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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...

Bernard
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Lostdawn
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  • The linearly independent vectors $(1,-1,0,0),(1,0,-1,0),(1,0,0,-1)$ are eigenvectors of the eigenvalue $x-a$ and the vector $(1,1,1,1)$ is an eigenvector of the eigenvalue $x+3a$. Therefore, the determinant is $(x-a)^3(x+3a)$. – MoonLightSyzygy Jan 02 '20 at 17:05
  • I never heard of that method of solving a determinant, cant to explain furthermore ? – Lostdawn Jan 02 '20 at 17:20
  • If you find out the eigenvalues of a matrix together with the dimensions of their corresponding invariant subspaces (which in this example turn out to be the corresponding eigenspaces), then you know the algebraic multiplicity of the eigenvalues. The determinant is the product of the eigenvalues taking each with exponent equal to their algebraic multiplicity. – MoonLightSyzygy Jan 02 '20 at 18:26

5 Answers5

2

HINT

Add the $3$ last rows to the first one.

1123581321
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HINT: After subtracting, say, 4th row from first three you have enough zeros to compute the determinant easily.

Przemysław Scherwentke
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  • I was thinking about subtracting the rows but it didnt seem so elegant due the fractions I might get – Lostdawn Jan 02 '20 at 17:14
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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 $x=-3a$.

Jack D'Aurizio
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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: $$\begin{vmatrix} 1 & -1 & 0 & 0\\ 0 & 1 & -1 & 0\\ 0 & 0 & 1 & -1\\ a&a&a&x \end{vmatrix}=\begin{vmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ a&2a&3a&x+3a \end{vmatrix}=x+3a.$$

Bernard
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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?

Bernard
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Arnaldo
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