Linear Algebra Quiz: Verifying Solutions And Identities
10 questions · exam conditions
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Verifying Solutions And IdentitiesQuestion 1 of 10

A student claims that x=(213)x = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} is a solution to the system Ax=bAx = b where A=(121011213)A = \begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & -1 \\ 2 & 1 & 3 \end{pmatrix} and $$b = \begin{pmatrix} 3 \ -4 \ 12 \end{pmatrix}

Solve the system using Gaussian elimination and compare the result
Check if Ax=bAx = b by computing the matrix-vector product AxAx
Verify that xx is in the column space of AA
Check if A1b=xA^{-1}b = x by computing the inverse of AA
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Linear Algebra Quiz

Linear Algebra Quiz: Verifying Solutions And Identities

Practice Verifying Solutions And Identities 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 Verifying Solutions And Identities, 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

A student claims that x=(213)x = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} is a solution to the system Ax=bAx = b where A=(121011213)A = \begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & -1 \\ 2 & 1 & 3 \end{pmatrix} and $$b = \begin{pmatrix} 3 \ -4 \ 12 \end{pmatrix}

  1. Solve the system using Gaussian elimination and compare the result
  2. Check if Ax=bAx = b by computing the matrix-vector product AxAx (correct answer)
  3. Verify that xx is in the column space of AA
  4. Check if A1b=xA^{-1}b = x by computing the inverse of AA
Explanation: To verify a proposed solution to Ax=bAx = b, the most direct and efficient method is to substitute and check: compute AxAx and verify it equals bb. Choice A requires solving the entire system, which is unnecessary. Choice C misses the point since we need bb in the column space, not xx. Choice D works but requires computing an inverse, which is more computationally expensive than a simple matrix-vector multiplication.

Question 2

To verify the identity (AB)T=BTAT(AB)^T = B^T A^T for matrices A=(1234)A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} and $$B = \begin{pmatrix} 0 & 1 \ -1 & 2 \end{pmatrix}

  1. Computing (AB)T(AB)^T by transposing AA and BB first, then multiplying
  2. Computing BTATB^T A^T as ATBTA^T B^T due to commutativity assumption (correct answer)
  3. Making arithmetic errors in the matrix multiplication ABAB
  4. Incorrectly transposing one of the matrices AA or BB
Explanation: The most common conceptual error when verifying (AB)T=BTAT(AB)^T = B^T A^T is computing ATBTA^T B^T instead of BTATB^T A^T, incorrectly assuming matrix multiplication is commutative. The order reversal in the transpose of a product is often forgotten. Choice A describes computing (AT)(BT)(A^T)(B^T) which is wrong but less common. Choices C and D are computational errors rather than conceptual misunderstandings of the identity.

Question 3

A student claims that matrices A=(1234)A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} and $$B = \begin{pmatrix} 4 & 3 \ 2 & 1 \end{pmatrix}

  1. Check if AA and BB have the same characteristic polynomial
  2. Verify that AA and BB have the same eigenvalues with same multiplicities
  3. Find an invertible matrix PP such that B=P1APB = P^{-1}AP (correct answer)
  4. Compare all invariants: determinant, trace, and eigenvalues of AA and BB
Explanation: Two matrices are similar if and only if there exists an invertible matrix PP such that B=P1APB = P^{-1}AP. While similar matrices have the same eigenvalues (choices A and B), having the same eigenvalues doesn't guarantee similarity - they must also have the same geometric multiplicities. Choice C provides the definitive test. Choice D lists necessary conditions but they're not sufficient for similarity.

Question 4

Consider the matrix equation A23A+2I=0A^2 - 3A + 2I = 0 where $$A = \begin{pmatrix} 2 & 1 \ 0 & 1 \end{pmatrix}

  1. Compute A2A^2 and verify that A2=3A2IA^2 = 3A - 2I
  2. Compute A2A^2 and verify that A23A=2IA^2 - 3A = -2I (correct answer)
  3. Verify that det(A23A+2I)=0\det(A^2 - 3A + 2I) = 0
  4. Check that the eigenvalues of AA satisfy λ23λ+2=0\lambda^2 - 3\lambda + 2 = 0
Explanation: To verify the matrix identity A23A+2I=0A^2 - 3A + 2I = 0, we need to compute A23A+2IA^2 - 3A + 2I and show it equals the zero matrix. This is equivalent to showing A23A=2IA^2 - 3A = -2I. Choice A rearranges correctly but doesn't match the verification process. Choice C is incorrect because a determinant being zero doesn't guarantee the matrix is zero. Choice D checks the characteristic polynomial but doesn't verify the matrix identity directly.

