Question

Solve the equation: \[ \left| \begin{matrix} x+a & x & x \\ x & x+a & x \\ x & x & x+a \end{matrix} \right| = 0 \] where \( a \neq 0 \).

Hint:

When calculating the determinant of a 3x3 matrix, expand along a row or column and calculate the minors and cofactors.

Answer 1

Step 1: Determining the determinant of the matrix. 

We are given the matrix: \[ A = \begin{bmatrix} x + a & x & x \\ x & x + a & x \\ x & x & x + a \end{bmatrix} \] We need to find the determinant of this matrix. The determinant of a 3x3 matrix is calculated as follows: \[ \text{det}(A) = (x + a) \left| \begin{matrix} x + a & x \\ x & x + a \end{matrix} \right| - x \left| \begin{matrix} x & x \\ x & x + a \end{matrix} \right| + x \left| \begin{matrix} x & x + a \\ x & x \end{matrix} \right| \]

Step 2: Expanding the determinant. 

Let's calculate each minor: - First minor: \[ \left| \begin{matrix} x + a & x \\ x & x + a \end{matrix} \right| = (x + a)(x + a) - x^2 = (x + a)^2 - x^2 \] \[ = x^2 + 2ax + a^2 - x^2 = 2ax + a^2 \] - Second minor: \[ \left| \begin{matrix} x & x \\ x & x + a \end{matrix} \right| = x(x + a) - x^2 = ax \] - Third minor: \[ \left| \begin{matrix} x & x + a \\ x & x \end{matrix} \right| = x(x) - x(x + a) = x^2 - x^2 - ax = -ax \] Now, substitute these into the determinant formula: \[ \text{det}(A) = (x + a)(2ax + a^2) - x(ax) + x(-ax) \] \[ = (x + a)(2ax + a^2) - ax^2 - ax^2 \] \[ = 2a x^2 + a^2 x + 2a^2 x + a^3 - 2a x^2 \] \[ = a^3 + 3a^2 x \]

Step 3: Solving the equation. 

We have the equation: \[ a^3 + 3a^2 x = 0 \] Factor the equation: \[ a^2(a + 3x) = 0 \] Since \( a \neq 0 \), we have: \[ a + 3x = 0 \] Thus, the solution is: \[ x = -\frac{a}{3} \]