Linear Algebra Quiz: Determinant Orientation And Sign
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Determinant Orientation And SignQuestion 1 of 19

A 3×33 \times 3 matrix AA has the property that det⁡(A)=−8\det(A) = -8. If B=2ATB = 2A^T, where ATA^T denotes the transpose of AA, what is the relationship between the orientations induced by transformations AA and BB?

Both transformations preserve orientation since det⁡(B)=−64<0\det(B) = -64 < 0
AA reverses orientation while BB preserves orientation since det⁡(B)=64>0\det(B) = 64 > 0
Both transformations reverse orientation since both determinants are negative
AA preserves orientation while BB reverses orientation since det⁡(B)=−64<0\det(B) = -64 < 0
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Linear Algebra Quiz

Linear Algebra Quiz: Determinant Orientation And Sign

Practice Determinant Orientation And Sign 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 Determinant Orientation And Sign, giving you a quick way to practice the rules, question types, and explanations that matter most for Linear Algebra.

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Question 1

A 3×33 \times 3 matrix AA has the property that det⁡(A)=−8\det(A) = -8. If B=2ATB = 2A^T, where ATA^T denotes the transpose of AA, what is the relationship between the orientations induced by transformations AA and BB?

  1. Both transformations preserve orientation since det⁡(B)=−64<0\det(B) = -64 < 0
  2. AA reverses orientation while BB preserves orientation since det⁡(B)=64>0\det(B) = 64 > 0
  3. Both transformations reverse orientation since both determinants are negative (correct answer)
  4. AA preserves orientation while BB reverses orientation since det⁡(B)=−64<0\det(B) = -64 < 0
Explanation: Given det(A) = -8, we have det(ATA^T) = det(A) = -8 (transpose preserves determinant). For B = 2A^T, we get det(B) = det(2AT2A^T) = 2³ det(ATA^T) = 8 × (-8) = -64. Since det(A) = -8 < 0, transformation A reverses orientation. Since det(B) = -64 < 0, transformation B also reverses orientation. Choice A is wrong because negative determinant means orientation reversal, not preservation. Choice B incorrectly states det(B) > 0. Choice D incorrectly claims A preserves orientation when det(A) < 0.

Question 2

A linear transformation T:R2→R2T: \mathbb{R}^2 \to \mathbb{R}^2 is represented by a matrix AA with det⁡(A)<0\det(A) < 0. If the standard basis vectors are i^=(1,0)\hat{i} = (1,0) and j^=(0,1)\hat{j} = (0,1), what is the geometric relationship between the transformed vectors T(i^)T(\hat{i}) and T(j^)T(\hat{j})?

  1. The angle from T(i^)T(\hat{i}) to T(j^)T(\hat{j}) is counter-clockwise and less than 180∘180^\circ.
  2. The angle from T(i^)T(\hat{i}) to T(j^)T(\hat{j}) is clockwise and less than 180∘180^\circ. (correct answer)
  3. The vectors T(i^)T(\hat{i}) and T(j^)T(\hat{j}) are collinear.
  4. The vectors T(i^)T(\hat{i}) and T(j^)T(\hat{j}) are orthogonal.
Explanation: The sign of the determinant indicates the orientation of the transformed basis vectors relative to the original orientation. The standard basis (i^,j^)(\hat{i}, \hat{j}) has a positive (counter-clockwise) orientation. A negative determinant means the transformation reverses the orientation. Therefore, the orientation of the ordered pair (T(i^),T(j^))(T(\hat{i}), T(\hat{j})) is negative, which corresponds to a clockwise direction from the first vector to the second through an angle less than 180∘180^\circ.

Question 3

Let u⃗,v⃗,w⃗\vec{u}, \vec{v}, \vec{w} be three vectors in R3\mathbb{R}^3 that form a left-handed system. Let AA be the matrix with these vectors as columns, A=[u⃗  v⃗  w⃗]A = [\vec{u} \; \vec{v} \; \vec{w}]. Which statement is true about the determinant of the matrix B=[w⃗  u⃗  v⃗]B = [\vec{w} \; \vec{u} \; \vec{v}]?

