Linear Algebra Quiz: Area Volume Scaling
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Area Volume ScalingQuestion 1 of 18

A triangle in the Cartesian plane has vertices at (0,0)(0,0), (4,0)(4,0), and (2,3)(2,3). This triangle is transformed by the linear transformation represented by the matrix A=(2113)A = \begin{pmatrix} 2 & 1 \\ -1 & 3 \end{pmatrix}. What is the area of the transformed triangle?

6
7
30
42
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Linear Algebra Quiz

Linear Algebra Quiz: Area Volume Scaling

Practice Area Volume Scaling 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 Area Volume Scaling, 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

A triangle in the Cartesian plane has vertices at (0,0)(0,0), (4,0)(4,0), and (2,3)(2,3). This triangle is transformed by the linear transformation represented by the matrix A=(2113)A = \begin{pmatrix} 2 & 1 \\ -1 & 3 \end{pmatrix}. What is the area of the transformed triangle?

  1. 6
  2. 7
  3. 30
  4. 42 (correct answer)
Explanation: First, calculate the area of the original triangle. The base along the x-axis is 4 units and the height is 3 units. The area is 12×base×height=12(4)(3)=6\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}(4)(3) = 6. The area scaling factor of the linear transformation is the absolute value of the determinant of its matrix. The determinant of AA is det(A)=(2)(3)(1)(1)=6+1=7\det(A) = (2)(3) - (1)(-1) = 6 + 1 = 7. The area of the transformed triangle is the original area multiplied by the scaling factor: 6×7=426 \times |7| = 42.

Question 2

Let T:R3R3T: \mathbb{R}^3 \to \mathbb{R}^3 be a linear transformation with standard matrix AA. If det(A)=2\det(A) = -2, which statement accurately describes the effect of TT on a solid region SS in R3\mathbb{R}^3?

  1. TT halves the volume of SS and preserves its orientation.
  2. TT doubles the volume of SS and preserves its orientation.
  3. TT doubles the volume of SS and reverses its orientation. (correct answer)
  4. TT subtracts 2 from the volume of SS and reverses its orientation.
Explanation: The scaling factor for volume under a linear transformation is the absolute value of the determinant of its matrix. Here, the scaling factor is det(A)=2=2|\det(A)| = |-2| = 2, which means the transformation doubles the volume. The sign of the determinant indicates its effect on orientation. A negative determinant means the transformation reverses the orientation of the space (e.g., it turns a right-handed system of vectors into a left-handed one). Therefore, TT doubles the volume and reverses the orientation.

Question 3

The linear transformation T(x)=AxT(\vec{x}) = A\vec{x} with A=(253k)A = \begin{pmatrix} 2 & -5 \\ 3 & k \end{pmatrix} maps a non-degenerate parallelogram in R2\mathbb{R}^2 to a line segment. What is the value of kk?

  1. -7.5 (correct answer)
  2. -7
  3. -2
  4. 7.5
Explanation: If a linear transformation maps a 2D region (a non-degenerate parallelogram) to a 1D object (a line segment), it means the area of the region is collapsed to zero. This occurs if and only if the determinant of the transformation matrix is zero. We must find the value of kk that makes det(A)=0\det(A) = 0. det(A)=(2)(k)(5)(3)=2k+15\det(A) = (2)(k) - (-5)(3) = 2k + 15. Setting the determinant to zero: 2k+15=0    2k=15    k=7.52k + 15 = 0 \implies 2k = -15 \implies k = -7.5.

Question 4

A solid object is transformed by the linear map represented by the matrix M=(200111312)M = \begin{pmatrix} 2 & 0 & 0 \\ 1 & 1 & 1 \\ 3 & -1 & 2 \end{pmatrix}. The resulting solid has a volume of 30 cubic units. What was the volume of the original object?

