Question

Show that det(A) = 0 without directly evaluating the determinant. A = [-4 1 1 1 1; 1 -4 1 1 1; 1 1 -4 1 1; 1 1 1 -4 1; 1 1 1 1 -4]

Answer 1

Alright, this might be a little complicated, so I apologize in advance.
So if the det(A) = 0 that means the matrix is not invertible. If the matrix isnt invertible, that means that there should be some vector 'x' that isnt 0 and Ax=0 (it would be in the null space of A) Just from inspection, that vector is x= (1,1,1,1,1) Multiplying the matrix A with x just adds each row up, which will give 0 for every row, making Ax = 0, which means A isnt invertible, which means the det(A) must be 0.

Answer 2

or you could interpret this as a system of equations:

-4x + y + z + w = 1
x - 4y + z + w = 1
x + y - 4z + w = 1
x + y + z - 4w = 1
x + y + z + w = -4

this is a system with five four unknowns and five equations. therefore a unique solution is not possible, so det(A) = 0

Answer 3

four unknowns*