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?

Question

For example, if in a 5x5 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?

Answer 1

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 $[4 \cdot 1 + (-4)](-4-1)^{4-1} = 0$.

Answer 2

No in general: a unit matrix $I$ has $1$ on the main diagonal, and $0$ outside, and is of full rank, with $\det I = 1$.

And with your example, the $2\times 2$ matrix:

$$ \begin{array}{r} -4 & 1 \\1 & -4 \end{array}$$

has determinant $(-4)^2-1^2=15$.

Your example with a $5\times 5$ matrix with $-4$ on the diagonal and $1$ elsewhere is quite specific: if you add all rows or columns, the resulting vector is null, thus your matrix is rank-deficient, and with null determinant. This will happen (with your constant + diagonal matrix structure) if your diagonal element equals the opposite of the sum of off-diagonal elements.