Linear Algebra Quiz: Matrix Representation
15 questions · exam conditions
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Matrix RepresentationQuestion 1 of 15

Consider the linear transformation T:R2→R3T: \mathbb{R}^2 \to \mathbb{R}^3 defined by T(x,y)=(2x−y,x+3y,−x+2y)T(x,y) = (2x-y, x+3y, -x+2y). If B1={(1,1),(1,−1)}B_1 = \{(1,1), (1,-1)\} is a basis for R2\mathbb{R}^2 and B2={(1,0,0),(0,1,0),(0,0,1)}B_2 = \{(1,0,0), (0,1,0), (0,0,1)\} is the standard basis for R3\mathbb{R}^3, what is the matrix representation [T]B1B2[T]_{B_1}^{B_2}?

(134−21−3)\begin{pmatrix} 1 & 3 \\ 4 & -2 \\ 1 & -3 \end{pmatrix}
(2−113−12)\begin{pmatrix} 2 & -1 \\ 1 & 3 \\ -1 & 2 \end{pmatrix}
(134−213)\begin{pmatrix} 1 & 3 \\ 4 & -2 \\ 1 & 3 \end{pmatrix}
(21−131−2)\begin{pmatrix} 2 & 1 \\ -1 & 3 \\ 1 & -2 \end{pmatrix}
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Linear Algebra Quiz

Linear Algebra Quiz: Matrix Representation

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

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

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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.

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Question 1

Consider the linear transformation T:R2→R3T: \mathbb{R}^2 \to \mathbb{R}^3 defined by T(x,y)=(2x−y,x+3y,−x+2y)T(x,y) = (2x-y, x+3y, -x+2y). If B1={(1,1),(1,−1)}B_1 = \{(1,1), (1,-1)\} is a basis for R2\mathbb{R}^2 and B2={(1,0,0),(0,1,0),(0,0,1)}B_2 = \{(1,0,0), (0,1,0), (0,0,1)\} is the standard basis for R3\mathbb{R}^3, what is the matrix representation [T]B1B2[T]_{B_1}^{B_2}?

  1. (134−21−3)\begin{pmatrix} 1 & 3 \\ 4 & -2 \\ 1 & -3 \end{pmatrix} (correct answer)
  2. (2−113−12)\begin{pmatrix} 2 & -1 \\ 1 & 3 \\ -1 & 2 \end{pmatrix}
  3. (134−213)\begin{pmatrix} 1 & 3 \\ 4 & -2 \\ 1 & 3 \end{pmatrix}
  4. (21−131−2)\begin{pmatrix} 2 & 1 \\ -1 & 3 \\ 1 & -2 \end{pmatrix}
Explanation: To find [T]B1B2[T]_{B_1}^{B_2}, we need to compute TT applied to each vector in B1B_1 and express the results in coordinates relative to B2B_2. First, T(1,1)=(2(1)−1,1+3(1),−1+2(1))=(1,4,1)T(1,1) = (2(1)-1, 1+3(1), -1+2(1)) = (1,4,1). Since B2B_2 is the standard basis, this is already in B2B_2-coordinates: (1,4,1)T(1,4,1)^T. Second, T(1,−1)=(2(1)−(−1),1+3(−1),−1+2(−1))=(3,−2,−3)T(1,-1) = (2(1)-(-1), 1+3(-1), -1+2(-1)) = (3,-2,-3). In B2B_2-coordinates: (3,−2,−3)T(3,-2,-3)^T. Therefore, [T]B1B2=(134−21−3)[T]_{B_1}^{B_2} = \begin{pmatrix} 1 & 3 \\ 4 & -2 \\ 1 & -3 \end{pmatrix} . Choice B gives the standard matrix representation [T]E2E3[T]_{E_2}^{E_3} where E2,E3E_2, E_3 are standard bases. Choice C has the wrong sign in the bottom-right entry. Choice D transposes some entries incorrectly.

Question 2

Let P2P_2 denote the vector space of polynomials of degree at most 22. Define the linear transformation T:P2→R3T: P_2 \to \mathbb{R}^3 by T(p(x))=(p(0),p(1),p(−1))T(p(x)) = (p(0), p(1), p(-1)). Using the basis {1,x,x2}\{1, x, x^2\} for P2P_2 and the standard basis for R3\mathbb{R}^3, what is the rank of the matrix representation of TT?

