lambda - \lambda_2)(\lambda - \lambda_3). p(λ)=−(λ−4)(λ−(−1))(λ−4)=−(λ−4)2(λ+1)p(\lambda) = -(\lambda - 4)(\lambda - (-1))(\lambda - 4) = -(\lambda-4)^2(\lambda+1). Distractor A corresponds to the polynomial det⁡(λI−M)\det(\lambda I - M), which is monic but not the standard definition det⁡(M−λI)\det(M - \lambda I). Distractor C has an incorrect sign on the λ\lambda term. Distractor D incorrectly suggests that all non-zero entries of the matrix contribute as roots to the polynomial.

Question 3

A 3×33 \times 3 matrix MM has characteristic polynomial p(λ)=−λ3+6λ2−11λ+6p(\lambda) = -\lambda^3 + 6\lambda^2 - 11\lambda + 6. If one eigenvalue of MM is known to be λ1=2\lambda_1 = 2, and the matrix M+kIM + kI (where kk is a constant) has λ=0\lambda = 0 as an eigenvalue, what are the possible values of kk?

  1. k=−1,−2,−3k = -1, -2, -3 (correct answer)
  2. k=1,2,3k = 1, 2, 3
  3. k=−1,−2k = -1, -2
  4. k=1,3k = 1, 3
Explanation: Since λ1=2\lambda_1 = 2 is an eigenvalue, we can factor the characteristic polynomial: p(λ)=−(λ−2)(λ2−4λ+3)=−(λ−2)(λ−1)(λ−3)p(\lambda) = -(\lambda - 2)(\lambda^2 - 4\lambda + 3) = -(\lambda - 2)(\lambda - 1)(\lambda - 3). So the eigenvalues of MM are 1,2,31, 2, 3. If M+kIM + kI has λ=0\lambda = 0 as an eigenvalue, then the eigenvalues of M+kIM + kI are the eigenvalues of MM shifted by kk: 1+k,2+k,3+k1+k, 2+k, 3+k. For λ=0\lambda = 0 to be an eigenvalue, we need one of 1+k=01+k = 0, 2+k=02+k = 0, or 3+k=03+k = 0. This gives k=−1,−2,k = -1, -2, or k=−3k = -3. Choice A is correct. Choice B gives positive values which would shift eigenvalues away from zero. Choice C is missing k=−3k = -3. Choice D has positive values and is missing k=−2k = -2.

Question 4

Let P(x)=x3−7x2+14x−8P(x) = x^3 - 7x^2 + 14x - 8 be the characteristic polynomial of a 3×33 \times 3 matrix AA. If matrix BB is obtained by applying the same similarity transformation to AA that would make the first column of the resulting matrix equal to (100)\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, which of the following statements about the characteristic polynomial of BB is true?

  1. The characteristic polynomial of BB is (x−1)(x2−6x+8)(x-1)(x^2 - 6x + 8)
  2. The characteristic polynomial of BB is the same as P(x)P(x) (correct answer)
  3. The characteristic polynomial of BB is x3−6x2+11x−6x^3 - 6x^2 + 11x - 6
  4. The characteristic polynomial of BB cannot be determined from the given information
Explanation: Similarity transformations preserve the characteristic polynomial. If B=Q−1AQB = Q^{-1}AQ for some invertible matrix QQ, then det⁡(B−λI)=det⁡(Q−1AQ−λI)=det⁡(Q−1(A−λI)Q)=det⁡(Q−1)det⁡(A−λI)det⁡(Q)=det⁡(A−λI)\det(B - \lambda I) = \det(Q^{-1}AQ - \lambda I) = \det(Q^{-1}(A - \lambda I)Q) = \det(Q^{-1})\det(A - \lambda I)\det(Q) = \det(A - \lambda I) since det⁡(Q−1)det⁡(Q)=1\det(Q^{-1})\det(Q) = 1. Therefore, BB has the same characteristic polynomial as AA, which is P(x)=x3−7x2+14x−8P(x) = x^3 - 7x^2 + 14x - 8. Choice A incorrectly assumes we can factor out (x−1)(x-1) based on the first column information. Choice C gives a different polynomial entirely. Choice D is incorrect because similarity invariance allows us to determine this.