Question 5

Consider the identity rank(AB)min(rank(A),rank(B))\text{rank}(AB) \leq \min(\text{rank}(A), \text{rank}(B)) for matrices AA and BB where the product ABAB is defined. A student wants to verify this with A=(101011112)A = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{pmatrix} and $$B = \begin{pmatrix} 1 & 1 \ 1 & 1 \ 0 & 0 \end{pmatrix}

  1. Compute rank(A)\text{rank}(A), rank(B)\text{rank}(B), and rank(AB)\text{rank}(AB), then check the inequality (correct answer)
  2. Show that col(AB)col(A)\text{col}(AB) \subseteq \text{col}(A) and row(AB)row(B)\text{row}(AB) \subseteq \text{row}(B)
  3. Verify that nullity(AB)max(nullity(A),nullity(B))\text{nullity}(AB) \geq \max(\text{nullity}(A), \text{nullity}(B))
  4. Compute determinants and check if det(AB)min(det(A),det(B))|\det(AB)| \leq \min(|\det(A)|, |\det(B)|)
Explanation: To verify the rank inequality for specific matrices, compute all three ranks and check that rank(AB)min(rank(A),rank(B))\text{rank}(AB) \leq \min(\text{rank}(A), \text{rank}(B)). Choice B explains why the inequality holds theoretically but doesn't verify it numerically. Choice C involves nullity which relates to rank but doesn't directly verify the stated inequality. Choice D is incorrect since determinants don't directly relate to the rank inequality, especially for non-square matrices.

Question 6

To verify that λ=3\lambda = 3 is an eigenvalue of matrix A=(210121012)A = \begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 2 \end{pmatrix} with eigenvector $$\mathbf{v} = \begin{pmatrix} 1 \ \sqrt{2} \ 1 \end{pmatrix}

  1. Show that det(A3I)=0\det(A - 3I) = 0 and Av=3vA\mathbf{v} = 3\mathbf{v}
  2. Verify only that Av=3vA\mathbf{v} = 3\mathbf{v} since this confirms both claims
  3. Compute the characteristic polynomial and check that 33 is a root
  4. Show that (A3I)v=0(A - 3I)\mathbf{v} = \mathbf{0} and v0\mathbf{v} \neq \mathbf{0} (correct answer)
Explanation: To verify an eigenvalue-eigenvector pair, we need to show (AλI)v=0(A - \lambda I)\mathbf{v} = \mathbf{0} and v0\mathbf{v} \neq \mathbf{0}. This is equivalent to Av=λvA\mathbf{v} = \lambda\mathbf{v} but emphasizes the key requirement that v\mathbf{v} be non-zero. Choice A is redundant since Av=3vA\mathbf{v} = 3\mathbf{v} implies det(A3I)=0\det(A - 3I) = 0. Choice B misses the non-zero requirement. Choice C only verifies the eigenvalue exists, not that v\mathbf{v} is a corresponding eigenvector.

Question 7

A student verifies that matrix P=(12121212)P = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix} is orthogonal by showing PTP=IP^T P = I. However, they claim this is sufficient to prove PP is orthogonal. What additional verification might be needed?

  1. Verify that PPT=IPP^T = I to ensure PP is square and invertible (correct answer)
  2. Check that det(P)=±1\det(P) = \pm 1 to confirm the orthogonal property
  3. Verify that all columns of PP have unit length independently
  4. Confirm that P1=PTP^{-1} = P^T by computing the inverse directly
Explanation: For a matrix to be orthogonal, both PTP=IP^T P = I and PPT=IPP^T = I must hold. While PTP=IP^T P = I is necessary, it's not sufficient for non-square matrices or in cases where careful verification is needed. Checking PPT=IPP^T = I confirms the matrix is both square and orthogonal. Choice B is a consequence, not a verification method. Choice C is redundant since PTP=IP^T P = I already confirms this. Choice D restates the definition rather than providing additional verification.

Question 8

A student attempts to verify the distributive property A(B+C)=AB+ACA(B + C) = AB + AC using matrices A=(2103)A = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix} , B=(1021)B = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} , and C=(0210)C = \begin{pmatrix} 0 & 2 \\ 1 & 0 \end{pmatrix} . They compute A(B+C)A(B + C) and get $$ \begin{pmatrix} 3 & 3 \ 9 & 3 \end{pmatrix}

  1. AB+ACAB + AC and verify it equals $$ \begin{pmatrix} 3 & 3 \ 9 & 3 \end{pmatrix} (correct answer)
  2. ABAB and ACAC separately, then check if their sum gives the expected result
  3. (B+C)A(B + C)A to verify that multiplication is commutative in this case
  4. BA+CABA + CA and compare with A(B+C)A(B + C) to check both distributive laws
Explanation: To complete the verification of A(B+C)=AB+ACA(B + C) = AB + AC, compute AB+ACAB + AC and confirm it equals the already computed value of A(B+C)=(3393)A(B + C) = \begin{pmatrix} 3 & 3 \\ 9 & 3 \end{pmatrix} . Choice B describes the same process but less efficiently. Choice C tests commutativity, which is irrelevant to distributivity. Choice D tests the other distributive law (B+C)A=BA+CA(B + C)A = BA + CA, which is different from what was started.