  1. 11 because TT only evaluates polynomials at specific points
  2. 22 because the evaluation at three points creates dependencies
  3. 33 because TT is injective on the finite-dimensional space P2P_2 (correct answer)
  4. 22 because dim⁡(P2)=3\dim(P_2) = 3 but the codomain constraint reduces rank
Explanation: To find the rank, we construct the matrix representation. For the basis polynomial 11: T(1)=(1,1,1)T(1) = (1, 1, 1). For xx: T(x)=(0,1,−1)T(x) = (0, 1, -1). For x2x^2: T(x2)=(0,1,1)T(x^2) = (0, 1, 1). So the matrix is A=(1001111−11)A = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{pmatrix} . To find the rank, we row reduce: (1001111−11)→(1000110−11)→(100011002)\begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -1 & 1 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \end{pmatrix}. This has 33 pivots, so rank is 33. Alternatively, TT is injective because if T(p)=0T(p) = 0, then p(0)=p(1)=p(−1)=0p(0) = p(1) = p(-1) = 0, which means the polynomial pp of degree at most 22 has three distinct roots, so p≡0p \equiv 0. Since TT is injective and dim⁡(P2)=3\dim(P_2) = 3, we have rank(T)=3\text{rank}(T) = 3. Choice A misunderstands that evaluation can still be injective. Choice B incorrectly assumes dependencies exist. Choice D incorrectly suggests codomain constraints reduce rank when the transformation is actually injective.

Question 3

Let T:R3→R3T: \mathbb{R}^3 \to \mathbb{R}^3 be a linear transformation with matrix representation A=(210021002)A = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{pmatrix} . If B={v1,v2,v3}B = \{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\} is a basis for R3\mathbb{R}^3 such that [T]B[T]_B is diagonal, which condition must the basis vectors satisfy?

  1. Each vi\mathbf{v}_i must be an eigenvector of AA with distinct eigenvalues 2,4,82, 4, 8
  2. The basis vectors must form a Jordan chain since AA has only one eigenvalue λ=2\lambda = 2
  3. No such basis BB exists because AA is not diagonalizable over R\mathbb{R} (correct answer)
  4. The basis vectors must satisfy Avi=2viA\mathbf{v}_i = 2\mathbf{v}_i and be orthogonal to each other
Explanation: The matrix AA has characteristic polynomial det⁡(A−λI)=(2−λ)3\det(A - \lambda I) = (2-\lambda)^3, so the only eigenvalue is λ=2\lambda = 2 with algebraic multiplicity 33. To determine if AA is diagonalizable, we check the geometric multiplicity by finding dim⁡(ker⁡(A−2I))\dim(\ker(A - 2I)). We have A−2I=(010001000)A - 2I = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} . The nullspace is spanned by (100)\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, so dim⁡(ker⁡(A−2I))=1\dim(\ker(A - 2I)) = 1. Since the geometric multiplicity (11) is less than the algebraic multiplicity (33), the matrix AA is not diagonalizable. Therefore, no basis BB exists such that [T]B[T]_B is diagonal. Choice A is wrong because AA doesn't have distinct eigenvalues. Choice B correctly identifies that we need Jordan chains, but the question asks for diagonalization, not Jordan form. Choice D is incorrect because even if we had three orthogonal eigenvectors with eigenvalue 22, we've shown only one linearly independent eigenvector exists.

Question 4

The linear transformation T:R2→R2T: \mathbb{R}^2 \to \mathbb{R}^2 represented by the matrix A=(02−20)A = \begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix} can be described as a composition of which two geometric operations?

  1. A reflection across the y-axis and a uniform scaling by a factor of 2.
  2. A rotation by 90∘90^\circ clockwise and a uniform scaling by a factor of 2. (correct answer)
  3. A rotation by 90∘90^\circ counterclockwise and a non-uniform scaling.
  4. A projection onto the y-axis and a rotation by 180∘180^\circ.
Explanation: The matrix can be factored as A=(2002)(01−10)A = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}. The first matrix represents a uniform scaling by a factor of 2. The second matrix represents a rotation by 270∘270^\circ counterclockwise, which is equivalent to a 90∘90^\circ clockwise rotation. The determinant of A is 4, which rules out reflections (det=-1) and projections (det=0).