Question 5

A 4×44 \times 4 matrix MM has characteristic polynomial q(λ)=λ4−10λ3+35λ2−50λ+24q(\lambda) = \lambda^4 - 10\lambda^3 + 35\lambda^2 - 50\lambda + 24. After factoring, it's found that q(λ)=(λ−1)(λ−2)(λ−3)(λ−4)q(\lambda) = (\lambda-1)(\lambda-2)(\lambda-3)(\lambda-4). If we construct matrix N=M2−5M+6IN = M^2 - 5M + 6I, and we know that NN is singular (non-invertible), which of the following statements must be true?

  1. MM must have eigenvalue 22 or eigenvalue 33 (correct answer)
  2. MM must have eigenvalue 11 or eigenvalue 44
  3. MM cannot have the eigenvalues given in the characteristic polynomial
  4. The characteristic polynomial must be incorrect
Explanation: For N=M2−5M+6IN = M^2 - 5M + 6I to be singular, it must have eigenvalue 00. If λ\lambda is an eigenvalue of MM, then the corresponding eigenvalue of NN is λ2−5λ+6\lambda^2 - 5\lambda + 6. For NN to be singular, we need λ2−5λ+6=0\lambda^2 - 5\lambda + 6 = 0 for at least one eigenvalue λ\lambda of MM. Factoring: λ2−5λ+6=(λ−2)(λ−3)=0\lambda^2 - 5\lambda + 6 = (\lambda - 2)(\lambda - 3) = 0. So λ=2\lambda = 2 or λ=3\lambda = 3. Since the characteristic polynomial tells us that MM has eigenvalues 1,2,3,41, 2, 3, 4, and we need MM to have eigenvalue 22 or 33 for NN to be singular, this condition is satisfied. Both eigenvalues 22 and 33 are indeed eigenvalues of MM. Therefore, NN will have at least one zero eigenvalue (in fact, it will have two zero eigenvalues corresponding to the eigenvalues 22 and 33 of MM), making NN singular. Choice A correctly identifies this requirement. Choice B gives the wrong eigenvalues. Choice C contradicts the given information. Choice D is unnecessary since the given eigenvalues are consistent with NN being singular.

Question 6

The characteristic polynomial of a 3×33 \times 3 matrix AA is given by p(λ)=−λ3−2λ2+5λ+6p(\lambda) = -\lambda^3 - 2\lambda^2 + 5\lambda + 6. What is the determinant of matrix AA?

  1. −2-2
  2. −1-1
  3. 66 (correct answer)
  4. 88
Explanation: The characteristic polynomial is defined as p(λ)=det⁡(A−λI)p(\lambda) = \det(A - \lambda I). To find the determinant of AA, we can evaluate the characteristic polynomial at λ=0\lambda = 0. p(0)=det⁡(A−0⋅I)=det⁡(A)p(0) = \det(A - 0 \cdot I) = \det(A). Substituting λ=0\lambda = 0 into the given polynomial: p(0)=−(0)3−2(0)2+5(0)+6=6p(0) = -(0)^3 - 2(0)^2 + 5(0) + 6 = 6. Therefore, det⁡(A)=6\det(A) = 6. Distractor A is the coefficient of λ2\lambda^2, which relates to the trace of the matrix, not the determinant. Distractor B is the leading coefficient. Distractor D is the sum of the coefficients, which is p(1)p(1), not p(0)p(0).

Question 7

A 3×33 \times 3 matrix BB has eigenvalues of $2, 2,$ and −3-3. What is the trace of matrix BB?

  1. −12-12
  2. 11 (correct answer)
  3. 22
  4. −1-1
Explanation: The trace of a matrix is the sum of its diagonal entries. A fundamental property of eigenvalues is that the sum of the eigenvalues of a matrix is equal to its trace. Given the eigenvalues λ1=2\lambda_1 = 2, λ2=2\lambda_2 = 2, and λ3=−3\lambda_3 = -3, the trace of BB is: tr(B)=λ1+λ2+λ3=2+2+(−3)=1\mathrm{tr}(B) = \lambda_1 + \lambda_2 + \lambda_3 = 2 + 2 + (-3) = 1. Distractor A is the product of the eigenvalues, which equals the determinant of the matrix. Distractor C is the repeated eigenvalue. Distractor D is the sum of the distinct eigenvalues, incorrectly ignoring the multiplicity of the eigenvalue 2.