  1. det⁡(B)>0\det(B) > 0 because BB is a permutation of the columns of AA.
  2. det⁡(B)<0\det(B) < 0 because an odd number of column swaps were performed.
  3. det⁡(B)<0\det(B) < 0 because the original system was left-handed and an even number of column swaps preserves orientation. (correct answer)
  4. det⁡(B)>0\det(B) > 0 because the original system was left-handed and an even number of column swaps reverses the orientation.
Explanation: Since (u⃗,v⃗,w⃗)(\vec{u}, \vec{v}, \vec{w}) is a left-handed system, the determinant of the matrix A=[u⃗  v⃗  w⃗]A = [\vec{u} \; \vec{v} \; \vec{w}] is negative. The matrix B=[w⃗  u⃗  v⃗]B = [\vec{w} \; \vec{u} \; \vec{v}] is obtained from AA by a cyclic permutation of its columns. This can be achieved with two column swaps: first swap columns 1 and 3 to get [w⃗  v⃗  u⃗][\vec{w} \; \vec{v} \; \vec{u}], then swap columns 2 and 3 to get [w⃗  u⃗  v⃗][\vec{w} \; \vec{u} \; \vec{v}]. An even number of swaps (two) means the sign of the determinant is preserved. Therefore, det⁡(B)=det⁡(A)<0\det(B) = \det(A) < 0.

Question 4

A linear transformation T:R2→R2T: \mathbb{R}^2 \to \mathbb{R}^2 consists of a rotation counter-clockwise by 45∘45^\circ followed by a reflection across the x-axis. What is the sign of the determinant of the matrix representing the transformation TT?

  1. Positive, because the rotation preserves orientation.
  2. Negative, because the reflection reverses orientation. (correct answer)
  3. Zero, because the operations cancel each other out.
  4. It cannot be determined without the specific matrix.
Explanation: The transformation TT is a composition of two transformations, T=Trefl∘TrotT = T_{refl} \circ T_{rot}. The determinant of the composite transformation's matrix is the product of the individual determinants. A rotation matrix always has a determinant of +1+1, which preserves orientation. A reflection matrix always has a determinant of −1-1, which reverses orientation. Thus, det⁡(T)=det⁡(Trefl)×det⁡(Trot)=(−1)×(1)=−1\det(T) = \det(T_{refl}) \times \det(T_{rot}) = (-1) \times (1) = -1. The resulting determinant is negative.

Question 5

Let AA be an invertible 3×33 \times 3 matrix such that its determinant is det⁡(A)=−2\det(A) = -2. Which of the following statements is correct regarding the determinants of its inverse, A−1A^{-1}, and the scaled matrix, −A-A?

  1. det⁡(A−1)>0\det(A^{-1}) > 0 and det⁡(−A)>0\det(-A) > 0.
  2. det⁡(A−1)<0\det(A^{-1}) < 0 and det⁡(−A)<0\det(-A) < 0.
  3. det⁡(A−1)>0\det(A^{-1}) > 0 and det⁡(−A)<0\det(-A) < 0.
  4. det⁡(A−1)<0\det(A^{-1}) < 0 and det⁡(−A)>0\det(-A) > 0. (correct answer)
Explanation: The determinant of the inverse matrix is the reciprocal of the original determinant: det⁡(A−1)=1/det⁡(A)=1/(−2)=−0.5\det(A^{-1}) = 1/\det(A) = 1/(-2) = -0.5. Since this value is negative, det⁡(A−1)<0\det(A^{-1}) < 0. For an n×nn \times n matrix, det⁡(kA)=kndet⁡(A)\det(kA) = k^n \det(A). Here, n=3n=3 and k=−1k=-1. So, det⁡(−A)=(−1)3det⁡(A)=(−1)(−2)=2\det(-A) = (-1)^3 \det(A) = (-1)(-2) = 2. Since this value is positive, det⁡(−A)>0\det(-A) > 0. Therefore, the correct statement is det⁡(A−1)<0\det(A^{-1}) < 0 and det⁡(−A)>0\det(-A) > 0.

