Geometry Quiz: Zero And Identity Matrices And Determinants

20 questions · exam conditions

0:00

Zero And Identity Matrices And DeterminantsQuestion 1 of 20

A linear transformation is represented by the matrix $A=\begin{pmatrix}0&-1\\1&0\end{pmatrix}.$ A unit square has vertices $(0,0)$ , $(1,0)$ , $(1,1)$ , and $(0,1)$ . Which claim about area scaling is correct?

The area becomes

0

because the matrix has a zero entry.

The area is multiplied by

-1

, so the square's area becomes negative.

The area is multiplied by

1

because

|\det(A)|=1

The area is multiplied by

2

because the matrix has two nonzero columns.

← Back to quizzes

Geometry Quiz

Geometry Quiz: Zero And Identity Matrices And Determinants

Practice Zero And Identity Matrices And Determinants in Geometry 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 Zero And Identity Matrices And Determinants, giving you a quick way to practice the rules, question types, and explanations that matter most for Geometry.

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

The area becomes $0$ because the matrix has a zero entry.
The area is multiplied by $-1$ , so the square's area becomes negative.
The area is multiplied by $1$ because $|\det(A)|=1$ . (correct answer)
The area is multiplied by $2$ because the matrix has two nonzero columns.

Explanation: Matrix interpretation geometrically includes rotations that preserve areas but change orientations based on

\det

sign. Identity preserves, zero collapses. Determinant absolute value scales areas, here

1

meaning no change in size. A rotates the unit square 90 degrees, keeping area

1

. Justified by

|\det(A)|=1

and matrix form for rotation. Misconception: area becomes negative from

\det

sign, but areas are positive, using absolute value. Read determinant as area change,

|\det|=1

preserving it.

Question 2

A linear transformation is represented by the matrix $Z=\begin{pmatrix}0&0\\0&0\end{pmatrix}.$ A point $V(4,-1)$ is transformed to $ZV$ . Which statement describes the geometric effect?

All points map to the origin, so $V$ goes to $(0,0)$ . (correct answer)
All points stay fixed, so $V$ remains $(4,-1)$ .
The point is reflected across the $y$ -axis.
The point's distance from the origin doubles.

Explanation: This question examines the zero matrix and its geometric interpretation. The zero matrix Z = [[0,0],[0,0]] maps every vector to the zero vector, effectively collapsing the entire plane to a single point at the origin. The determinant of the zero matrix is 0, which geometrically means all areas become 0—the transformation collapses 2D regions into lower dimensions. When we apply Z to the point V(4,-1), we get Z·V = [[0,0],[0,0]]·[[4],[-1]] = [[0],[0]], so V maps to the origin (0,0). Students might incorrectly think points stay fixed because they see zeros as "doing nothing," but zeros in a matrix mean "multiply by zero." The strategy is to recognize that det(Z) = 0 signals dimensional collapse—all points converge to one location.

Question 3

A matrix $A$ sends the basis vectors $\vec e_1=(1,0)$ and $\vec e_2=(0,1)$ to the points shown on the coordinate plane. Which statement describes the geometric effect of $A$ on area?

Matrix: $A=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$

It collapses all areas to zero because the determinant is $0$ .
It scales all areas by a factor of $2$ because lengths double.
It preserves area because $|\det(A)|=1$ . (correct answer)
It reverses area by making all areas negative in magnitude.

Explanation: This question examines area preservation in rotational transformations. The matrix

A = \begin{pmatrix}0&-1\\1&0\end{pmatrix}

represents a 90° counterclockwise rotation, sending

(1,0)

(0,1)

and

(0,1)

(-1,0)

. The determinant is

\det(A) = 0(0) - (-1)(1) = 1

, and since

|\det(A)| = 1

, areas are preserved exactly. This matrix rotates shapes without changing their size, demonstrating that rotations are area-preserving transformations. The positive determinant also tells us the transformation preserves orientation (no reflection occurs). A misconception might be thinking that because we see a zero in the matrix, areas become zero, but the determinant calculation shows otherwise. The key insight is that

|\det(A)| = 1

always means area preservation, whether through rotation, reflection, or their combination.