Question 8

Let AA be a 2×22 \times 2 matrix with tr(A)=7\mathrm{tr}(A) = 7 and det⁡(A)=10\det(A) = 10. What is the characteristic polynomial of AA?

  1. λ2+7λ+10\lambda^2 + 7\lambda + 10
  2. λ2−10λ+7\lambda^2 - 10\lambda + 7
  3. λ2−7λ+10\lambda^2 - 7\lambda + 10 (correct answer)
  4. λ2+10λ−7\lambda^2 + 10\lambda - 7
Explanation: For any 2×22 \times 2 matrix MM, the characteristic polynomial is given by the formula p(λ)=λ2−tr(M)λ+det⁡(M)p(\lambda) = \lambda^2 - \mathrm{tr}(M)\lambda + \det(M). Given tr(A)=7\mathrm{tr}(A) = 7 and det⁡(A)=10\det(A) = 10, we can substitute these values directly into the formula: p(λ)=λ2−(7)λ+(10)=λ2−7λ+10p(\lambda) = \lambda^2 - (7)\lambda + (10) = \lambda^2 - 7\lambda + 10. Distractor A has an incorrect sign for the trace term. Distractor B incorrectly swaps the roles of the trace and the determinant. Distractor D swaps the roles and also has an incorrect sign for the determinant term.

Question 9

For what value of kk does the matrix A=(37k5)A = \begin{pmatrix} 3 & 7 \\ k & 5 \end{pmatrix} have an eigenvalue of λ=10\lambda = 10?

  1. k=−5k = -5
  2. k=5/7k = 5/7
  3. k=5k = 5 (correct answer)
  4. k=7k = 7
Explanation: If λ=10\lambda = 10 is an eigenvalue of AA, then the matrix A−λIA - \lambda I must be singular, meaning its determinant is zero. Here, A−10IA - 10I. Now, we set the determinant of this matrix to zero: det⁡(A−10I)=(−7)(−5)−(7)(k)=0\det(A - 10I) = (-7)(-5) - (7)(k) = 0 35−7k=035 - 7k = 0 35=7k35 = 7k k=5k = 5 Distractor A results from a sign error in the calculation (−35−7k=0 -35 - 7k = 0). Distractor B occurs if one incorrectly sets the determinant of the original matrix AA equal to the eigenvalue (15−7k=1015 - 7k = 10). Distractor D results from a simple calculation error.

Question 10

A square matrix AA is singular. Which statement must be true about its characteristic polynomial p(λ)=det⁡(A−λI)p(\lambda) = \det(A - \lambda I)?

  1. The constant term of p(λ)p(\lambda) is zero. (correct answer)
  2. The leading coefficient of p(λ)p(\lambda) is zero.
  3. The sum of the coefficients of p(λ)p(\lambda) is zero.
  4. p(λ)p(\lambda) has no real roots.
Explanation: A matrix AA is singular if and only if its determinant is zero, i.e., det⁡(A)=0\det(A) = 0. The characteristic polynomial is p(λ)=det⁡(A−λI)p(\lambda) = \det(A - \lambda I). The constant term of this polynomial is found by setting λ=0\lambda = 0: p(0)=det⁡(A−0⋅I)=det⁡(A)p(0) = \det(A - 0 \cdot I) = \det(A). Since AA is singular, det⁡(A)=0\det(A) = 0. Therefore, the constant term of its characteristic polynomial, p(0)p(0), must be zero. This is equivalent to saying λ=0\lambda=0 is an eigenvalue. Distractor B is impossible; the leading coefficient of the characteristic polynomial of an n×nn \times n matrix is (−1)n(-1)^n. Distractor C implies p(1)=0p(1)=0, which means λ=1\lambda=1 is an eigenvalue, which is not necessarily true for all singular matrices. Distractor D is false; λ=0\lambda=0 is a real root.

Question 11

Given the matrix A=(4123)A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}, what is the characteristic polynomial of the matrix A2A^2?

