Determine if the statement is true or false and justify your

Determine if the statement is true or false, and justify your answer.

If A is an n × n matrix with all entries equal to 1, then det(A) = n

True for all n > 1. In these cases, the columns will be equal so linearly dependent, and hence det(A) = n.

True for all n > 1. In these cases, the columns will be equal so linearly independent, and hence det(A) = n.

False for all n > 1. In these cases, the columns will be equal so linearly dependent, and hence det(A) = n.

False for all n > 1. In these cases, the columns will be equal so linearly independent, and hence det(A) = 0.

False for all n > 1. In these cases, the columns will be equal so linearly dependent, and hence det(A) = 0.

Determine if the statement is true or false, and justify your answer.

Suppose that A is a 4 × 4 matrix and that B is the matrix obtained by multiplying the third column of A by 5. Then det(B) = 5det(A).

True. B^T is obtained by multiplying the third row of A^T by 5, sodet(B) = det(B^T) = 5det(A^T) = (5det(A)) = 5det(A).

True. B^T is obtained by multiplying the third row of A^T by 5, so det(B) = det(B^T) = 5det(A^T) = 5det(A).   

False. B^T is obtained by multiplying the third row of A^T by 5, sodet(B) = det(B^T) = (1/5)det(A^T) = ((1/5)det(A)) = (1/5)det(A).

False. B^T is obtained by multiplying the third row of A^T by 5, so det(B) = det(B^T) = (det(A^T))⁵ = (det(A))⁵.

False. B^T is obtained by multiplying the third row of A^T by 5, so det(B) = det(B^T) = (1/5)det(A^T) = (1/5)det(A).

Determine if the statement is true or false, and justify your answer.

Suppose A, B, and S are

n × n

matrices, and that S is invertible. If

B = S¹AS,

then

det(A) = det(B).

True, since det(B) = det(S¹AS) = det(SS¹A) = det(A).

True, since det(B) = det(S¹AS) = det(S¹) det(A) det(S) =

det(A) det(S) = det(A).    

False, since det(B) = det(S¹AS) = det(SAS) = det(S) det(A) det(S) = det(A) det(S²) = det(A).

False, since det(B) = det(S¹AS) = det(S) det(A) det(S) = det(SAS).

False, since det(B) = det(S¹AS) = det(SAS)¹ =

Solution

(1) False for all n > 1. In these cases, the columns will be equal so linearly dependent, and hence det(A) = 0.

(2) True. BT is obtained by multiplying the third row of AT by 5, so det(B) = det(BT) = 5det(AT) = 5det(A).   

(3) True, since det(B) = det(S^-1AS) = det(S^-1) det(A) det(S) = (1/det(S)) * det(A) * det(S) = det(A)