Determinants and inverses
1.[1p] What is the determinant of ?
What is the determinant of ?
CorrectNot quite: 10
2.[2p] Geometrically, the determinant of a two by two matrix is
Geometrically, the determinant of a two by two matrix is
Correct
The answer is: the signed area of the parallelogram the unit square becomes
The answer is: the signed area of the parallelogram the unit square becomes
The answer is: the signed area of the parallelogram the unit square becomes
3.[2p] A matrix with determinant zero has dependent columns and cannot be inverted.
A matrix with determinant zero has dependent columns and cannot be inverted.
Correct
The answer is: True
4.[2p] If and , what is ?
If and , what is ?
CorrectNot quite: 40
5.[2p] What does a negative determinant tell you?
What does a negative determinant tell you?
Correct
The answer is: The transformation reverses orientation, flipping the plane over
The answer is: The transformation reverses orientation, flipping the plane over
The answer is: The transformation reverses orientation, flipping the plane over
6.[3p] What is the determinant of the matrix with rows , and ?
What is the determinant of the matrix with rows , and ?
CorrectNot quite: 1
7.[3p] The inverse of is . What is ?
The inverse of is . What is ?
CorrectNot quite: -5
8.[3p] Which of these are equivalent to a square matrix being invertible?
Which of these are equivalent to a square matrix being invertible?
Select all that apply
Correct
Correct
Correct
The answer is: Its determinant is non-zero, Its columns are linearly independent, $A\mathbf{x} = \mathbf{0}$ has only the solution $\mathbf{x} = \mathbf{0}$
9.[3p] Why is a small determinant a poor test for near-singularity in floating point?
Why is a small determinant a poor test for near-singularity in floating point?
Correct
The answer is: Scaling a well-behaved matrix by $0.1$ multiplies its determinant by $0.1^n$, which underflows without any loss of invertibility
The answer is: Scaling a well-behaved matrix by $0.1$ multiplies its determinant by $0.1^n$, which underflows without any loss of invertibility
The answer is: Scaling a well-behaved matrix by $0.1$ multiplies its determinant by $0.1^n$, which underflows without any loss of invertibility