  1. λ2−7λ+10\lambda^2 - 7\lambda + 10
  2. λ2−25λ+140\lambda^2 - 25\lambda + 140
  3. λ2−49λ+100\lambda^2 - 49\lambda + 100
  4. λ2−29λ+100\lambda^2 - 29\lambda + 100 (correct answer)
Explanation: First, find the eigenvalues of AA. The characteristic polynomial of AA is pA(λ)=λ2−tr(A)λ+det⁡(A)p_A(\lambda) = \lambda^2 - \mathrm{tr}(A)\lambda + \det(A). tr(A)=4+3=7\mathrm{tr}(A) = 4+3=7. det⁡(A)=(4)(3)−(1)(2)=10\det(A) = (4)(3)-(1)(2)=10. So, pA(λ)=λ2−7λ+10=(λ−2)(λ−5)=0p_A(\lambda) = \lambda^2 - 7\lambda + 10 = (\lambda-2)(\lambda-5) = 0. The eigenvalues of AA are λ1=2\lambda_1=2 and λ2=5\lambda_2=5. The eigenvalues of A2A^2 are the squares of the eigenvalues of AA, which are μ1=22=4\mu_1 = 2^2=4 and μ2=52=25\mu_2 = 5^2=25. The characteristic polynomial of A2A^2 is pA2(λ)=λ2−tr(A2)λ+det⁡(A2)p_{A^2}(\lambda) = \lambda^2 - \mathrm{tr}(A^2)\lambda + \det(A^2). The new trace is the sum of the new eigenvalues: tr(A2)=4+25=29\mathrm{tr}(A^2) = 4+25=29. The new determinant is the product of the new eigenvalues: det⁡(A2)=4×25=100\det(A^2) = 4 \times 25 = 100. Thus, the characteristic polynomial of A2A^2 is λ2−29λ+100\lambda^2 - 29\lambda + 100. Distractor A is the characteristic polynomial of AA, not A2A^2. Distractor B is derived from incorrectly squaring each element of AA to get A2A^2. Distractor C is derived from squaring the trace and determinant of AA to find the new coefficients.

Question 12

Consider the matrix A=(010001k−12)A = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ k & -1 & 2 \end{pmatrix}. For what value of kk is λ=1\lambda=1 an eigenvalue of AA?

  1. k=−2k = -2
  2. k=−1k = -1
  3. k=0k = 0 (correct answer)
  4. k=1k = 1
Explanation: If λ=1\lambda=1 is an eigenvalue, then by definition det⁡(A−λI)=0\det(A - \lambda I) = 0. In this case, we need to solve det⁡(A−1⋅I)=0\det(A - 1 \cdot I) = 0. Now, we compute the determinant: =−1((−1)(1)−(1)(−1))−1((0)(1)−(1)(k))+0= -1((-1)(1) - (1)(-1)) - 1((0)(1) - (1)(k)) + 0 =−1(−1+1)−1(−k)= -1(-1 + 1) - 1(-k) =−1(0)+k=k= -1(0) + k = k For λ=1\lambda=1 to be an eigenvalue, we must have det⁡(A−I)=0\det(A-I)=0, which means k=0k=0. Distractor A, B, and D result from various calculation errors or misconceptions, such as incorrectly calculating the determinant or confusing the eigenvalue with the determinant.

Question 13

A 3×33 \times 3 matrix BB has the characteristic polynomial p(λ)=(λ−3)2(λ+1)p(\lambda) = (\lambda-3)^2(\lambda+1). Which of the following statements about BB must be true?

  1. The determinant of BB is 9.
  2. The trace of BB is 5. (correct answer)
  3. BB is not invertible.
  4. BB is guaranteed to have three linearly independent eigenvectors.
Explanation: The roots of the characteristic polynomial are the eigenvalues of the matrix. The roots of p(λ)=(λ−3)2(λ+1)=0p(\lambda) = (\lambda-3)^2(\lambda+1)=0 are λ=3\lambda=3 (with algebraic multiplicity 2) and λ=−1\lambda=-1. The eigenvalues are {3,3,−1}\{3, 3, -1\}. The trace of a matrix is the sum of its eigenvalues: tr(B)=3+3+(−1)=5\mathrm{tr}(B) = 3 + 3 + (-1) = 5. This statement is true. Distractor A is false; the determinant is the product of the eigenvalues: det⁡(B)=3×3×(−1)=−9\det(B) = 3 \times 3 \times (-1) = -9. Distractor C is false; a matrix is not invertible (singular) if and only if 0 is an eigenvalue. Since 0 is not an eigenvalue, BB is invertible. Distractor D is not guaranteed to be true; the eigenvalue 3 has algebraic multiplicity 2, but its geometric multiplicity (the number of linearly independent eigenvectors) could be 1 or 2. If it is 1, the matrix is not diagonalizable and only has two linearly independent eigenvectors.

