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How to calculate the determinant of $$ A=(a_{i,j})_{n \times n}=\left( \begin{array}{ccccc} a&b&b& \cdots & b\\ b& a& b& \cdots& b\\ \vdots& \vdots& \vdots& \ddots&\cdots\\ b&b&b & \cdots&a \end{array} \right)? $$

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marked as duplicate by Martin Sleziak, Misha Lavrov, Jonas Meyer, user91500, Dario Apr 20 '17 at 8:30

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We note that the sum of the elements of each column is $a+(n-1)b$ for each column, hence $$\det A=(a+(n-1)b)\det\pmatrix{1&1&\dots& 1&1\\ b&a&\dots&b&b\\ \vdots&\ddots&\ddots&\vdots&\vdots\\ b&b&\dots&b&a}.$$ To each row of index $\geq 2$, we do $R_j\leftarrow bR_1$ to get $$\det A=(a+(n-1)b)\det\pmatrix{1&1&\dots& 1&1\\ 0&a-b&\dots&0&0\\ \vdots&\ddots&\ddots&\vdots&\vdots\\ 0&0&\dots&0&a-b}.$$ Finally, we obtain $$\det A=(a+(n-1)b)(a-b)^{n-1}.$$

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HINT 1: Your matrix $A$ is $$(a-b)I + b e e^T$$

Can you now compute the determinant?

Move your mouse over the gray area below for another hint.

HINT 2: Make use of the fact that $\det(\lambda A) = \lambda^n \det(A)$

Move your mouse over the gray area below for another hint.

HINT 3: $\det(I + \alpha e e^T) = (1+n \alpha)$ where $e$ is a column vector of ones.

Move your mouse over the gray area for the complete answer.

$$\det ((a-b)I + b e e^T) = (a-b)^n \det \left( I + \dfrac{b}{a-b} e e^T\right)$$ Hence, all we need is to find the determinant of $I + \alpha ee^T$, where $\alpha = \dfrac{b}{a-b}$ in our case. Note that $ee^T$ is a rank one matrix and its eigen values are $e^Te = n$ and $n-1$ zeros. If $\lambda$ is an eigen value of $I + \alpha ee^T$, then $$\det (I + \alpha ee^T - \lambda I) = \alpha^n \det \left(ee^T + \dfrac{(1-\lambda)}{\alpha}I \right) = 0$$ This means that $-\dfrac{(1-\lambda)}{\alpha}$ are the eigenvalues of $ee^T$. Hence, we get that $$-\dfrac{(1-\lambda)}{\alpha} = n \text{ or }0 \text{ ($n-1$ times)}.$$ Hence, we get that $$\lambda = 1 + n \alpha, 1 \text{ ($n-1$ times)}$$ Hence, $$\det ((a-b)I + b e e^T) = (a-b)^n \times \left( 1 + n \dfrac{b}{a-b} \right) = (a-b)^{n-1} (a+(n-1)b)$$

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