Question 6

Let T:R3→R3T: \mathbb{R}^3 \to \mathbb{R}^3 be a linear transformation represented by matrix AA. If TT maps the standard right-handed basis (i^,j^,k^)(\hat{i}, \hat{j}, \hat{k}) to a set of vectors (T(i^),T(j^),T(k^))(T(\hat{i}), T(\hat{j}), T(\hat{k})) that form a left-handed system, which statement about det⁡(A)\det(A) must be true?

  1. det⁡(A)=−1\det(A) = -1
  2. det⁡(A)=0\det(A) = 0
  3. det⁡(A)>0\det(A) > 0
  4. det⁡(A)<0\det(A) < 0 (correct answer)
Explanation: The determinant of a transformation matrix indicates how orientation is changed. A positive determinant means orientation is preserved (e.g., a right-handed system maps to a right-handed system). A negative determinant means orientation is reversed (e.g., a right-handed system maps to a left-handed system). Since the transformation T maps a right-handed basis to a left-handed system, it reverses orientation. Therefore, the determinant of its matrix A must be negative.

Question 7

A linear transformation on R2\mathbb{R}^2 is defined as a reflection across the line y=xy=x followed by a clockwise rotation of 90∘90^\circ. What is the determinant of the matrix representing this composite transformation?

  1. −1-1 (correct answer)
  2. 11
  3. 00
  4. 22
Explanation: When you encounter composite transformations, remember that the determinant of the composition equals the product of the individual determinants, and geometric properties give you shortcuts to find these values. Let's work through each transformation. A reflection across the line y=xy = x has matrix (0110)\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, which has determinant 0⋅0−1⋅1=−10 \cdot 0 - 1 \cdot 1 = -1. All reflections have determinant −1-1 because they reverse orientation (flip the plane). A clockwise rotation by 90°90° has matrix (01−10)\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} with determinant 0⋅0−1⋅(−1)=10 \cdot 0 - 1 \cdot (-1) = 1. All rotations have determinant 11 because they preserve orientation and area. The composite transformation has determinant (−1)×1=−1(-1) \times 1 = -1, confirming answer (A). (B) 11 would be correct if both transformations preserved orientation, but reflection reverses it. (C) 00 would mean the transformation collapses the plane to a line or point, which neither reflection nor rotation does. (D) 22 suggests the transformation doubles areas, but geometric transformations like reflections and rotations preserve areas (though they may reverse orientation). Study tip: Memorize that reflections always have determinant −1-1 (orientation-reversing) while rotations always have determinant 11 (orientation-preserving). For composite transformations, multiply the determinants rather than computing the full matrix product—it's much faster and less error-prone.

Question 8

Let AA be a 3×33 \times 3 matrix with columns c⃗1,c⃗2,c⃗3\vec{c}_1, \vec{c}_2, \vec{c}_3 such that A=[c⃗1  c⃗2  c⃗3]A = [\vec{c}_1 \; \vec{c}_2 \; \vec{c}_3] and det⁡(A)=−4\det(A) = -4. What is the determinant of the matrix B=[c⃗2  c⃗3  c⃗1]B = [\vec{c}_2 \; \vec{c}_3 \; \vec{c}_1]?

  1. 44
  2. −4-4 (correct answer)
  3. 88
  4. −8-8
Explanation: The matrix BB is formed by a cyclic permutation of the columns of AA. A cyclic permutation of three items can be achieved by two swaps. For example, to get from [c⃗1,c⃗2,c⃗3][\vec{c}_1, \vec{c}_2, \vec{c}_3] to [c⃗2,c⃗3,c⃗1][\vec{c}_2, \vec{c}_3, \vec{c}_1], we can first swap columns 1 and 2 to get [c⃗2,c⃗1,c⃗3][\vec{c}_2, \vec{c}_1, \vec{c}_3], and then swap columns 2 and 3 to get [c⃗2,c⃗3,c⃗1][\vec{c}_2, \vec{c}_3, \vec{c}_1]. Each column swap multiplies the determinant by −1-1. Since there are two swaps, the determinant is multiplied by (−1)2=1(-1)^2 = 1. Thus, det⁡(B)=det⁡(A)=−4\det(B) = \det(A) = -4.

