I was given this $4$ by $4$ determinant
$$ \begin{vmatrix} x & a & a & a \\ a & x & a & a \\ a & a & x & a \\ a & a & a & x \end{vmatrix} = 0 $$
Clearly one of the answers is $x=a,$ how do i find the other answer?
I've tried splitting it into $4$ "$3$ by $3$" determinants but that didnt work well...
On
HINT: After subtracting, say, 4th row from first three you have enough zeros to compute the determinant easily.
On
Your matrix is the sum between a rank-$1$ matrix $M$ made by $a$s only and $(x-a)I$.
The spectrum of $M$ is $\{4a,0,0,0\}$ and the spectrum of $(x-a)I$ is $\{x-a,x-a,x-a,x-a\}$, so the spectrum of your matrix is $\{3a+x,(x-a),(x-a),(x-a)\}$ and the determinant (as the product of the eigenvalues) equals $(3a+x)(x-a)^3$. This is zero if $x=a$ or $x=-3a$.
On
Hint
There are many properties for solving determinant problems. One of them is that:
If you switch any column by a linear combination of any others, you don't change the determinant.
Using that, you can change the first column by the sum of all columns:
$$ \begin{vmatrix} 3a+x & a & a & a \\ 3a+x & x & a & a \\ 3a+x & a & x & a \\ 3a+x & a & a & x \end{vmatrix} = 0 = (3a+x)\begin{vmatrix} 1 & a & a & a \\ 1 & x & a & a \\ 1 & a & x & a \\ 1 & a & a & x \end{vmatrix}$$
Now you can use Laplace or Chió.
Can you finish?
On
Subtracting the 2nd row from the first, the 3rd from the second, the 4th from the third, tou obtain the determinant $$\begin{vmatrix} x-a&a-x&0&0\\0&x-a&a-x&0\\0&0&x-a&a-x\\ a&a&a&x \end{vmatrix}=(x-a)^3\begin{vmatrix} 1 & -1 & 0 & 0\\ 0 & 1 & -1 & 0\\ 0 & 0 & 1 & -1\\ a&a&a&x \end{vmatrix}$$ Then, adding successively the 1st column to the first, the 2nd to the third, the 3rd to the fourth, we get an upper triangular determinant: $$\begin{vmatrix} 1 & -1 & 0 & 0\\ 0 & 1 & -1 & 0\\ 0 & 0 & 1 & -1\\ a&a&a&x \end{vmatrix}=\begin{vmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ a&2a&3a&x+3a \end{vmatrix}=x+3a.$$
HINT
Add the $3$ last rows to the first one.