Question 5

A linear transformation T:R2→R2T: \mathbb{R}^2 \to \mathbb{R}^2 first reflects a vector across the line y=xy=x and then projects the resulting vector orthogonally onto the x-axis. What is the standard matrix for this transformation TT?

  1. (0100)\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} (correct answer)
  2. (0010)\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}
  3. (1000)\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}
  4. (0110)\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}
Explanation: Let T1T_1 be the reflection across y=xy=x, and T2T_2 be the projection onto the x-axis. The matrix for T1T_1 is A1=(0110)A_1 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}. The matrix for T2T_2 is A2=(1000)A_2 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}. The composite transformation is T=T2∘T1T = T_2 \circ T_1, so its matrix is A=A2A1=(1000)(0110)=(0100)A = A_2 A_1 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}.

Question 6

A linear transformation T:R2→R2T: \mathbb{R}^2 \to \mathbb{R}^2 is represented by the invertible standard matrix A=(3−5−12)A = \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}. Which matrix represents the inverse transformation T−1T^{-1}?

  1. (2513)\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix} (correct answer)
  2. (2−1−53)\begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}
  3. (3512)\begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix}
Explanation: The matrix of the inverse transformation T−1T^{-1} is the inverse of the matrix AA. For a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix}, the inverse is 1ad−bc(d−b−ca)\frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}. The determinant of AA is (3)(2)−(−5)(−1)=6−5=1(3)(2) - (-5)(-1) = 6 - 5 = 1. Therefore, A−1=11(2513)=(2513)A^{-1} = \frac{1}{1}\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix} = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}.

Question 7

Consider the linear transformation T:P2→P2T: P_2 \to P_2 on the space of polynomials of degree at most 2, defined by T(p(x))=xp′(x)−p(x)T(p(x)) = x p'(x) - p(x). What is the matrix representation of TT with respect to the standard basis B={1,x,x2}\mathcal{B} = \{1, x, x^2\}?

  1. (010002000)\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{pmatrix}
  2. (000010002)\begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix}
  3. (−100000001)\begin{pmatrix} -1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} (correct answer)
  4. (−1−1−1010002)\begin{pmatrix} -1 & -1 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix}
Explanation: We apply the transformation TT to each basis vector and write the result in coordinates with respect to the basis B\mathcal{B}. T(1)=x⋅(0)−1=−1=−1⋅1+0⋅x+0⋅x2  ⟹  [−1,0,0]TT(1) = x \cdot (0) - 1 = -1 = -1 \cdot 1 + 0 \cdot x + 0 \cdot x^2 \implies [-1, 0, 0]^T. T(x)=x⋅(1)−x=0=0⋅1+0⋅x+0⋅x2  ⟹  [0,0,0]TT(x) = x \cdot (1) - x = 0 = 0 \cdot 1 + 0 \cdot x + 0 \cdot x^2 \implies [0, 0, 0]^T. T(x2)=x⋅(2x)−x2=2x2−x2=x2=0⋅1+0⋅x+1⋅x2  ⟹  [0,0,1]TT(x^2) = x \cdot (2x) - x^2 = 2x^2 - x^2 = x^2 = 0 \cdot 1 + 0 \cdot x + 1 \cdot x^2 \implies [0, 0, 1]^T. These coordinate vectors form the columns of the matrix, which is (−100000001)\begin{pmatrix} -1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}.

Question 8

Let T:R2→R2T: \mathbb{R}^2 \to \mathbb{R}^2 be a linear transformation defined by T(xy)=(3x−yx+2y)T\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3x - y \\ x + 2y \end{pmatrix}. Let B={(11),(1−1)}\mathcal{B} = \left\{ \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ -1 \end{pmatrix} \right\} be a basis for R2\mathbb{R}^2. Find the matrix representation of TT with respect to the basis B\mathcal{B}, denoted [T]B[T]_{\mathcal{B}}.

