Quiz 10: Determinants

The determinant of a square matrix is always defined, so (4) is wrong.
Either:

det(A)	=
	= 1- (-3) + (-4)
	= 1(1 × 5 - 0 × 3) - (-3)(2 × 5 - 0 × 3) + (-4)(2 × (-2) - 1 × 3)
	= 1 × 5 - (-3) × 10 + (-4) × (-7) = 63,

or, using elementary row operations R₂ := R₂ - 2R₁; R₃ := R₃ - 3R₁ R₃ := R₃ -R₂ (all of which give matrices with the same determinant) to reduce to (upper) triangular form: Hence the correct answer is (2).

The determinant of a non-square matrix is undefined, so (4) is correct.

The determinant of a square matrix is always defined, so (4) is wrong.
A is not triangular (which means all zero on one side of the main top-left to bottom-right diagonal, not the top-right to bottom-left diagonal), so det(A) is not necessarily the product of the “wrong” diagonal elements, i.e., not necessarily 6 ⋅ 2 ⋅ 6 ⋅ 9 ⋅ 2 ⋅ 3 = 3888.
The three elementary row operations R₁ ↔ R₆; R₂ ↔ R₅; R₃ ↔ R₄; change the sign of the determinant three times, i.e., So the correct answer is (5).

It is not invertible since det A = 0.

	= -
	= -
	= -
	= -(+13 + 22 - 42 + 11 - 14 + 10)
	= 0.

Try expanding the determinant along the third row.

det(A)	=
	= a- b + c
	= a(ei - fh) - b(di - fg) + c(dh - eg)
	= aei - afh + bdi - bgf + cdh - ceg.

Hence the correct answer is (4).

The determinant of A has the opposite sign of the determinant of the matrix

since a row swap has been applied. The determinant of a triangular matrix is the product of the diagonal entries hence detA = -1 × 1 × 6 = -6.

Call the new matrix A′. Now detA′ = - detA since A′ has had a row swap to change the sign and A′ has had a row divided by 2. Now detA′ = -6 since it is triangular. Thus detA = -2 × 6 = -12.

Since det(A) = 0, A is not invertible, so we cannot solve the system by left multiplying by A^-1 both sides of Ax^T = b^T (x = [x y z], b = [5 1 4]) to get a unique solution x^T = A^-1b^T.
In fact, if we solve by reducing the augmented matrix [A∣b^T], the reduced form will have third row [0 0 0 ∣*]. Not all variables can correspond to a leading 1, so there can be no unique solution, so (3) cannot possibly be true.
If *0, (1) will be true. If * = 0, (2) will be true.
However we do not have enough information to decide if * = 0 or not. So neither (1) nor (2) is necessarily true. Hence the correct answer is (4)!!

1 = detI = det(AA^-1) = detAdetA^-1.

1 = detI = det(AA^-1) = detAdetA^-1.

This is correct because detA^T = detA.

As = 2³, this is incorrect.