  1. 5 (correct answer)
  2. 6
  3. 30
  4. 180
Explanation: Let V0V_0 be the original volume and VfV_f be the final volume. The relationship is Vf=det(M)×V0V_f = |\det(M)| \times V_0. We are given Vf=30V_f = 30. We need to calculate det(M)\det(M). Expanding along the first row: det(M)=2((1)(2)(1)(1))0+0=2(2+1)=6\det(M) = 2((1)(2) - (1)(-1)) - 0 + 0 = 2(2 + 1) = 6. So, 30=6×V030 = |6| \times V_0. Solving for V0V_0, we get V0=30/6=5V_0 = 30 / 6 = 5.

Question 5

A unit cube, defined by the vectors e1,e2,e3\vec{e_1}, \vec{e_2}, \vec{e_3} in R3\mathbb{R}^3, has a volume of 1 cubic unit. It is transformed by the linear transformation T(x)=AxT(\vec{x}) = A\vec{x}, where A=(120031214)A = \begin{pmatrix} 1 & 2 & 0 \\ 0 & -3 & 1 \\ 2 & 1 & 4 \end{pmatrix}. What is the volume of the resulting parallelepiped?

  1. -17
  2. -9
  3. 9 (correct answer)
  4. 17
Explanation: The volume of the transformed solid is the volume of the original solid multiplied by the absolute value of the determinant of the transformation matrix. The original volume is 1. We calculate the determinant of AA using cofactor expansion along the first row: det(A)=1((3)(4)(1)(1))2((0)(4)(1)(2))+0=1(13)2(2)=13+4=9\det(A) = 1((-3)(4) - (1)(1)) - 2((0)(4) - (1)(2)) + 0 = 1(-13) - 2(-2) = -13 + 4 = -9. The volume scaling factor is det(A)=9=9|\det(A)| = |-9| = 9. The new volume is 1×9=91 \times 9 = 9.

Question 6

A polygon with an area of 10 square units undergoes a linear transformation T1T_1 represented by matrix A=(3111)A = \begin{pmatrix} 3 & 1 \\ 1 & 1 \end{pmatrix}, followed by a second linear transformation T2T_2 represented by matrix B=(2012)B = \begin{pmatrix} 2 & 0 \\ -1 & 2 \end{pmatrix}. What is the area of the final polygon after both transformations are applied?

  1. 20
  2. 40
  3. 60
  4. 80 (correct answer)
Explanation: The scaling factor for a composite transformation T2T1T_2 \circ T_1 is the product of the determinants of the individual transformation matrices. First, calculate the determinants: det(A)=(3)(1)(1)(1)=2\det(A) = (3)(1) - (1)(1) = 2. det(B)=(2)(2)(0)(1)=4\det(B) = (2)(2) - (0)(-1) = 4. The total scaling factor is det(B)×det(A)=4×2=8\det(B) \times \det(A) = 4 \times 2 = 8. The final area is the original area multiplied by this scaling factor: 10×8=8010 \times 8 = 80.

Question 7

A parallelogram with an area of 4 square units is transformed by the matrix A=(k324)A = \begin{pmatrix} k & 3 \\ -2 & 4 \end{pmatrix}. The area of the transformed parallelogram is 40 square units. Which of the following is a possible value for kk?

  1. -4
  2. -0.5
  3. 1 (correct answer)
  4. 8.5
Explanation: The area scaling factor is the ratio of the final area to the original area, which is 40/4=1040 / 4 = 10. This factor must be equal to the absolute value of the determinant of the transformation matrix. det(A)=(k)(4)(3)(2)=4k+6\det(A) = (k)(4) - (3)(-2) = 4k + 6. We set det(A)=10|\det(A)| = 10, which gives 4k+6=10|4k + 6| = 10. This leads to two possible equations: 4k+6=104k + 6 = 10 or 4k+6=104k + 6 = -10. For the first equation, 4k=44k = 4, so k=1k = 1. For the second equation, 4k=164k = -16, so k=4k = -4. Of the choices provided, k=1k=1 is a possible value.

