Geometry Quiz: Zero And Identity Matrices And Determinants

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

1. All areas become $0$ because the determinant is negative.
2. All lengths are multiplied by $-1$ because the determinant is $-1$.
3. Areas keep the same size, but orientation is reversed.
4. The transformation is the identity because $|\det(M)|=1$.

1. The area becomes negative, so the square disappears.
2. The area is multiplied by $\left|\det(D)\right|=1$, so the area stays the same.
3. The area is multiplied by $-1$, so the area doubles in size.
4. The area is multiplied by $2$ because one axis is reflected.

1. It collapses the plane to the origin because all off-diagonal entries are $0$.
2. It leaves every point fixed because it is the identity matrix.
3. It reverses orientation because the determinant is $-1$.
4. It changes area by a factor of $2$ because there are two ones on the diagonal.

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

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

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

1. It leaves every vector unchanged.
2. It sends every vector to the origin.
3. It rotates every vector $180^\circ$ about the origin.
4. It doubles the length of every vector without changing direction.

1. Areas are scaled by a factor of $5$.
2. Areas are scaled by a factor of $6$.
3. Lengths are scaled by a factor of $6$ in every direction.
4. Areas are scaled by a factor of $\sqrt{6}$.

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

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

1. Every vector is sent to the origin.
2. All areas become zero after the transformation.
3. The transformation is the same as the identity on all points.
4. The determinant is $0$, so the transformation is not invertible.

1. It maps both endpoints to $(0,0)$.
2. It keeps the segment the same length and location.
3. It reflects the segment across the $x$-axis.
4. It doubles the segment’s length without changing direction.

1. It preserves the unit square exactly.
2. It collapses the square into a line segment, so the image has zero area.
3. It scales the square’s area by a factor of $3$.
4. It rotates the square $180^\circ$ about the origin.

1. Triangle areas are multiplied by $0$ because $\det(M)=0$.
2. Triangle areas are multiplied by $5$ because $1+4=5$.
3. Triangle areas are multiplied by $6$ because $1\cdot4+2\cdot2=8$.
4. Triangle areas are unchanged because the entries are integers.

1. Areas are multiplied by $3$ because each coordinate is multiplied by $3$.
2. Areas are multiplied by $6$ because $3+3=6$.
3. Areas are multiplied by $9$ because $\left|\det(A)\right|=9$.
4. Areas are unchanged because the determinant is positive.

1. The area is multiplied by $1$ because one factor is negative.
2. The area is multiplied by $-6$, so the region has negative area.
3. The area is multiplied by $6$ because $|\det(A)|=6$.
4. The area is multiplied by $5$ because $3+2=5$.

1. It collapses the triangle to a single point at the origin.
2. It leaves the triangle unchanged in position and shape.
3. It doubles the area of the triangle but keeps its orientation.
4. It reflects the triangle across the $x$-axis.

1. A reflection across the $y$-axis.
2. A collapse of all points to the origin.
3. A $90^\circ$ rotation about the origin.
4. No change to the segment’s location or length.

1. The vector is unchanged because one diagonal entry is $1$.
2. The vector is reflected across the $x$-axis because a diagonal entry is $0$.
3. The vector collapses onto the $y$-axis, so the $x$-component becomes $0$.
4. The vector’s length is multiplied by $0$, so it becomes the zero vector.