Question 4

A linear transformation $T$ in the plane is represented by the matrix $A=\begin{pmatrix}2&0\\0&\tfrac{1}{2}\end{pmatrix}.$ A rectangle in the coordinate plane has one corner at the origin and adjacent sides along the positive axes, with vertices $(0,0)$ , $(4,0)$ , $(4,2)$ , and $(0,2)$ . Which claim about area scaling is correct?

The area is multiplied by $2$ because the $x$ -direction is stretched by $2$ .
The area is multiplied by $1$ because $|\det(A)|=1$ . (correct answer)
The area is multiplied by $\tfrac{1}{2}$ because the $y$ -direction is shrunk by $\tfrac{1}{2}$ .
The area becomes $0$ because one direction is reduced.

Explanation: The skill of matrix interpretation allows us to see how transformations affect shapes like rectangles through scaling in different directions. Identity matrices preserve shapes, while zero matrices squash them to points. The determinant's absolute value geometrically measures the area scaling factor of the transformation. For this rectangle, the matrix A stretches the x-direction by

2

and shrinks y by

\frac{1}{2}

, but the overall area remains the same since

|\det(A)|=1

. This is because the scalings compensate each other, maintaining the original area of

8

. A misconception is thinking area multiplies only by the x-stretch of

2

, ignoring the y-shrink, but determinant combines them. To apply elsewhere, read the determinant as the net area change, here preserving it at

1

Question 5

A transformation is represented by the matrix $A=\begin{pmatrix}2&0\\0&3\end{pmatrix}$ . Which claim about area scaling is correct for a square region in the plane?

Areas are scaled by a factor of $5$ .
Areas are scaled by a factor of $6$ . (correct answer)
Lengths are scaled by a factor of $6$ in every direction.
Areas are scaled by a factor of $\sqrt{6}$ .

Explanation: This question tests understanding of how diagonal matrices scale areas through their determinant. The matrix

A = \begin{pmatrix}2&0\\0&3\end{pmatrix}

is a diagonal scaling matrix that stretches the x-direction by factor 2 and the y-direction by factor 3. The determinant is

\det(A) = 2 \times 3 = 6

, which tells us that areas are scaled by a factor of 6. For a unit square, the transformed shape becomes a rectangle with dimensions 2 × 3, giving area 6 times the original. This illustrates the fundamental principle that determinant measures area scaling factor. A common misconception is adding the diagonal entries (2 + 3 = 5) instead of multiplying them, or confusing length scaling with area scaling. The key insight is that determinant equals the product of eigenvalues for diagonal matrices, directly giving the area scaling factor.

Question 6

A transformation is given by $A=\begin{pmatrix}4&0\\0&0\end{pmatrix}.$ Which statement describes the geometric effect on the unit square (area 1)?

It preserves the square's area because one entry is 4.
It multiplies the area by $4$ because the $x$ -direction scales by 4.
It collapses the square to a line segment (area becomes 0). (correct answer)
It leaves the square unchanged because the determinant is not needed.

Explanation: This question examines a singular matrix that collapses one dimension. The matrix

A=\begin{pmatrix}4&0\\0&0\end{pmatrix}

stretches the

x

-direction by 4 but completely collapses the

y

-direction to 0. The determinant is

4 \times 0 - 0 \times 0 = 0

, indicating total area collapse. When applied to the unit square, all points get projected onto the

x

-axis: vertices like

(1,1)

map to

(4,0)

, creating a line segment from

(0,0)

(4,0)

. This line segment has zero area, confirming the determinant's prediction. Students might focus on the "4" and think area quadruples, but the zero in the second diagonal entry dominates—any factor times zero is zero. The geometric insight: if any eigenvalue is 0, the transformation collapses at least one dimension, making all areas zero.

Question 7

A linear transformation is represented by $G=\begin{pmatrix}1&1\\1&1\end{pmatrix}.$ The parallelogram formed by the vectors $\vec a=(1,0)$ and $\vec b=(0,1)$ (the unit square) is shown on the coordinate plane. Which conclusion follows from the determinant value?