  1. (3−112)\begin{pmatrix} 3 & -1 \\ 1 & 2 \end{pmatrix}
  2. (243−1)\begin{pmatrix} 2 & 4 \\ 3 & -1 \end{pmatrix}
  3. (5/23/2−1/25/2)\begin{pmatrix} 5/2 & 3/2 \\ -1/2 & 5/2 \end{pmatrix} (correct answer)
  4. (5/2−1/23/25/2)\begin{pmatrix} 5/2 & -1/2 \\ 3/2 & 5/2 \end{pmatrix}
Explanation: Let b1=(11)\mathbf{b}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix} and b2=(1−1)\mathbf{b}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}. First, find the images of the basis vectors: T(b1)=(23)T(\mathbf{b}_1) = \begin{pmatrix} 2 \\ 3 \end{pmatrix} and T(b2)=(4−1)T(\mathbf{b}_2) = \begin{pmatrix} 4 \\ -1 \end{pmatrix}. Next, express these images in B\mathcal{B}-coordinates. For T(b1)T(\mathbf{b}_1), solve c1b1+c2b2=(23)c_1\mathbf{b}_1 + c_2\mathbf{b}_2 = \begin{pmatrix} 2 \\ 3 \end{pmatrix}, which yields c1=5/2,c2=−1/2c_1=5/2, c_2=-1/2. This is the first column. For T(b2)T(\mathbf{b}_2), solve d1b1+d2b2=(4−1)d_1\mathbf{b}_1 + d_2\mathbf{b}_2 = \begin{pmatrix} 4 \\ -1 \end{pmatrix}, which yields d1=3/2,d2=5/2d_1=3/2, d_2=5/2. This is the second column. So, [T]B=(5/23/2−1/25/2)[T]_{\mathcal{B}} = \begin{pmatrix} 5/2 & 3/2 \\ -1/2 & 5/2 \end{pmatrix}.

Question 9

A linear transformation T:R2→R2T: \mathbb{R}^2 \to \mathbb{R}^2 maps the vector u=(11)\mathbf{u} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} to T(u)=(52)T(\mathbf{u}) = \begin{pmatrix} 5 \\ 2 \end{pmatrix} and the vector v=(0−1)\mathbf{v} = \begin{pmatrix} 0 \\ -1 \end{pmatrix} to T(v)=(−1−1)T(\mathbf{v}) = \begin{pmatrix} -1 \\ -1 \end{pmatrix}. What is the standard matrix AA for this transformation?

  1. (5−12−1)\begin{pmatrix} 5 & -1 \\ 2 & -1 \end{pmatrix}
  2. (4111)\begin{pmatrix} 4 & 1 \\ 1 & 1 \end{pmatrix} (correct answer)
  3. (6−13−1)\begin{pmatrix} 6 & -1 \\ 3 & -1 \end{pmatrix}
  4. (4−11−1)\begin{pmatrix} 4 & -1 \\ 1 & -1 \end{pmatrix}
Explanation: The columns of the standard matrix AA are T(e1)T(\mathbf{e}_1) and T(e2)T(\mathbf{e}_2). We must express e1\mathbf{e}_1 and e2\mathbf{e}_2 as linear combinations of u\mathbf{u} and v\mathbf{v}. We see that e1=u+v\mathbf{e}_1 = \mathbf{u} + \mathbf{v} and e2=−v\mathbf{e}_2 = -\mathbf{v}. By linearity: T(e1)=T(u)+T(v)=(52)+(−1−1)=(41)T(\mathbf{e}_1) = T(\mathbf{u}) + T(\mathbf{v}) = \begin{pmatrix} 5 \\ 2 \end{pmatrix} + \begin{pmatrix} -1 \\ -1 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}. And T(e2)=−T(v)=−(−1−1)=(11)T(\mathbf{e}_2) = -T(\mathbf{v}) = -\begin{pmatrix} -1 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}. Therefore, the matrix AA is (4111)\begin{pmatrix} 4 & 1 \\ 1 & 1 \end{pmatrix}.

Question 10

Let T1:R2→R2T_1: \mathbb{R}^2 \to \mathbb{R}^2 be a rotation counterclockwise by 90∘90^\circ, and let T2:R2→R2T_2: \mathbb{R}^2 \to \mathbb{R}^2 be a shear transformation that maps (x,y)(x,y) to (x+2y,y)(x+2y, y). What is the standard matrix for the composite transformation T1∘T2T_1 \circ T_2?