Question 8

A linear transformation T:R2R2T: \mathbb{R}^2 \to \mathbb{R}^2 is defined by a reflection across the line y=xy=x followed by a projection onto the x-axis. What is the area of the image of the unit square (with vertices at (0,0), (1,0), (0,1), and (1,1)) under this transformation?

  1. 0 (correct answer)
  2. 1
  3. 2\sqrt{2}
  4. 2
Explanation: We find the matrices for each transformation. The reflection across y=xy=x has matrix A1=(0110)A_1 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}. The projection onto the x-axis has matrix A2=(1000)A_2 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}. The composite transformation is T=A2A1=(1000)(0110)=(0100)T = 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}. The area of the transformed shape is the original area (which is 1 for the unit square) times det(T)|\det(T)|. Here, det(T)=(0)(0)(1)(0)=0\det(T) = (0)(0) - (1)(0) = 0. Therefore, the new area is 1×0=01 \times 0 = 0. The projection collapses the 2D square onto a 1D line segment, which has zero area.

Question 9

A linear transformation T:R3R3T: \mathbb{R}^3 \to \mathbb{R}^3 is represented by an invertible matrix AA. The transformation TT is found to triple the volume of any solid. What is the volume scaling factor of the transformation represented by the matrix 2A2A?

  1. 6
  2. 12
  3. 18
  4. 24 (correct answer)
Explanation: The statement that TT triples the volume of any solid means that its volume scaling factor is 3. Therefore, det(A)=3|\det(A)| = 3. We want the scaling factor for the transformation with matrix 2A2A, which is det(2A)|\det(2A)|. For an n×nn \times n matrix, the property of determinants states that det(cA)=cndet(A)\det(cA) = c^n \det(A). In this case, n=3n=3 (since T:R3R3T: \mathbb{R}^3 \to \mathbb{R}^3) and c=2c=2. So, det(2A)=23det(A)=8det(A)\det(2A) = 2^3 \det(A) = 8 \det(A). The new scaling factor is 8det(A)=8det(A)=8×3=24|8 \det(A)| = 8 |\det(A)| = 8 \times 3 = 24.

Question 10

A triangle PQRPQR has vertices P=(1,1)P=(1,1), Q=(5,2)Q=(5,2), and R=(2,4)R=(2,4). This triangle is transformed by the matrix M=(3121)M = \begin{pmatrix} 3 & -1 \\ 2 & 1 \end{pmatrix}. What is the area of the transformed triangle?

  1. 5.5
  2. 22
  3. 27.5 (correct answer)
  4. 55
Explanation: This is a two-step problem. First, find the area of the original triangle PQRPQR. We can form two vectors from vertex PP: PQ=QP=(4,1)\vec{PQ} = Q - P = (4, 1) and PR=RP=(1,3)\vec{PR} = R - P = (1, 3). The area of the parallelogram spanned by these vectors is det((4113))=4(3)1(1)=11=11|\det(\begin{pmatrix} 4 & 1 \\ 1 & 3 \end{pmatrix})| = |4(3) - 1(1)| = |11| = 11. The triangle's area is half of this, which is 11/2=5.511/2 = 5.5. Second, find the area scaling factor from matrix MM. det(M)=(3)(1)(1)(2)=3+2=5\det(M) = (3)(1) - (-1)(2) = 3 + 2 = 5. The area of the transformed triangle is the original area multiplied by the scaling factor: 5.5×5=27.55.5 \times 5 = 27.5.

Question 11

A parallelepiped in R3\mathbb{R}^3 with a volume of 12 is first scaled by a factor of 2 in the z-direction only. Then, it is sheared according to the transformation with matrix S=(102010001)S = \begin{pmatrix} 1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}. What is the volume of the final solid?

  1. 12
  2. 24 (correct answer)
  3. 36
  4. 48
Explanation: The total transformation is a composition of two transformations. The first is a scaling in the z-direction by 2, represented by matrix D=(100010002)D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix}. Its determinant is det(D)=2\det(D) = 2. The second is a shear, represented by matrix SS. Its determinant is det(S)=1((1)(1)(0)(0))=1\det(S) = 1((1)(1) - (0)(0)) = 1. The total volume scaling factor is the product of the individual determinants: det(S)×det(D)=1×2=2\det(S) \times \det(D) = 1 \times 2 = 2. The final volume is the original volume multiplied by this factor: 12×2=2412 \times 2 = 24.

