Given $a, b ∈\Bbb R$, evaluate Given $a, b ∈\Bbb R$, evaluate $\det(aI_n +bJ_n)$, where $J_n ∈ M_n(\Bbb R)$ has every entry equal to $1$.
My try is $\det(aI_n +bJ_n)= a^n+ na^{n-1}b$ for $n > 1$, Now. How do I test it by Mathematical induction?
I have that doubt, who can answer it please.
 A: There is a really easy way to do this by using eigenvalues.  However you asked for induction so here we go.  Call your determinant $D_n$.  First, subtract the final column of the matrix from the first column, then add the first row to the last row: this gives
$$D_n=\det\pmatrix{a&b&\cdots&b&b\cr 0&a+b&\cdots&b&b\cr \vdots&\vdots&\ddots&\vdots&\vdots\cr 0&b&\cdots&a+b&b\cr 0&2b&\cdots&2b&a+2b\cr}\ .$$
Expanding along the first column,
$$D_n=a\det\pmatrix{a+b&\cdots&b&b\cr \vdots&\ddots&\vdots&\vdots\cr b&\cdots&a+b&b\cr 2b&\cdots&2b&a+2b\cr}\ .$$
Now we'll use the following very important fact.  Take matrices $A$ and $B$ which differ in one row only, and let $C$ be the matrix you get by adding these rows and leaving all the others the same.  For example
$$A=\pmatrix{p&q&r\cr s&t&u\cr v_1&w_1&x_1\cr}\ ,\quad
  B=\pmatrix{p&q&r\cr s&t&u\cr v_2&w_2&x_2\cr}\ ,\quad
  C=\pmatrix{p&q&r\cr s&t&u\cr v_1+v_2&w_1+w_2&x_1+x_2\cr}.$$
Then $\det C=\det A+\det B$.  Applied to the above this gives
$$D_n=a\det\pmatrix{a+b&\cdots&b&b\cr \vdots&\ddots&\vdots&\vdots\cr b&\cdots&a+b&b\cr 2b&\cdots&2b&2a+2b\cr}+a\det\pmatrix{a+b&\cdots&b&b\cr \vdots&\ddots&\vdots&\vdots\cr b&\cdots&a+b&b\cr 0&\cdots&0&-a\cr}\ .$$
Remember these are matrices of size $n-1$.  Taking out a factor of $2$ in the first one and expanding along the last row in the second,
$$D_n=2aD_{n-1}-a^2D_{n-2}\ .$$
Using your formula for $n-1$ and $n-2$, we get
$$D_n=2a(a^{n-1}+(n-1)a^{n-2}b)-a^2(a^{n-2}+(n-2)a^{n-3}b)=a^n+na^{n-1}b\ ,$$
so the formula also works for $D_n$.  Check that it is true for $D_1$ and $D_2$, and you're done.

The eigenvalue method: your matrix has an eigenvalue $a$ with geometric multiplicity $n-1$; the other eigenvalue is the trace minus $(n-1)$ times $a$, which is $a+nb$; the determinant is the product of the eigenvalues,
$$D_n=a^{n-1}(a+nb)\ .$$
If you haven't studied eigenvalues yet I hope this will encourage you to learn them!!
