Why isn't the determinant of a squared matrix negative?

Question

When a square matrix is squared, then why isn't its determinant negative?

For example, in the $2 \times 2$ matrix

$$\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}^2 = \begin{pmatrix} 7 & 10 \\ 15 & 22 \end{pmatrix}$$

$$\begin{vmatrix} 7 & 10 \\ 15 & 22 \\ \end{vmatrix} = 154 - 150 = +4$$ i.e in 2*2 matrix, the determinant of the squared matrix, the product of the left diagonal always become greater than the product of the right diagonal.

I just can't figured out, why?

Thanks.

[edit]: If the determinant is not zero or it has a solution.

Answer 1

That is because the determinant of a matrix product of square matrices equals the product of their determinants. $$\text{det}(AB)=\text{det}(A)\text{det}(B).$$ More on this can be found here. So the determinant of $A^2$ becomes $(\text{det}(A))^2,$ which is of course non-negative.

Answer 2

Because the determinant function is multiplicative (i.e. $\operatorname{det}(AB)=\operatorname{det}(A)\operatorname{det}(B))$ and the square of any real number is nonnegative.