Question 9

A linear transformation T(x⃗)=Ax⃗T(\vec{x}) = A\vec{x} maps the unit square in R2\mathbb{R}^2 to a parallelogram. The vertices of this parallelogram, starting from the origin and listed in counter-clockwise order, are (0,0)(0,0), (3,1)(3,1), (2,5)(2,5), and (−1,4)(-1,4). What can be concluded about the determinant of AA?

  1. det⁡(A)>0\det(A) > 0 (correct answer)
  2. det⁡(A)<0\det(A) < 0
  3. det⁡(A)=0\det(A) = 0
  4. det⁡(A)=1\det(A) = 1
Explanation: When you encounter a linear transformation problem involving area and orientation, focus on how the determinant encodes both the scaling factor and orientation of the transformation. The key insight is that a linear transformation T(x⃗)=Ax⃗T(\vec{x}) = A\vec{x} maps the unit square with vertices (0,0)(0,0), (1,0)(1,0), (0,1)(0,1), and (1,1)(1,1) to a parallelogram. The columns of matrix AA are exactly where the unit vectors are mapped: A=[T(e1⃗)∣T(e2⃗)]A = [T(\vec{e_1}) \mid T(\vec{e_2})]. Since the transformed vertices listed counter-clockwise are (0,0)(0,0), (3,1)(3,1), (2,5)(2,5), and (−1,4)(-1,4), we can determine that T(e1⃗)=(3,1)T(\vec{e_1}) = (3,1) and T(e2⃗)=(2,5)T(\vec{e_2}) = (2,5). Therefore, $$A = \begin{bmatrix} 3 & 2 \ 1 & 5 \end{bmatrix} Computing $$\det(A) = 3(5) - 2(1) = 15 - 2 = 13 > 0$$. The positive determinant tells us the transformation preserves orientation (counter-clockwise remains counter-clockwise). Looking at the wrong answers: **(B)** $$\det(A) < 0$$ would mean orientation reversal, but the vertices are still listed counter-clockwise. **(C)** $$\det(A) = 0$$ would mean the parallelogram is degenerate (collapsed to a line), which contradicts having four distinct vertices. **(D)** $$\det(A) = 1$$ would mean area preservation, but our parallelogram has area 13, not 1. **Study tip:** Remember that $$\det(A)$$ gives both magnitude (area scaling factor) and sign (orientation preservation). Positive determinant means orientation is preserved.

Question 10

Let AA be a 3×33 \times 3 matrix with det⁡(A)=5\det(A) = 5. A new matrix BB is created from AA by first swapping columns 1 and 3, and then multiplying row 2 by the scalar −2-2. What is the determinant of BB and how does the transformation represented by BB affect orientation?

  1. det⁡(B)=−10\det(B) = -10, and it preserves orientation.
  2. det⁡(B)=10\det(B) = 10, and it reverses orientation.
  3. det⁡(B)=−10\det(B) = -10, and it reverses orientation.
  4. det⁡(B)=10\det(B) = 10, and it preserves orientation. (correct answer)
Explanation: The initial determinant is positive, so AA preserves orientation. Swapping two columns negates the determinant, so after the swap, the determinant is −5-5. Multiplying a single row by −2-2 multiplies the determinant by −2-2. So, det⁡(B)=(−5)×(−2)=10\det(B) = (-5) \times (-2) = 10. Since det⁡(B)=10>0\det(B) = 10 > 0, the transformation represented by BB preserves orientation.

Question 11

Consider the linear transformation represented by matrix A=(cos⁡θ−sin⁡θsin⁡θcos⁡θ)A = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} followed by matrix B=(100−1)B = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} . For which values of θ\theta does the composite transformation BABA preserve orientation?