Because $\det(G)=0$ , the unit square's area becomes $0$ under the transformation. (correct answer)
Because $\det(G)=0$ , the unit square's perimeter becomes $0$ under the transformation.
Because $\det(G)=0$ , the unit square keeps its area but changes orientation.
Because $\det(G)=0$ , the unit square's area is multiplied by $-1$ and flips.

Explanation: Geometric matrix interpretation reveals transformations that preserve or destroy dimensionality in shapes. Zero matrices collapse to points, identity preserves, but this matrix projects onto a line due to dependent rows. Determinant zero geometrically signals area collapse, as the image loses a dimension. For the unit square (parallelogram from basis vectors), G maps it to a line segment with zero area. This is justified by linearly dependent columns, flattening the shape. A distractor might think det=0 affects perimeter instead, but it's area that's nullified. To transfer, read determinant as area multiplier: zero means degeneration, helpful for any basis-defined parallelogram.

Question 8

A transformation $T$ is defined by $M=\begin{pmatrix}0&0\\0&0\end{pmatrix}.$ On the coordinate plane, segment $\overline{PQ}$ has endpoints $P(2,1)$ and $Q(4,3)$ . Which statement describes the geometric effect the matrix produces on $\overline{PQ}$ ?

It maps both endpoints to $(0,0)$ . (correct answer)
It keeps the segment the same length and location.
It reflects the segment across the $x$ -axis.
It doubles the segment's length without changing direction.

Explanation: The skill involves interpreting matrices, especially identity and zero types, for their geometric impact on segments in the plane. The identity matrix preserves all positions and lengths, whereas the zero matrix maps every point to the origin, eliminating distances. The determinant geometrically captures area scaling, where 0 denotes a transformation that squashes areas to nothing by reducing dimensionality. For the segment from

P(2,1)

Q(4,3)

, the zero matrix transforms both endpoints to

(0,0)

, collapsing the entire segment to a point. This is justified as matrix multiplication by zeros yields zero vectors for any input, merging distinct points. A common distractor is thinking it doubles length, perhaps misreading zeros as scaling factors. To transfer this, view the determinant as area change: 0 here explains the loss of any enclosed area.

Question 9

A linear transformation in the plane is represented by the matrix $A=\begin{pmatrix}1&0\\0&1\end{pmatrix}.$ Which statement describes the geometric effect of applying $A$ to the triangle with vertices $(0,0)$ , $(2,0)$ , and $(0,3)$ ?

It collapses the triangle to a single point at the origin.
It leaves the triangle unchanged in position and shape. (correct answer)
It doubles the area of the triangle but keeps its orientation.
It reflects the triangle across the $x$ -axis.

Explanation: This question tests understanding of the identity matrix and its geometric interpretation. The identity matrix

\begin{pmatrix}1&0\\0&1\end{pmatrix}

leaves every vector unchanged: it multiplies the

x

-component by 1 and the

y

-component by 1. The determinant of the identity matrix is

1 \times 1 - 0 \times 0 = 1

, which means areas are multiplied by 1 (unchanged). When applied to the triangle with vertices

(0,0)

(2,0)

, and

(0,3)

, each vertex maps to itself, so the triangle remains in the same position with the same shape. The correct answer recognizes that the identity transformation preserves both position and shape. A common misconception is thinking that a matrix with 1s must double something, but the identity matrix is the "do nothing" transformation—like multiplying by 1 in arithmetic.

Question 10

On the coordinate plane, the segment from $(-2,1)$ to $(2,1)$ is shown. A transformation $T$ is defined by the matrix $I=\begin{pmatrix}1&0\\0&1\end{pmatrix}.$ Which transformation does the matrix represent?

A reflection across the $y$ -axis.
A collapse of all points to the origin.
A $90^\circ$ rotation about the origin.
No change to the segment's location or length. (correct answer)

Explanation: Geometric interpretation of matrices helps understand transformations like identities that maintain segments' positions. Identity matrix leaves points fixed, zero sends to origin. Determinant of 1 preserves area and orientation without scaling. For this segment, I keeps it unchanged in location and length. Justification is that identity multiplication returns the same coordinates. Misconception: confusing with zero matrix causing collapse, but identity does nothing. Read determinant as area change, here 1 meaning no alteration.