Question 9

To verify that the set {(110),(101),(011)}\left\{\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\right\} forms a basis for R3\mathbb{R}^3, a student checks that the vectors are linearly independent. What additional verification is needed?

  1. Verify that the vectors span R3\mathbb{R}^3 by showing any vector can be written as their linear combination
  2. Check that each vector has unit length to ensure orthogonality
  3. No additional verification needed since three linearly independent vectors in R3\mathbb{R}^3 automatically span R3\mathbb{R}^3 (correct answer)
  4. Verify that the vectors are orthogonal to each other
Explanation: When you're asked to verify that a set of vectors forms a basis for R3\mathbb{R}^3, you need to understand what makes a basis: the vectors must be both linearly independent AND span the entire space. However, there's a powerful dimensional shortcut that applies here. Since R3\mathbb{R}^3 is a 3-dimensional space, any set of exactly 3 linearly independent vectors will automatically span the entire space. This is because the dimension of a vector space equals the number of vectors in any basis for that space. If you have 3 linearly independent vectors in a 3-dimensional space, they must form a basis - no additional verification needed. Let's examine why the other options miss the mark: Choice A suggests verifying that the vectors span R3\mathbb{R}^3. While spanning is indeed required for a basis, this additional check is unnecessary given that we already have 3 linearly independent vectors in R3\mathbb{R}^3. Choice B incorrectly focuses on unit length and orthogonality. Basis vectors don't need to be unit vectors or orthogonal - those properties define an orthonormal basis, which is a special type of basis but not required for the general definition. Choice D also mentions orthogonality, which again is not required for vectors to form a basis. Linear independence is sufficient when combined with the correct count. Study tip: Remember the dimension rule: in an n-dimensional space, exactly n linearly independent vectors automatically form a basis. This saves you from doing extra work to verify spanning.

Question 10

Consider the vector equation c1v1+c2v2+c3v3=bc_1 \mathbf{v_1} + c_2 \mathbf{v_2} + c_3 \mathbf{v_3} = \mathbf{b} where v1=(121)\mathbf{v_1} = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} , v2=(011)\mathbf{v_2} = \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix} , v3=(231)\mathbf{v_3} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} , and b=(471)\mathbf{b} = \begin{pmatrix} 4 \\ 7 \\ 1 \end{pmatrix} . A proposed solution is c1=2c_1 = 2, c2=1c_2 = 1, c3=1c_3 = 1. Which verification step would immediately reveal if this solution is incorrect?

  1. Check if the coefficient matrix has full rank
  2. Check if {v1,v2,v3}\{\mathbf{v_1}, \mathbf{v_2}, \mathbf{v_3}\} are linearly independent
  3. Determine if b\mathbf{b} is in the span of {v1,v2,v3}\{\mathbf{v_1}, \mathbf{v_2}, \mathbf{v_3}\}
  4. Verify that 2v1+1v2+1v3=b2\mathbf{v_1} + 1\mathbf{v_2} + 1\mathbf{v_3} = \mathbf{b} component-wise (correct answer)
Explanation: When you encounter a proposed solution to a vector equation, the most direct approach is to test whether that specific solution actually works by substituting the values and checking if both sides are equal. The correct approach is D: directly verify that 2v1+1v2+1v3=b2\mathbf{v_1} + 1\mathbf{v_2} + 1\mathbf{v_3} = \mathbf{b} component-wise. Let's compute: $$2\begin{pmatrix} 1 \ 2 \ 1 \end{pmatrix} + 1\begin{pmatrix} 0 \ 1 \ -1 \end{pmatrix} + 1\begin{pmatrix} 2 \ 3 \ 1 \end{pmatrix} = \begin{pmatrix} 2 \ 4 \ 2 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \ -1 \end{pmatrix} + \begin{pmatrix} 2 \ 3 \ 1 \end{pmatrix} = \begin{pmatrix} 4 \ 8 \ 2 \end{pmatrix} Since this equals $$ \begin{pmatrix} 4 \\ 8 \\ 2 \end{pmatrix} \neq \begin{pmatrix} 4 \\ 7 \\ 1 \end{pmatrix} = \mathbf{b}$$, the proposed solution is incorrect. The other options address important theoretical questions but don't directly test the given solution. Option A (checking if the coefficient matrix has full rank) tells you about the existence and uniqueness of solutions generally, but not whether this specific solution works. Option B (linear independence of the vectors) is essentially the same as A—it's about the system's properties, not this particular answer. Option C (whether $$\mathbf{b}$$ is in the span) tells you if any solution exists, but again doesn't verify this specific one. Remember: when given a proposed solution to any equation, always test it directly first. This immediate verification can save time and provides definitive answers about that specific solution's correctness.