Verifying Solutions & Identities - Linear Algebra

1. A student claims that matrices $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 4 & 3 \\ 2 & 1 \end{pmatrix}$ are similar because they have the same determinant and trace. To definitively verify or refute this claim, what is the most reliable approach?
   Answer: Find an invertible matrix $P$ such that $B = P^{-1}AP$ Two matrices are similar if and only if there exists an invertible matrix $P$ such that $B = P^{-1}AP$. 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.
2. To verify the identity $(AB)^T = B^T A^T$ for matrices $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 1 \\ -1 & 2 \end{pmatrix}$, a student computes both sides and gets different results. Which error is most likely?
   Answer: Computing $B^T A^T$ as $A^T B^T$ due to commutativity assumption The most common conceptual error when verifying $(AB)^T = B^T A^T$ is computing $A^T B^T$ instead of $B^T A^T$, incorrectly assuming matrix multiplication is commutative. The order reversal in the transpose of a product is often forgotten. Choice A describes computing $(A^T)(B^T)$ which is wrong but less common. Choices C and D are computational errors rather than conceptual misunderstandings of the identity.
3. A student attempts to verify the distributive property $A(B + C) = AB + AC$ using matrices $A = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}$, $B = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}$, and $C = \begin{pmatrix} 0 & 2 \\ 1 & 0 \end{pmatrix}$. They compute $A(B + C)$ and get $\begin{pmatrix} 3 & 3 \\ 9 & 3 \end{pmatrix}$. What should they compute next to complete the verification?
   Answer: $AB + AC$ and verify it equals $\begin{pmatrix} 3 & 3 \\ 9 & 3 \end{pmatrix}$ To complete the verification of $A(B + C) = AB + AC$, compute $AB + AC$ and confirm it equals the already computed value of $A(B + C) = \begin{pmatrix} 3 & 3 \\ 9 & 3 \end{pmatrix}$. 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 $(B + C)A = BA + CA$, which is different from what was started.
4. A student verifies that matrix $P = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix}$ is orthogonal by showing $P^T P = I$. However, they claim this is sufficient to prove $P$ is orthogonal. What additional verification might be needed?
   Answer: Verify that $PP^T = I$ to ensure $P$ is square and invertible For a matrix to be orthogonal, both $P^T P = I$ and $PP^T = I$ must hold. While $P^T P = I$ is necessary, it's not sufficient for non-square matrices or in cases where careful verification is needed. Checking $PP^T = I$ confirms the matrix is both square and orthogonal. Choice B is a consequence, not a verification method. Choice C is redundant since $P^T P = I$ already confirms this. Choice D restates the definition rather than providing additional verification.
5. Consider the vector equation $c_1 \mathbf{v_1} + c_2 \mathbf{v_2} + c_3 \mathbf{v_3} = \mathbf{b}$ where $\mathbf{v_1} = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$, $\mathbf{v_2} = \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}$, $\mathbf{v_3} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix}$, and $\mathbf{b} = \begin{pmatrix} 4 \\ 7 \\ 1 \end{pmatrix}$. A proposed solution is $c_1 = 2$, $c_2 = 1$, $c_3 = 1$. Which verification step would immediately reveal if this solution is incorrect?
   Answer: Verify that $2\mathbf{v_1} + 1\mathbf{v_2} + 1\mathbf{v_3} = \mathbf{b}$ component-wise 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 $2\mathbf{v_1} + 1\mathbf{v_2} + 1\mathbf{v_3} = \mathbf{b}$ 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.
6. Consider the matrix equation $A^2 - 3A + 2I = 0$ where $A = \begin{pmatrix} 2 & 1 \\ 0 & 1 \end{pmatrix}$. To verify this identity, which of the following computations is both necessary and sufficient?
   Answer: Compute $A^2$ and verify that $A^2 - 3A = -2I$ To verify the matrix identity $A^2 - 3A + 2I = 0$, we need to compute $A^2 - 3A + 2I$ and show it equals the zero matrix. This is equivalent to showing $A^2 - 3A = -2I$. 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.
7. To verify that $\lambda = 3$ is an eigenvalue of matrix $A = \begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 2 \end{pmatrix}$ with eigenvector $\mathbf{v} = \begin{pmatrix} 1 \\ \sqrt{2} \\ 1 \end{pmatrix}$, which computation provides the most complete verification?
   Answer: Show that $(A - 3I)\mathbf{v} = \mathbf{0}$ and $\mathbf{v} \neq \mathbf{0}$ To verify an eigenvalue-eigenvector pair, we need to show $(A - \lambda I)\mathbf{v} = \mathbf{0}$ and $\mathbf{v} \neq \mathbf{0}$. This is equivalent to $A\mathbf{v} = \lambda\mathbf{v}$ but emphasizes the key requirement that $\mathbf{v}$ be non-zero. Choice A is redundant since $A\mathbf{v} = 3\mathbf{v}$ implies $\det(A - 3I) = 0$. Choice B misses the non-zero requirement. Choice C only verifies the eigenvalue exists, not that $\mathbf{v}$ is a corresponding eigenvector.
8. A student claims that $x = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$ is a solution to the system $Ax = b$ where $A = \begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & -1 \\ 2 & 1 & 3 \end{pmatrix}$ and $b = \begin{pmatrix} 3 \\ -4 \\ 12 \end{pmatrix}$. What is the most efficient way to verify this claim?
   Answer: Check if $Ax = b$ by computing the matrix-vector product $Ax$ To verify a proposed solution to $Ax = b$, the most direct and efficient method is to substitute and check: compute $Ax$ and verify it equals $b$. Choice A requires solving the entire system, which is unnecessary. Choice C misses the point since we need $b$ in the column space, not $x$. Choice D works but requires computing an inverse, which is more computationally expensive than a simple matrix-vector multiplication.
9. To verify that the set $\left\{\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\right\}$ forms a basis for $\mathbb{R}^3$, a student checks that the vectors are linearly independent. What additional verification is needed?
   Answer: No additional verification needed since three linearly independent vectors in $\mathbb{R}^3$ automatically span $\mathbb{R}^3$ When you're asked to verify that a set of vectors forms a basis for $\mathbb{R}^3$, 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 $\mathbb{R}^3$ 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 $\mathbb{R}^3$. While spanning is indeed required for a basis, this additional check is unnecessary given that we already have 3 linearly independent vectors in $\mathbb{R}^3$. 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.
10. Consider the identity $\text{rank}(AB) \leq \min(\text{rank}(A), \text{rank}(B))$ for matrices $A$ and $B$ where the product $AB$ is defined. A student wants to verify this with $A = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 1 \\ 1 & 1 \\ 0 & 0 \end{pmatrix}$. What would constitute a complete verification?
    Answer: Compute $\text{rank}(A)$, $\text{rank}(B)$, and $\text{rank}(AB)$, then check the inequality To verify the rank inequality for specific matrices, compute all three ranks and check that $\text{rank}(AB) \leq \min(\text{rank}(A), \text{rank}(B))$. 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.

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