Question 11

A linear transformation is represented by the matrix $A=\begin{pmatrix}0&0\\0&1\end{pmatrix}.$ A vector $\vec{v}$ is drawn from the origin to the point $(2,-3)$ on the coordinate plane. Which statement describes the geometric effect?

The vector is unchanged because one diagonal entry is $1$ .
The vector is reflected across the $x$ -axis because a diagonal entry is $0$ .
The vector collapses onto the $y$ -axis, so the $x$ -component becomes $0$ . (correct answer)
The vector's length is multiplied by $0$ , so it becomes the zero vector.

Explanation: This question tests understanding of projection onto coordinate axes. The matrix A = [[0,0],[0,1]] projects vectors onto the y-axis: it zeros out x-components while preserving y-components. For the vector from origin to (2,-3), we compute A·[[2],[-3]] = [[0,0],[0,1]]·[[2],[-3]] = [[0],[-3]], so the vector collapses from (2,-3) to (0,-3), lying entirely on the y-axis. The determinant is 0, confirming dimensional collapse. Students might think the vector becomes the zero vector because they see zeros in the matrix, but the preserved y-component prevents total collapse. The key insight is that this matrix selectively preserves one dimension while eliminating the other—it's a projection, not a complete annihilation.

Question 12

On the coordinate plane, consider the triangle with vertices $A(0,0)$ , $B(2,0)$ , and $C(0,1)$ . A transformation $T$ is defined by $M=\begin{pmatrix}1&0\\0&1\end{pmatrix}.$ Which statement describes the geometric effect?

The triangle's area becomes $0$ because the determinant is $0$ .
The triangle is unchanged; its shape and position stay the same. (correct answer)
The triangle is stretched so its area doubles.
The triangle is mapped to the origin because the matrix has zeros off the diagonal.

Explanation: This question applies the identity matrix to a specific triangle. The identity matrix

\begin{pmatrix}1&0\\0&1\end{pmatrix}

leaves every point unchanged, so the triangle with vertices A(0,0), B(2,0), and C(0,1) remains exactly where it is. Each vertex maps to itself: A stays at (0,0), B stays at (2,0), and C stays at (0,1), preserving the triangle's shape, size, and position. The determinant of 1 confirms that the area remains unchanged—the original triangle has area ½(base × height) = ½(2 × 1) = 1, and it stays area 1. This is the defining property of the identity transformation: nothing moves. Students might think that a matrix must "do something," but the identity matrix's job is precisely to do nothing. The transfer strategy is: identity matrix means complete preservation—no translation, rotation, reflection, or scaling occurs.

Question 13

On the coordinate plane, the vector $\vec v$ is drawn from $O(0,0)$ to $P(2,1)$ . A transformation $T$ is defined by $M=\begin{pmatrix}0&0\\0&0\end{pmatrix}.$ Which statement describes the geometric effect on $\vec v$ ?

$\vec v$ is unchanged because both coordinates stay the same.
$\vec v$ is reflected across the $x$ -axis.
$\vec v$ becomes the zero vector at the origin. (correct answer)
$\vec v$ is rotated $90^\circ$ because the matrix has zeros.

Explanation: This question applies the zero matrix to a specific vector. The zero matrix

\begin{pmatrix}0&0\\0&0\end{pmatrix}

transforms every vector to the zero vector, so the vector from O(0,0) to P(2,1) becomes the zero vector at the origin. Mathematically,

\begin{pmatrix}0&0\\0&0\end{pmatrix}\begin{pmatrix}2\\1\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}

. The geometric effect is that the arrow representing

\vec{v}

shrinks to a point at the origin—it loses all length and direction. This happens to every vector in the plane, not just this particular one. A common misconception is thinking that zeros in a matrix always mean rotation (since rotation matrices have some zero entries), but the all-zero matrix is unique in annihilating every vector. The transfer principle is: the zero matrix is the "destroy everything" transformation that sends all vectors to the origin.

Question 14

A linear transformation in the plane is represented by the matrix $A=\begin{pmatrix}1&0\\0&1\end{pmatrix}.$ Consider the triangle with vertices $P(0,0)$ , $Q(2,0)$ , and $R(0,3)$ . Which statement describes the geometric effect?