  1. (2−110)\begin{pmatrix} 2 & -1 \\ 1 & 0 \end{pmatrix}
  2. (01−1−2)\begin{pmatrix} 0 & 1 \\ -1 & -2 \end{pmatrix}
  3. (0−112)\begin{pmatrix} 0 & -1 \\ 1 & 2 \end{pmatrix} (correct answer)
  4. (−2−110)\begin{pmatrix} -2 & -1 \\ 1 & 0 \end{pmatrix}
Explanation: The matrix for the rotation T1T_1 is A1=(cos⁡(90∘)−sin⁡(90∘)sin⁡(90∘)cos⁡(90∘))=(0−110)A_1 = \begin{pmatrix} \cos(90^\circ) & -\sin(90^\circ) \\ \sin(90^\circ) & \cos(90^\circ) \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}. The matrix for the shear T2T_2 is found by transforming the basis vectors: T2(1,0)=(1,0)T_2(1,0)=(1,0) and T2(0,1)=(2,1)T_2(0,1)=(2,1), so A2=(1201)A_2 = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}. The matrix for the composition T1∘T2T_1 \circ T_2 is A1A2=(0−110)(1201)=(0−112)A_1 A_2 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & 2 \end{pmatrix}.

Question 11

A linear transformation T:R2→R2T: \mathbb{R}^2 \to \mathbb{R}^2 has eigenvectors v1=(11)\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix} with eigenvalue λ1=3\lambda_1 = 3, and v2=(1−1)\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix} with eigenvalue λ2=−1\lambda_2 = -1. What is the standard matrix AA for this transformation?

  1. (300−1)\begin{pmatrix} 3 & 0 \\ 0 & -1 \end{pmatrix}
  2. (3−131)\begin{pmatrix} 3 & -1 \\ 3 & 1 \end{pmatrix}
  3. (1120)\begin{pmatrix} 1 & 1 \\ 2 & 0 \end{pmatrix}
  4. (1221)\begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} (correct answer)
Explanation: When you're given eigenvectors and eigenvalues of a linear transformation, you need to reconstruct the standard matrix using the relationship Av=λvA\mathbf{v} = \lambda\mathbf{v} for each eigenvector-eigenvalue pair. To find matrix AA, use the fact that it must satisfy both eigenvalue equations simultaneously. From the given information: A(11)=3(11)=(33)A\begin{pmatrix} 1 \\ 1 \end{pmatrix} = 3\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \end{pmatrix} and $$A\begin{pmatrix} 1 \ -1 \end{pmatrix} = (-1)\begin{pmatrix} 1 \ -1 \end{pmatrix} = \begin{pmatrix} -1 \ 1 \end{pmatrix} Let $$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} $$. From the first equation: $$a + b = 3$$ and $$c + d = 3$$. From the second equation: $$a - b = -1$$ and $$c - d = 1$$. Solving these systems gives $$a = 1$$, $$b = 2$$, $$c = 2$$, $$d = 1$$, so $$A = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} $$. Option A is a diagonal matrix with the eigenvalues on the diagonal, but this would only be correct if the eigenvectors were the standard basis vectors, which they're not. Option B incorrectly places eigenvalues as entries without proper construction. Option C fails both eigenvalue tests—multiplying this matrix by either eigenvector doesn't yield the correct scalar multiple. Option D correctly satisfies both eigenvalue equations when you multiply it out. **Study tip:** Always verify your answer by checking that $$A\mathbf{v}_i = \lambda_i\mathbf{v}_i$$ for each given eigenvector-eigenvalue pair. The diagonal form only works when eigenvectors are orthonormal basis vectors.

Question 12

Let T:R2→R2T: \mathbb{R}^2 \to \mathbb{R}^2 be a linear transformation whose standard matrix is AA. If T(e1+e2)=(35)T(\mathbf{e}_1 + \mathbf{e}_2) = \begin{pmatrix} 3 \\ 5 \end{pmatrix} and T(e1−e2)=(1−1)T(\mathbf{e}_1 - \mathbf{e}_2) = \begin{pmatrix} 1 \\ -1 \end{pmatrix}, what is the matrix AA?