Question 14

What is the characteristic polynomial of the matrix A=(201131102)A = \begin{pmatrix} 2 & 0 & 1 \\ 1 & 3 & 1 \\ 1 & 0 & 2 \end{pmatrix}?

  1. p(λ)=−λ3+7λ2−15λ+9p(\lambda) = -\lambda^3 + 7\lambda^2 - 15\lambda + 9 (correct answer)
  2. p(λ)=λ3−7λ2+15λ−9p(\lambda) = \lambda^3 - 7\lambda^2 + 15\lambda - 9
  3. p(λ)=−λ3+7λ2−11λ+3p(\lambda) = -\lambda^3 + 7\lambda^2 - 11\lambda + 3
  4. p(λ)=−λ3+5λ2−6λ+9p(\lambda) = -\lambda^3 + 5\lambda^2 - 6\lambda + 9
Explanation: The characteristic polynomial is given by p(λ)=det⁡(A−λI)p(\lambda) = \det(A - \lambda I). We compute the determinant by cofactor expansion along the second column: p(λ)=(3−λ)[(2−λ)2−1]p(\lambda) = (3-\lambda) [ (2-\lambda)^2 - 1 ] p(λ)=(3−λ)[(4−4λ+λ2)−1]p(\lambda) = (3-\lambda) [ (4 - 4\lambda + \lambda^2) - 1 ] p(λ)=(3−λ)(λ2−4λ+3)p(\lambda) = (3-\lambda) (\lambda^2 - 4\lambda + 3) p(λ)=3λ2−12λ+9−λ3+4λ2−3λp(\lambda) = 3\lambda^2 - 12\lambda + 9 - \lambda^3 + 4\lambda^2 - 3\lambda p(λ)=−λ3+7λ2−15λ+9p(\lambda) = -\lambda^3 + 7\lambda^2 - 15\lambda + 9 Distractor B represents the polynomial det⁡(λI−A)\det(\lambda I - A), which is a common alternative definition but results in opposite signs. Distractor C results from an error in calculating the determinant, specifically in the expansion of (2−λ)2−1(2-\lambda)^2-1. Distractor D incorrectly uses the diagonal entries as coefficients.

Question 15

The characteristic polynomial of a matrix AA is p(λ)=−λ3+5λ2−8λ+4p(\lambda) = -\lambda^3 + 5\lambda^2 - 8\lambda + 4. What are the eigenvalues of the matrix A+2IA+2I?

  1. {1,2,2}\{1, 2, 2\}
  2. {3,4,4}\{3, 4, 4\} (correct answer)
  3. {−1,0,0}\{-1, 0, 0\}
  4. {2,4,4}\{2, 4, 4\}
Explanation: First, we find the eigenvalues of AA by finding the roots of its characteristic polynomial p(λ)=−λ3+5λ2−8λ+4=0p(\lambda) = -\lambda^3 + 5\lambda^2 - 8\lambda + 4 = 0. By the Rational Root Theorem, we test integer factors of 4. We find that p(1)=−1+5−8+4=0p(1) = -1+5-8+4=0 and p(2)=−8+20−16+4=0p(2) = -8+20-16+4=0. Since the sum of the roots must be −5/(−1)=5-5/(-1)=5, and we have found roots 1 and 2, the third root λ3\lambda_3 must satisfy 1+2+λ3=51+2+\lambda_3=5, so λ3=2\lambda_3=2. The eigenvalues of AA are {1,2,2}\{1, 2, 2\}. If λ\lambda is an eigenvalue of AA, then λ+k\lambda+k is an eigenvalue of A+kIA+kI. For this problem, k=2k=2. So, the eigenvalues of A+2IA+2I are obtained by adding 2 to each eigenvalue of AA: {1+2,2+2,2+2}={3,4,4}\{1+2, 2+2, 2+2\} = \{3, 4, 4\}. Distractor A lists the original eigenvalues of AA, forgetting to perform the shift. Distractor C represents the eigenvalues of A−2IA-2I, a shift in the wrong direction. Distractor D represents the eigenvalues of 2A2A, a scaling instead of a shift.