It reflects the triangle across the $x$ -axis.
It collapses the triangle to the origin.
It leaves the triangle unchanged. (correct answer)
It doubles the area of the triangle.

Explanation: This question tests understanding of the identity matrix and its geometric effect. The identity matrix I = [[1,0],[0,1]] has special properties: it leaves every vector unchanged when multiplied. The determinant of the identity matrix is 1, which geometrically means areas are preserved (multiplied by 1). When we apply this transformation to the triangle with vertices P(0,0), Q(2,0), and R(0,3), each vertex maps to itself, so the entire triangle remains unchanged. Some students might think the matrix reflects across the x-axis because they confuse it with [[1,0],[0,-1]], while others might think it doubles the area by adding the diagonal entries. The key insight is that the identity matrix is the "do nothing" transformation—read its determinant of 1 as preserving all areas and shapes.

Question 15

On the coordinate plane, the unit square has vertices $(0,0)$ , $(1,0)$ , $(1,1)$ , and $(0,1)$ . It is transformed by $M=\begin{pmatrix}0&0\\0&0\end{pmatrix}.$ Which statement describes the geometric effect the matrix produces on the square?

It becomes a larger square with the same center.
It becomes a line segment on the $x$ -axis.
It becomes a single point at the origin. (correct answer)
It stays the same because all entries are symmetric.

Explanation: The skill involves interpreting matrices, such as zero and identity, for their geometric transformations on shapes like the unit square. The identity matrix keeps everything intact, but the zero matrix uniformly maps all points to the origin, nullifying structures. Geometrically, the determinant indicates area scaling, with 0 signifying a complete loss of area through dimensional reduction. Applied to the unit square, the zero matrix transforms all vertices to (0,0), resulting in a single point. This is justified because every coordinate vector multiplied by zero yields the zero vector, collapsing the shape. A common distractor is assuming symmetry in entries prevents change, confusing it with identity. To transfer this, read the determinant as area change: 0 here predicts the degeneration to a point.

Question 16

A linear transformation sends the standard basis vectors $\mathbf e_1=(1,0)$ and $\mathbf e_2=(0,1)$ to the points shown: $\mathbf e_1$ maps to $(1,0)$ and $\mathbf e_2$ maps to $(0,1)$ . The matrix given is $I=\begin{pmatrix}1&0\\0&1\end{pmatrix}.$ Which reasoning correctly interprets the matrix?

It is the zero transformation because it sends basis vectors to points on the axes.
It is the identity transformation because it leaves $\mathbf e_1$ and $\mathbf e_2$ unchanged. (correct answer)
It is a reflection because the determinant is negative.
It is a rotation because the determinant equals $1$ .

Explanation: Interpreting matrices geometrically reveals their effects on basis vectors, which define the transformation for the whole space. The identity matrix maps basis vectors to themselves, preserving the coordinate grid, while the zero matrix sends them to origin, collapsing everything. A determinant of 1 signifies area preservation with possible rotation or shear, but here it's exactly the identity's det=1. The given mapping of e1 to (1,0) and e2 to (0,1) directly matches the identity matrix I, leaving the basis unchanged. This is justified as the columns of I are precisely the standard basis vectors. A misconception might label it a rotation due to det=1, but no angle change occurs here. To transfer, view determinant as area scaling, helping identify if transformations preserve volumes in higher dimensions too.

Question 17

A linear transformation $T$ is defined by the matrix $Z=\begin{pmatrix}0&0\\0&0\end{pmatrix}.$ On the coordinate plane, consider the triangle with vertices $(1,1)$ , $(3,1)$ , and $(2,4)$ . Which statement describes the geometric effect of applying $T$ to this triangle?

It sends every vertex to the origin, so the triangle collapses to a single point. (correct answer)
It leaves the triangle unchanged because the matrix does nothing.
It keeps the base length the same but makes the height zero, forming a line segment.
It rotates the triangle $90^\circ$ counterclockwise about the origin.