  1. (315−1)\begin{pmatrix} 3 & 1 \\ 5 & -1 \end{pmatrix}
  2. (4246)\begin{pmatrix} 4 & 2 \\ 4 & 6 \end{pmatrix}
  3. (1232)\begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix}
  4. (2123)\begin{pmatrix} 2 & 1 \\ 2 & 3 \end{pmatrix} (correct answer)
Explanation: When you encounter a linear transformation problem where you're given the outputs for specific input vectors, your goal is to find what the transformation does to the standard basis vectors e1=(10)\mathbf{e}_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix} and $$\mathbf{e}_2 = \begin{pmatrix} 0 \ 1 \end{pmatrix} You have two equations: $$T(\mathbf{e}_1 + \mathbf{e}_2) = \begin{pmatrix} 3 \\ 5 \end{pmatrix} $$ and $$T(\mathbf{e}_1 - \mathbf{e}_2) = \begin{pmatrix} 1 \\ -1 \end{pmatrix} $$. Since $$T$$ is linear, you can write these as $$T(\mathbf{e}_1) + T(\mathbf{e}_2) = \begin{pmatrix} 3 \\ 5 \end{pmatrix} $$ and $$T(\mathbf{e}_1) - T(\mathbf{e}_2) = \begin{pmatrix} 1 \\ -1 \end{pmatrix} $$. Adding these equations: $$2T(\mathbf{e}_1) = \begin{pmatrix} 4 \\ 4 \end{pmatrix} $$, so $$T(\mathbf{e}_1) = \begin{pmatrix} 2 \\ 2 \end{pmatrix} $$. Subtracting the second from the first: $$2T(\mathbf{e}_2) = \begin{pmatrix} 2 \\ 6 \end{pmatrix} $$, so $$T(\mathbf{e}_2) = \begin{pmatrix} 1 \\ 3 \end{pmatrix} $$. Therefore, $$A = \begin{pmatrix} 2 & 1 \\ 2 & 3 \end{pmatrix} $$, which is choice D. Choice A simply places the given output vectors as columns, ignoring that these aren't the images of the standard basis vectors. Choice B contains a computational error in solving the system. Choice C reverses some entries, likely from incorrectly setting up the linear system. **Study tip:** When given transformations of non-standard vectors, always use linearity to set up a system of equations to find $$T(\mathbf{e}_1)$$ and $$T(\mathbf{e}_2)$$ first—these become your matrix columns.

Question 13

Let T:R3→R2T: \mathbb{R}^3 \to \mathbb{R}^2 be a linear transformation. If the kernel of TT is the line spanned by the vector (1,1,1)(1, 1, 1), and T(1,0,0)=(1,2)T(1, 0, 0) = (1, 2), which of the following could be the standard matrix for TT?

  1. (11−220−1)\begin{pmatrix} 1 & 1 & -2 \\ 2 & 0 & -1 \end{pmatrix}
  2. (2−1−111−2)\begin{pmatrix} 2 & -1 & -1 \\ 1 & 1 & -2 \end{pmatrix}
  3. (1−102−20)\begin{pmatrix} 1 & -1 & 0 \\ 2 & -2 & 0 \end{pmatrix}
  4. (11−22−1−1)\begin{pmatrix} 1 & 1 & -2 \\ 2 & -1 & -1 \end{pmatrix} (correct answer)
Explanation: When you encounter questions about linear transformations with given kernel and image conditions, you need to use both pieces of information as constraints to identify the correct matrix. Since the kernel of TT is spanned by (1,1,1)(1,1,1), this vector must map to the zero vector. Additionally, T(1,0,0)=(1,2)T(1,0,0) = (1,2) gives us the first column of the standard matrix. Let's call the matrix A=(abcdef)A = \begin{pmatrix} a & b & c \\ d & e & f \end{pmatrix}. From T(1,0,0)=(1,2)T(1,0,0) = (1,2), we know the first column is (12)\begin{pmatrix} 1 \\ 2 \end{pmatrix}, so a=1a = 1 and d=2d = 2. From the kernel condition, A(1,1,1)=(0,0)A(1,1,1) = (0,0), which means:
  • a+b+c=0a + b + c = 0, so 1+b+c=01 + b + c = 0, giving us b+c=−1b + c = -1
  • d+e+f=0d + e + f = 0, so 2+e+f=02 + e + f = 0, giving us e+f=−2e + f = -2
Now let's check each option: Option A: (11−220−1)\begin{pmatrix} 1 & 1 & -2 \\ 2 & 0 & -1 \end{pmatrix} - First column is correct, but 1+(−2)=−11 + (-2) = -1 ✓ and 0+(−1)=−1≠−20 + (-1) = -1 \neq -2 ✗ Option B: (2−1−111−2)\begin{pmatrix} 2 & -1 & -1 \\ 1 & 1 & -2 \end{pmatrix} - First column is (2,1)≠(1,2)(2,1) \neq (1,2) ✗ Option C: (1−102−20)\begin{pmatrix} 1 & -1 & 0 \\ 2 & -2 & 0 \end{pmatrix} - First column is correct, but (−1)+0=−1(-1) + 0 = -1 ✓ and (−2)+0=−2(-2) + 0 = -2 ✓, but this gives kernel dimension 2, not 1 ✗ Option D: (11−22−1−1)\begin{pmatrix} 1 & 1 & -2 \\ 2 & -1 & -1 \end{pmatrix} - First column is correct, 1+(−2)=−11 + (-2) = -1 ✓, and (−1)+(−1)=−2(-1) + (-1) = -2 ✓ Strategy tip: Always verify both the given transformation values and kernel conditions systematically. The kernel constraint often provides the most restrictive conditions for elimination.

