For example, if in a 5×5 square matrix all the entries on the main diagonal are -4, and everywhere else the entries are 1, the determinant is 0. Why is this?
[Math] If a square matrix has the same number on the main diagonal and all other entries are the same (but different) number, the determinant is 0. Why
determinantlinear algebra
Best Answer
In general, if $$A = \begin{bmatrix}a&a&\cdots&a\\a&a&\cdots&a\\ \vdots&\vdots&\ddots&\vdots \\ a&a&\cdots&a\end{bmatrix}$$ is an $n \times n$ matrix has all entries equal to some constant $a$, then $A$ has rank $1$. Hence, $A$ will have at most one non-zero eigenvalue and $n-1$ eigenvalues of $0$. By inspection, this eigenvalue is $na$ (one eigenvector corresponding to this eigenvalue is the vector of all $1$'s).
Then, by adding $(b-a)I_{n \times n}$, the eigenvalues all increase by $b-a$. So $$A' = A+(b-a)I_{n \times n} = \begin{bmatrix}b&a&\cdots&a\\a&b&\cdots&a\\ \vdots&\vdots&\ddots&\vdots \\ a&a&\cdots&b\end{bmatrix}$$ has one eigenvalue of $na+(b-a) = (n-1)a+b$ and $n-1$ eigenvalues of $0+(b-a) = b-a$.
Since the determinant of a matrix is the product of its eigenvalues, the determinant of $A'$ is simply $[(n-1)a+b](b-a)^{n-1}$.
For your case, we have a $5 \times 5$ matrix with $-4$'s on the diagonal, and $1$'s elsewhere. Hence, the determinant is $[4 \cdot 1 + (-4)](-4-1)^{4-1} = 0$.