Explanation: Matrix interpretation in geometry helps us visualize how linear transformations alter figures like triangles by mapping their vertices. The zero matrix collapses all points to the origin, whereas the identity matrix keeps everything fixed. Geometrically, the determinant tells us about area scaling; a zero determinant means the transformation flattens shapes to lower dimensions, resulting in zero area. Applying the zero matrix Z to the triangle sends each vertex to (0,0), collapsing it to a point. This occurs because every vector is multiplied by zero, erasing its magnitude and direction. A distractor might suggest it leaves the triangle unchanged, confusing zero with identity, but zero indeed maps to origin. For broader application, interpret the determinant as the area change factor, here

det(Z)=0

indicating complete collapse.

Question 18

A linear transformation $T$ is represented by the matrix $A=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}.$ On the coordinate plane, the triangle with vertices $(0,0)$ , $(2,0)$ , and $(0,1)$ is shown. Which statement describes the geometric effect of applying $T$ to the triangle?

It rotates the triangle $180^\circ$ about the origin and keeps the area the same. (correct answer)
It collapses the triangle to a line because the determinant is negative.
It leaves the triangle unchanged because the determinant equals $1$ .
It scales the triangle's area by a factor of $-2$ because both entries are $-1$ .

Explanation: Geometric matrix interpretation reveals rotations like 180 degrees that invert positions but preserve areas. Identity fixes, zero collapses. Positive determinant of 1 means area preserved with orientation. A rotates the triangle 180 degrees, mapping to opposite quadrant with same area. Justified by A scaling by -1 in both directions, equivalent to 180 rotation, |det|=1. Distractor: negative det collapses, but det=1 is positive, no collapse. Interpret determinant as area change, here 1 keeping it same.

Question 19

A linear transformation $T$ is represented by the matrix $A=\begin{pmatrix}1&0\0&-1\end{pmatrix}.$ Consider the vector $\vec v$ from the origin to the point $(2,3)$ shown on the coordinate plane. Which statement describes the geometric effect of applying $T$ to $\vec v$ ?

It reflects $\vec v$ across the $x$ -axis, mapping $(2,3)$ to $(2,-3)$ . (correct answer)
It reflects $\vec v$ across the line $y=x$ , swapping coordinates to $(3,2)$ .
It leaves $\vec v$ unchanged because $|\det(A)|=1$ .
It collapses $\vec v$ to the origin because the determinant is negative.

Explanation: Geometric matrix interpretation reveals how vectors are transformed, such as through reflections or scalings. The identity matrix fixes vectors, while zero maps them to origin. Determinants indicate area scaling and orientation; a negative determinant often signals a reflection that reverses orientation. Applying A to vector

v=(2,3)

reflects it over the x-axis to

(2,-3)

, changing the y-component's sign. This is justified by the matrix's structure, which preserves x and flips y, with

\det(A)=-1

confirming the reflection. A common distractor is assuming collapse to origin due to negative det, but negative det means orientation flip, not collapse. Transfer by viewing determinant as area scaler, here

|\det(A)|=1

preserving magnitude but flipping sign.

Question 20

A transformation in the plane is represented by $B=\begin{pmatrix}1&2\\2&4\end{pmatrix}.$ The unit square is drawn on the coordinate plane. Which conclusion follows from the determinant value?

Since $\det(B)=0$ , the square's area collapses to $0$ under the transformation. (correct answer)
Since $\det(B)=0$ , the square's side lengths are unchanged but the area flips sign.
Since $\det(B)=0$ , the square's area is multiplied by $0$ but each point keeps its direction from the origin.
Since $\det(B)=0$ , the square's area is multiplied by $1$ because the matrix has ones in it.

Explanation: Geometric matrix interpretation helps us see how transformations affect shapes, such as preserving or altering their dimensions. The zero matrix collapses all to the origin, contrasting with the identity which changes nothing, but singular matrices like this one project onto lower dimensions.

\det = 0

geometrically means the transformation flattens areas to zero, indicating non-invertibility and loss of full plane coverage. For the unit square, this matrix with

\det=0

maps it to a line segment, effectively making its area zero. This follows because the columns are linearly dependent, causing the image to degenerate. A common distractor is thinking

\det=0

keeps area but flips sign, but actually, it eliminates area entirely. Transfer by reading determinant as area change:

0

implies collapse, useful for any parallelogram or shape.