Question 14

Let T:R3→R3T: \mathbb{R}^3 \to \mathbb{R}^3 be the linear transformation that reflects vectors across the plane x+y−z=0x + y - z = 0. If AA is the standard matrix representation of TT, which of the following statements about AA is correct?

  1. AA has eigenvalues 1,1,−11, 1, -1 and det⁡(A)=−1\det(A) = -1 (correct answer)
  2. AA has eigenvalues 1,−1,−11, -1, -1 and det⁡(A)=1\det(A) = 1
  3. AA is symmetric and A2=I3A^2 = I_3, but AA is not orthogonal
  4. AA has trace 11 and all eigenvalues are real and distinct
Explanation: For a reflection across a plane, vectors in the plane are unchanged (eigenvalue 11) and vectors perpendicular to the plane are reversed (eigenvalue −1-1). Since the plane has dimension 22, there are two linearly independent eigenvectors with eigenvalue 11 and one with eigenvalue −1-1. The normal vector to the plane x+y−z=0x + y - z = 0 is (1,1,−1)(1,1,-1), which corresponds to eigenvalue −1-1. Two vectors in the plane (eigenvalue 11) can be found, such as (1,−1,0)(1,-1,0) and (1,0,1)(1,0,1). Therefore the eigenvalues are 1,1,−11, 1, -1 and det⁡(A)=1⋅1⋅(−1)=−1\det(A) = 1 \cdot 1 \cdot (-1) = -1. Choice B has the wrong eigenvalue multiplicities. Choice C is incorrect because AA is orthogonal (reflections preserve lengths and angles). Choice D is wrong because not all eigenvalues are distinct (11 has multiplicity 22).

Question 15

What is the standard matrix for the linear transformation T:R3→R3T: \mathbb{R}^3 \to \mathbb{R}^3 that orthogonally projects vectors onto the plane defined by the equation x+2y+2z=0x + 2y + 2z = 0?

  2. (correct answer)
Explanation: The transformation is given by T(x)=x−projnxT(\mathbf{x}) = \mathbf{x} - \text{proj}_{\mathbf{n}}\mathbf{x}, where n=(1,2,2)\mathbf{n} = (1, 2, 2) is the normal vector to the plane. The matrix for the projection onto the normal vector is P=1nTnnnT=19(122244244)P = \frac{1}{\mathbf{n}^T\mathbf{n}}\mathbf{n}\mathbf{n}^T = \frac{1}{9}\begin{pmatrix} 1 & 2 & 2 \\ 2 & 4 & 4 \\ 2 & 4 & 4 \end{pmatrix}. The matrix for TT is A=I−P=(100010001)−19(122244244)=19(8−2−2−25−4−2−45)A = I - P = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} - \frac{1}{9}\begin{pmatrix} 1 & 2 & 2 \\ 2 & 4 & 4 \\ 2 & 4 & 4 \end{pmatrix} = \frac{1}{9}\begin{pmatrix} 8 & -2 & -2 \\ -2 & 5 & -4 \\ -2 & -4 & 5 \end{pmatrix}. Choice A is the matrix for projection onto the normal line, not the plane.