I was asked to evaluate the determinant of the

Question

I was asked to evaluate the determinant of the $n$x$n$ matrix $$ A= \begin{bmatrix} x & 2 & 2 & \cdots & 2 \\ 2 & x & 2 & \cdots & 2 \\ 2 & 2 & x & \cdots & 2 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 2 & 2 & 2 & \cdots & x \\ \end{bmatrix} $$ I tried starting from a 1x1 matrix to 5x5 matrix, and saw the pattern $$\det{A}=(x-2)^{n-1}(x+2(n-1)).$$ Now I need to prove that this is the case for any nxn matrix of type A, so I tried using induction.The base case (n=1) holds. $$A= \begin{bmatrix} x \end{bmatrix} $$ And clearly $\det{A}=x$ using both the definition of a determinant and the explicit formula above. So now assume it holds for any $n$, and test for $n+1$. $$\det{A}=(x-2)^n(x+2n).$$ What do I do now?

Answer 1

First subtract the last row from the first $n-1$ rows. Then add first $n-1$ columns to the last one. \begin{align} \det A &= \begin{vmatrix} x & 2 & 2 & \cdots & 2 \\ 2 & x & 2 & \cdots & 2 \\ 2 & 2 & x & \cdots & 2 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 2 & 2 & 2 & \cdots & x \\ \end{vmatrix} \\ &= \begin{vmatrix} x-2 & 0 & 0 & \cdots & 2-x \\ 0 & x-2 & 0 & \cdots & 2-x \\ 0 & 0 & x-2 & \cdots & 2-x \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 2 & 2 & 2 & \cdots & x \\ \end{vmatrix} \\ &= \begin{vmatrix} x-2 & 0 & 0 & \cdots & 0 \\ 0 & x-2 & 0 & \cdots & 0 \\ 0 & 0 & x-2 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 2 & 2 & 2 & \cdots & x + 2(n-1) \end{vmatrix}\\ &= (x-2)^{n-1}(x-2(n-1)) \end{align} The resulting determinant is lower-triangular so it is equal to the product of diagonal elements.

Answer 2

We add all rows to the first one, then factor $x+2(n-1)$ from this first row, than use it (doubled) to eliminate in the other rows. With this strategy we are done quickly: $$ \begin{aligned} \det A &= \begin{vmatrix} x & 2 & 2 & \cdots & 2 \\ 2 & x & 2 & \cdots & 2 \\ 2 & 2 & x & \cdots & 2 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 2 & 2 & 2 & \cdots & x \\ \end{vmatrix} \\ &= \begin{vmatrix} x+2(n-1) & x+2(n-1) & x+2(n-1) & \cdots & x+2(n-1) \\ 2 & x & 2 & \cdots & 2 \\ 2 & 2 & x & \cdots & 2 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 2 & 2 & 2 & \cdots & x \\ \end{vmatrix} \\ &= (x+2(n-1)) \begin{vmatrix} 1 & 1 & 1 & \cdots & 1 \\ 2 & x & 2 & \cdots & 2 \\ 2 & 2 & x & \cdots & 2 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 2 & 2 & 2 & \cdots & x \\ \end{vmatrix} \\ &= (x+2(n-1)) \begin{vmatrix} 1 & 1 & 1 & \cdots & 1 \\ 2-2 & x-2 & 2-2 & \cdots & 2-2 \\ 2-2 & 2-2 & x-2 & \cdots & 2-2 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 2-2 & 2-2 & 2-2 & \cdots & x-2 \\ \end{vmatrix} \\ &= (x+2(n-1)) \begin{vmatrix} 1 & 1 & 1 & \cdots & 1 \\ 0 & x-2 & 0 & \cdots & 0 \\ 0 & 0 & x-2 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & x-2 \\ \end{vmatrix} \\ &=(x+2(n-1))(x-2)^{n-1}\ . \end{aligned} $$

Answer 3

One way to proceed is to find the eigenvalues of $A$.

The matrix $$ A - (x-2)I = \begin{bmatrix}2 & 2 & \cdots & 2 \\ 2 & 2 & \cdots & 2 \\ \vdots & \vdots & \ddots & \vdots \\ 2 & 2 & \cdots & 2\end{bmatrix}$$ has rank $1$, which tells us that $x-2$ is an eigenvalue with multiplicity $n-1$. To find the remaining eigenvalue, there are two approaches: we can

• spot the eigenvector $\mathbf v = (1,1,\dots,1)^{\mathsf T}$, and notice that $A\mathbf v = (x + 2(n-1))\mathbf v$, or
• use the fact that $\operatorname{tr}(A) = nx$ is the sum of all the eigenvalues, and therefore the remaining eigenvalue is $nx - (n-1)(x-2) = x + 2(n-1)$.

Either way, now that we know all the eigenvalues, we conclude that $\det(A)$ is the product of all of them: $$ \det(A) = (x-2)^{n-1} (x + 2(n-1)). $$