  1. For all values of θ\theta, since rotations always preserve orientation
  2. Never, since det⁡(BA)=−1<0\det(BA) = -1 < 0 for all θ\theta (correct answer)
  3. Only when cos⁡θ>0\cos\theta > 0, which determines the orientation effect
  4. Only when sin⁡θ=0\sin\theta = 0, making the transformation purely diagonal
Explanation: Matrix A is a rotation matrix with det(A) = cos²θ + sin²θ = 1 > 0. Matrix B is a reflection across the x-axis with det(B) = -1 < 0. The composite transformation BA has det(BA) = det(B)det(A) = (-1)(1) = -1 < 0 for all values of θ. Since the determinant is always negative, the composite transformation always reverses orientation, regardless of θ. Choice A ignores the effect of B. Choice C incorrectly focuses on cosθ when the determinant doesn't depend on the specific angle. Choice D incorrectly suggests sinθ = 0 changes the orientation behavior.

Question 12

Let u=(2,−1,3)\mathbf{u} = (2, -1, 3), v=(1,4,−2)\mathbf{v} = (1, 4, -2), and w=(−3,2,1)\mathbf{w} = (-3, 2, 1) be vectors in R3\mathbb{R}^3. If the scalar triple product u⋅(v×w)=−42\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = -42, what does this tell us about the orientation of the ordered triple (u,v,w)(\mathbf{u}, \mathbf{v}, \mathbf{w})?

  1. The vectors form a left-handed coordinate system with volume 42 cubic units (correct answer)
  2. The vectors form a right-handed coordinate system with volume 42 cubic units
  3. The vectors are coplanar and cannot form a coordinate system
  4. The vectors form a right-handed coordinate system with volume -42 cubic units
Explanation: The scalar triple product gives both the signed volume of the parallelepiped and orientation information. Since the result is negative (-42), the ordered triple (u,v,w) forms a left-handed coordinate system, and the volume is |−42| = 42 cubic units. Choice B is wrong because negative scalar triple product indicates left-handed orientation. Choice C is wrong because non-zero scalar triple product means vectors are linearly independent. Choice D is wrong because volume is always positive; it's the absolute value of the scalar triple product.

Question 13

A 2×22 \times 2 matrix TT transforms the unit square [0,1]×[0,1][0,1] \times [0,1] into a parallelogram with area 32\frac{3}{2} and vertices traced in counterclockwise order when starting from T(0,0)T(0,0). If T(1,0)=(2,1)T(1,0) = (2,1) and T(0,1)=(a,b)T(0,1) = (a,b), what must be true about aa and bb?

  1. a+2b=32a + 2b = \frac{3}{2} and the transformation is orientation-preserving
  2. 2b−a=−322b - a = -\frac{3}{2} and the transformation reverses orientation
  3. a−2b=32a - 2b = \frac{3}{2} and the transformation preserves orientation
  4. 2b−a=322b - a = \frac{3}{2} and the transformation preserves orientation (correct answer)
Explanation: When you see a matrix transforming a geometric shape, you're dealing with linear transformations where the determinant tells you both the area scaling factor and orientation. The matrix TT has the form T=(2a1b)T = \begin{pmatrix} 2 & a \\ 1 & b \end{pmatrix} since T(1,0)=(2,1)T(1,0) = (2,1) gives the first column and T(0,1)=(a,b)T(0,1) = (a,b) gives the second column. The determinant is det⁡(T)=2b−a\det(T) = 2b - a. Since the unit square has area 1 and transforms to a parallelogram with area 32\frac{3}{2}, the absolute value of the determinant must equal 32\frac{3}{2}. The problem states the vertices are traced counterclockwise starting from T(0,0)T(0,0), which means the transformation preserves orientation. For orientation-preserving transformations, the determinant is positive, so det⁡(T)=+32\det(T) = +\frac{3}{2}. Therefore, 2b−a=322b - a = \frac{3}{2}. Option D correctly identifies both conditions: 2b−a=322b - a = \frac{3}{2} and orientation preservation. Option A uses the wrong determinant formula (a+2ba + 2b instead of 2b−a2b - a) but correctly identifies orientation preservation. Option B has the correct determinant magnitude but incorrectly claims the transformation reverses orientation—if it did, the vertices would be traced clockwise. Option C uses yet another incorrect determinant formula (a−2ba - 2b) and gets the wrong value. Remember: for 2×22 \times 2 matrices, the determinant formula is ad−bcad - bc, and positive determinants preserve orientation while negative ones reverse it.