Question 12

Consider the transformation that first reflects vectors across the line y=xy = x, then scales by factor 3 in the x-direction and factor 2 in the y-direction. If this composite transformation is applied to a triangle with area 5, what is the area of the transformed triangle?

  1. 1515
  2. 3030 (correct answer)
  3. 1010
  4. 66
Explanation: The reflection across y = x has matrix R = [[0,1],[1,0]] with det(R) = -1. The scaling has matrix S = [[3,0],[0,2]] with det(S) = 6. The composite transformation has matrix SR = [[3,0],[0,2]][[0,1],[1,0]] = [[0,3],[2,0]] with determinant det(SR) = (0)(0) - (3)(2) = -6. The area scales by |det(SR)| = |-6| = 6, so the final area is 5 × 6 = 30. Choice A incorrectly uses only the x-scaling factor. Choice C uses only the y-scaling factor. Choice D forgets the absolute value.

Question 13

A linear transformation TT maps the unit square [0,1]×[0,1][0,1] \times [0,1] to a parallelogram with vertices at (0,0)(0,0), (3,1)(3,1), (2,4)(2,4), and (5,5)(5,5). What is the area of the image of any region under transformation TT if the original region has area 7?

  1. 7070 (correct answer)
  2. 7777
  3. 8484
  4. 9191
Explanation: The unit square has vertices (0,0), (1,0), (0,1), (1,1). Under transformation T, these map to (0,0), (3,1), (2,4), (5,5) respectively. The transformation matrix has columns equal to T(e₁) = (3,1) and T(e₂) = (2,4), so T = [[3,2],[1,4]]. The determinant is det(T) = 3(4) - 2(1) = 12 - 2 = 10. Area scales by |det(T)| = 10, so a region with area 7 maps to area 7 × 10 = 70. Choice B adds the determinant to the area instead of multiplying. Choice C uses det = 12. Choice D incorrectly computes det = 13.

Question 14

A shear transformation in R2\mathbb{R}^2 is followed by a rotation of 90°90° counterclockwise. If the original parallelogram had area 12 and the final parallelogram has area 36, what is the determinant of the shear transformation matrix?

  1. 33
  2. 3-3
  3. ±3\pm 3 (correct answer)
  4. 11
Explanation: A 90° counterclockwise rotation has matrix R = [[0,-1],[1,0]] with det(R) = 1. Let the shear transformation have determinant d. The composite transformation has determinant det(R)·d = 1·d = d. Since area scales by |determinant|, we have |d|·12 = 36, so |d| = 3. Therefore d = ±3. Choice A and B each give only one possible value, missing that shears can have positive or negative determinants. Choice D incorrectly assumes shear transformations always have determinant 1 (which is only true for elementary shears along coordinate directions).

Question 15

A transformation TT in R3\mathbb{R}^3 has the property that it maps every unit cube to a parallelepiped of volume 6. If TT is applied to a tetrahedron with volume 20, what is the volume of the resulting solid?

  1. 2626
  2. 3.333.33
  3. 1414
  4. 120120 (correct answer)
Explanation: When you encounter transformation problems involving volume changes, you're dealing with how linear transformations scale areas and volumes through their determinant. The key insight is that if a transformation changes the volume of one shape by a certain factor, it changes the volume of ALL shapes by that same factor. Since the unit cube has volume 1 and transforms to a parallelepiped with volume 6, the transformation TT has a scaling factor of 6. This means det(T)=6|\det(T)| = 6. The determinant tells us how much any volume gets multiplied under the transformation. Therefore, when TT is applied to the tetrahedron with volume 20, the resulting volume is 20×6=12020 \times 6 = 120, which is answer D. Let's examine why the other options are wrong. Answer A (2626) appears to come from incorrectly adding the volumes: 20+6=2620 + 6 = 26, but transformations multiply volumes, they don't add them. Answer B (3.333.33) seems to result from dividing: 20÷63.3320 \div 6 ≈ 3.33, which reverses the scaling relationship. Answer C (1414) might come from subtracting: 206=1420 - 6 = 14, again misunderstanding how transformations work. Remember this key principle: linear transformations scale all volumes uniformly by the absolute value of their determinant. If you know how one shape's volume changes, you immediately know how any shape's volume will change under the same transformation. This makes volume scaling problems much more straightforward once you identify the scaling factor.

