If $J$ is the $n×n$ matrix of all ones, and $A = (l−b)I +bJ$, then $\det(A) = (l − b)^{n−1}(l + (n − 1)b)$ I am stuck on how to prove this by induction. 

Let $J$ be the $n×n$ matrix of all ones, and let $A = (l−b)I +bJ$. Show that $$\det(A) = (l − b)^{n−1}(l + (n − 1)b).$$

I have shown that it holds for $n=2$, and I'm assuming that it holds for the $n=k$ case, $$(l-b)^{k-1}(a+(k-1)b)$$ but I'm having trouble proving that it holds for the $k+1$ case. Please help. 
 A: Hint:
$$A=(l−b)I +bJ=lI+b(J-I)=\begin{pmatrix}l&b&b&...&b\\b&l&b&...&b\\...&...&...&...&...\\...&...&...&...&...\\b&b&b&...&l\end{pmatrix}_{n\times n}$$
A: I think that it would be better to use $J_n$ for the $n \times n$ matrix of all ones, (and similarly $A_n, I_n$) so it is clear what the dimensions of the matrices are.
Proof by induction on $n$ that $\det(A_n)=(l-b)^n+nb(l-b)^{n-1}$:
When $n=1, 2$, this is easy to verify. We have $\det(A_1)=\det(l)=l=(l-b)^1+b(l-b)^0$ and $\det(A_2)=\det(\begin{array}{ccc} l & b \\ b & l \end{array})=l^2-b^2=(l-b)^2+2b(l-b)$.
Suppose that the statement holds for $n=k$. Consider $$A_{k+1}=(l-b)I_{k+1}+bJ_{k+1}=\left(\begin{array}{ccccc} l & b & b & \ldots & b \\ b & l & b & \ldots & b \\ \ldots & \ldots & \ldots &\ldots &\ldots \\ b & b & b & \ldots & l \end{array}\right)$$
Now subtracting the second row from the first gives 
\begin{align}
\det(A_{k+1})& =\det\left(\begin{array}{ccccc} l & b & b & \ldots & b \\ b & l & b & \ldots & b \\ \ldots & \ldots & \ldots &\ldots &\ldots \\ b & b & b & \ldots & l \end{array}\right) \\
& =\det\left(\begin{array}{ccccc} l-b & b-l & 0 & \ldots & 0 \\ b & l & b & \ldots & b \\ \ldots & \ldots & \ldots &\ldots &\ldots \\ b & b & b & \ldots & l \end{array}\right) \\
& =(l-b)\det(A_k)-(b-l)\det\left(\begin{array}{cccccc} b & b & b & b & \ldots & b \\ b & l & b & b & \ldots & b \\ b & b & l & b & \ldots & b \\ \ldots & \ldots & \ldots &\ldots &\ldots & \ldots \\ b & b & b & b & \ldots & l \end{array}\right)
\end{align}
Now taking the matrix in the last line above and subtracting the first row from all other rows gives an upper triangular matrix:
\begin{align}
\det\left(\begin{array}{cccccc} b & b & b & b & \ldots & b \\ b & l & b & b & \ldots & b \\ b & b & l & b & \ldots & b \\ \ldots & \ldots & \ldots &\ldots &\ldots & \ldots \\ b & b & b & b & \ldots & l \end{array}\right) & =\det\left(\begin{array}{cccccc} b & b & b & b & \ldots & b \\ 0 & l-b & 0 & 0 & \ldots & 0 \\ 0 & 0 & l-b & 0 & \ldots & 0 \\ \ldots & \ldots & \ldots &\ldots &\ldots & \ldots \\ 0 & 0 & 0 & 0 & \ldots & l-b \end{array}\right) \\
& =b(l-b)^k
\end{align}
Therefore we have (using the induction hypothesis) 
\begin{align}
\det(A_{k+1}) & =(l-b)\det(A_k)-(b-l)(b(l-b)^k) \\
& =(l-b)((l-b)^k+kb(l-b)^{k-1})+(l-b)(b(l-b)^{k-1}) \\
& =(l-b)^{k+1}+(k+1)b(l-b)^k
\end{align}
We are thus done by induction.
A: If you really have to do this by induction, see Ivan Loh's answer. I'll give an induction-free argument using diagonalization. 
The matrix $J$ has rank $1$. Therefore $0$ is an eigenvalue of multiplicity $n-1$. Hence the remaining eigenvalue is given by the trace: it is $n$. It follows that $J$ is diagonalizable. So there exists an invertible matrix $P$ such that 
$$
PJP^{-1}=\mbox{diagonal}(n,0,\ldots,0)
$$
whence
$$
PAP^{-1}=(l-b)I_n+bPJP^{-1}=\mbox{diagonal}((l-b)+nb,l-b,\ldots,l-b).
$$
Therefore
$$
\det A=\det PAP^{-1}=((l-b)+nb)(l-b)^{n-1}=(l-b)^n+nb(l-b)^{n-1}.
$$
So you have a typo and your expression should be

$$
\det A=(l-b)^{n-1}(l+(n-1)b).
$$