Question 14

Let P=(010100001)P = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} and Q=(1000−10001)Q = \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix} . If a coordinate system has positive orientation, what is the orientation of the coordinate system after applying first PP, then QQ?

  1. Positive orientation, since the composition involves two orientation-preserving transformations
  2. Negative orientation, since det⁡(QP)=1>0\det(QP) = 1 > 0 but QQ reverses orientation
  3. Positive orientation, since det⁡(QP)=1>0\det(QP) = 1 > 0 indicating orientation preservation overall (correct answer)
  4. Negative orientation, since det⁡(QP)=−1<0\det(QP) = -1 < 0 indicating orientation reversal overall
Explanation: First, det(P) = -1 (swapping first two rows/columns of identity) and det(Q) = -1 (one negative diagonal entry). The composition QP has det(QP) = det(Q)det(P) = (-1)(-1) = 1 > 0, indicating overall orientation preservation. Even though each individual transformation reverses orientation, their composition preserves it (two reversals make a preservation). Choice A is wrong because both P and Q individually reverse orientation. Choice B has correct reasoning about individual transformations but wrong conclusion about the composition. Choice D has the wrong value for det(QP).

Question 15

Consider three vectors a\mathbf{a}, b\mathbf{b}, and c\mathbf{c} in R3\mathbb{R}^3 arranged as columns of matrix MM. If swapping vectors b\mathbf{b} and c\mathbf{c} results in matrix M′M' with det⁡(M′)=15\det(M') = 15, and then scaling vector a\mathbf{a} by factor −2-2 gives matrix M′′M'', what is det⁡(M′′)\det(M'')?

  1. −30-30, indicating the final configuration reverses the original orientation (correct answer)
  2. 3030, indicating the final configuration preserves the original orientation
  3. −15-15, indicating the final configuration reverses the original orientation
  4. 1515, indicating the final configuration preserves the original orientation
Explanation: Starting from the original matrix MM, swapping two columns changes the sign, so det⁡(M)=−det⁡(M′)=−15\det(M) = -\det(M') = -15. When we scale vector a\mathbf{a} by −2-2 in matrix M′M', we get det⁡(M′′)=−2×det⁡(M′)=−2×15=−30\det(M'') = -2 \times \det(M') = -2 \times 15 = -30. Comparing to the original: det⁡(M)=−15\det(M) = -15 and det⁡(M′′)=−30\det(M'') = -30 have the same sign, but since we're asked about the final configuration, det⁡(M′′)=−30<0\det(M'') = -30 < 0 indicates orientation reversal compared to the standard orientation.

Question 16

Consider the transformation T:R3→R3T: \mathbb{R}^3 \to \mathbb{R}^3 defined by T(x)=AxT(\mathbf{x}) = A\mathbf{x} where A=(2−1304−2105)A = \begin{pmatrix} 2 & -1 & 3 \\ 0 & 4 & -2 \\ 1 & 0 & 5 \end{pmatrix} . If det⁡(A)<0\det(A) < 0, what can be concluded about the geometric effect of this transformation?

  1. The transformation preserves orientation and volume is scaled by ∣det⁡(A)∣|\det(A)|
  2. The transformation reverses orientation and volume is scaled by ∣det⁡(A)∣|\det(A)| (correct answer)
  3. The transformation preserves orientation but reduces all volumes to zero
  4. The transformation reverses orientation and all volumes become negative
Explanation: When det(A) < 0, the transformation reverses orientation (changes the handedness of coordinate systems) and scales volumes by the absolute value |det(A)|. The sign of the determinant indicates orientation change, while its absolute value gives the volume scaling factor. Choice A is wrong because negative determinant means orientation reversal. Choice C is wrong because det(A) < 0 doesn't mean det(A) = 0. Choice D is wrong because volumes are always non-negative; the negative determinant affects orientation, not volume sign.

Question 17

In R2\mathbb{R}^2, vectors u=(a,3)\mathbf{u} = (a, 3) and v=(2,b)\mathbf{v} = (2, b) form the columns of matrix MM. If the parallelogram spanned by u\mathbf{u} and v\mathbf{v} has area 10 and the ordered pair (u,v)(\mathbf{u}, \mathbf{v}) creates a clockwise orientation, what constraint must abab satisfy?