Question 16

The linear transformation represented by matrix B=(abcd)B = \begin{pmatrix} a & b \\ c & d \end{pmatrix} satisfies det(B)=12\det(B) = -12. If this transformation is applied three times consecutively to a triangle with area 4, what is the area of the final triangle?

  1. 17281728
  2. 6912-6912
  3. 51845184
  4. 69126912 (correct answer)
Explanation: When you encounter problems involving linear transformations and area changes, remember that the determinant controls how areas scale. The key insight is that when you apply a transformation represented by matrix BB, any area gets multiplied by det(B)|\det(B)| (the absolute value of the determinant). Since det(B)=12\det(B) = -12, each application of this transformation multiplies areas by det(B)=12=12|\det(B)| = |-12| = 12. The negative sign indicates the transformation reverses orientation (flips the figure), but doesn't affect area calculations. Applying the transformation three times consecutively means the final area scaling factor is 123=172812^3 = 1728. Starting with a triangle of area 4, the final area becomes 4×1728=69124 \times 1728 = 6912. Looking at the wrong answers: Choice A (17281728) represents just the scaling factor without multiplying by the original area of 4 - a common oversight when students forget to include the initial area. Choice B (6912-6912) incorrectly applies the negative sign from the determinant to the final area, but areas are always positive regardless of orientation changes. Choice C (51845184) appears to result from incorrectly calculating the scaling factor, possibly confusing how multiple applications compound. The correct answer is D (69126912). Study tip: For consecutive transformations, raise the absolute value of the determinant to the power equal to the number of applications, then multiply by the original area. The sign of the determinant only affects orientation, never the magnitude of area.

Question 17

Consider the transformation matrix A=(210032101)A = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 3 & 2 \\ 1 & 0 & 1 \end{pmatrix} . If a parallelepiped undergoes this transformation followed by the transformation represented by A1A^{-1}, what happens to its volume?

  1. The volume increases by a factor equal to det(A)\det(A)
  2. The volume decreases by a factor equal to det(A)\det(A)
  3. The volume remains unchanged (correct answer)
  4. The volume becomes zero
Explanation: When applying transformation A followed by A⁻¹, the composite transformation is A⁻¹∘A = I (the identity matrix). Since det(I) = 1, volume scales by |det(I)| = 1, meaning volume is unchanged. This is true regardless of the specific value of det(A). Choice A incorrectly suggests volume increases. Choice B incorrectly suggests volume decreases. Choice D would only be true if det(A) = 0, which would make A⁻¹ undefined.

Question 18

The unit circle in R2\mathbb{R}^2, which has area π\pi, is transformed by the matrix A=(4213)A = \begin{pmatrix} 4 & -2 \\ 1 & 3 \end{pmatrix}. What is the area of the resulting ellipse?

  1. π\pi
  2. 7π7\pi
  3. 10π10\pi
  4. 14π14\pi (correct answer)
Explanation: A linear transformation maps a circle to an ellipse. The area of the transformed shape is the original area multiplied by the absolute value of the determinant of the transformation matrix. The area of the unit circle is π\pi. The determinant of AA is det(A)=(4)(3)(2)(1)=12+2=14\det(A) = (4)(3) - (-2)(1) = 12 + 2 = 14. The area of the resulting ellipse is 14×π=14π|14| \times \pi = 14\pi.