659 views

Let $a$ be a non-zero real number. Define

$$f(x) = \begin{vmatrix} x & a & a & a \\ a & x & a & a \\ a & a & x & a \\ a & a & a & x \end{vmatrix}$$ for $x \in \mathbb{R}$. Then, the number of distinct real roots of $f(x) =0$ is

1. $1$
2. $2$
3. $3$
4. $4$

a?

No. Its B.

Here we need to find the determinant of this matrix.

notice here it is block matrix.

This is a $2×2$ block matrix where the first and last and the second and third elements are the same. So, applying the formula for determinant of a block matrix as given here

When A = D and B = C, the blocks are square matrices of the same order and the following formula holds (even if A and B do not commute)

$det \begin{pmatrix} A & B\\ C& D \end{pmatrix} = det (A-B) \ det(A+B)$

https://en.wikipedia.org/wiki/Determinant#Block_matrices

here A =  $\begin{pmatrix} x & a\\ a& x \end{pmatrix}$

and B =  $\begin{pmatrix} a & a\\ a& a \end{pmatrix}$

using above formula determinant is

$f(x) = \begin{vmatrix} x-a & 0\\ 0 & x-a \end{vmatrix}\times \begin{vmatrix} x+a & 2a \\ 2a & x+a \end{vmatrix}$$= (x-a)^2[(x+a)^2 - 4a^2] = 0$

$\Rightarrow (x-a)^2(x+a)^2 = 4a^2(x-a)^2$

$\Rightarrow (x+a)^2 = 4a^2$

so, the given equation have two real roots $x= -3a$ or $x = a$

### 1 comment

Although I get the same roots $-3a$ and $a$, I get the determinant expansion polynomial as $(x-a)^3(x+3a)$, which indeed is a quartic as $f(x)$ suggests. Not sure how this block determinant formula results in a quadratic, cannot find anything much about that formula. Please correct me here but I think that this is a case where there are 4 real roots (The matrix of $f(x)$ is symmetric) but only two distinct roots as suggested by whatever determinant I got, $a$ having multiplicity $3$ and $-3a$ with multiplicity $1$. I only confirmed the multiplicites with wolfram cloud

1
442 views