  1. ab−6=10ab - 6 = 10, so ab=16ab = 16 for proper orientation and area
  2. ∣ab−6∣=10|ab - 6| = 10 with ab<6ab < 6 for the clockwise constraint
  3. ∣ab−6∣=10|ab - 6| = 10 with ab>6ab > 6 for the clockwise constraint
  4. ab−6=−10ab - 6 = -10, so ab=−4ab = -4 for clockwise orientation and area 10 (correct answer)
Explanation: This problem combines two key concepts: the determinant as area and orientation in the plane. When you see questions about parallelograms formed by vectors, think about the determinant of the matrix whose columns are those vectors. The area of a parallelogram spanned by vectors u\mathbf{u} and v\mathbf{v} equals the absolute value of the determinant of matrix M=(a23b)M = \begin{pmatrix} a & 2 \\ 3 & b \end{pmatrix} . The determinant is det⁡(M)=ab−6\det(M) = ab - 6, so the area condition gives us ∣ab−6∣=10|ab - 6| = 10. However, orientation matters crucially here. In R2\mathbb{R}^2, when vectors are ordered (u,v)(\mathbf{u}, \mathbf{v}), a positive determinant indicates counterclockwise orientation, while a negative determinant indicates clockwise orientation. Since we need clockwise orientation, we require det⁡(M)=ab−6<0\det(M) = ab - 6 < 0. Combining the area and orientation constraints: ab−6=−10ab - 6 = -10, which gives ab=−4ab = -4. Choice A incorrectly uses ab−6=10ab - 6 = 10, which would give counterclockwise orientation. Choice B correctly identifies ∣ab−6∣=10|ab - 6| = 10 for the area but wrongly claims ab<6ab < 6 ensures clockwise orientation—this is insufficient since we need the specific negative value. Choice C makes the opposite orientation error, suggesting ab>6ab > 6 for clockwise when we actually need the determinant to be negative. Remember: the sign of the determinant determines orientation in R2\mathbb{R}^2. Negative determinant means clockwise, positive means counterclockwise. Don't just focus on the absolute value when orientation is specified.

Question 18

Let A=(k28k)A = \begin{pmatrix} k & 2 \\ 8 & k \end{pmatrix}. For what range of values of kk does the linear transformation represented by AA reverse orientation?

  1. k>4k > 4
  2. k<−4k < -4 or k>4k > 4
  3. −4<k<4-4 < k < 4 (correct answer)
  4. k=−4k = -4 or k=4k=4
Explanation: A linear transformation reverses orientation if the determinant of its matrix is negative. The determinant of matrix AA is det⁡(A)=(k)(k)−(2)(8)=k2−16\det(A) = (k)(k) - (2)(8) = k^2 - 16. To find when the orientation is reversed, we must solve the inequality det⁡(A)<0\det(A) < 0. This gives k2−16<0k^2 - 16 < 0, or k2<16k^2 < 16. Taking the square root of both sides, we find ∣k∣<4|k| < 4, which is equivalent to the interval −4<k<4-4 < k < 4.

Question 19

Which of the following row operations on an invertible 3×33 \times 3 matrix AA is guaranteed not to change the sign of its determinant?

  1. Swapping row 1 and row 2.
  2. Multiplying the entire matrix by −1-1.
  3. Replacing row 2 with R2−5R3R_2 - 5R_3. (correct answer)
  4. Multiplying row 1 by −3-3.
Explanation: Let's analyze the effect of each operation on det⁡(A)\det(A). (A) Swapping two rows multiplies the determinant by −1-1, which changes its sign. (B) Multiplying a 3×33 \times 3 matrix by −1-1 is equivalent to multiplying each of the 3 rows by −1-1. This multiplies the determinant by (−1)3=−1(-1)^3 = -1, changing its sign. (C) Replacing a row RiR_i with Ri+kRjR_i + kR_j (a shear) does not change the value of the determinant at all. Therefore, it does not change the sign. (D) Multiplying a single row by −3-3 multiplies the determinant by −3-3, which changes its sign.