Let A be a 3 times 3 matrix and let v vector 1 2 1 If Av ve

Let A be a 3 times 3 matrix and let v vector = [1 2 1]. If Av vector = 0 vector, what is det(-A)? Justify your answer.

Solution

det(A) = 0.

Suppose det(A) is non zero. Then we know that A is invertible and A^{-1} = 1/det(A) adj(A).
Now consider the equation Av = 0. Multiply A^{-1} to either side of this equation, then we have
A^{-1}A v = A^{-1}0. This implies that v = 0. But v is a nonzero vector.

Thus we see that our initial assumption that det(A) is non-zero leads to a contradiction. So it must be false. That is det(A) cannot be non-zero. So it is zero.