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Step-by-Step Solution
Step 1: Write down the determinant
We are given the determinant of a 3×3 matrix equal to zero:
$
\left|
\begin{matrix}
x & -6 & -1 \\
2 & -3x & x - 3 \\
-3 & 2x & x + 2
\end{matrix}
\right| = 0.
$
Step 2: Expand the determinant
Use the standard expansion for a 3×3 determinant along the first row (for example). For the matrix
$
\begin{pmatrix}
x & -6 & -1 \\
2 & -3x & x - 3 \\
-3 & 2x & x + 2
\end{pmatrix},
$
the expansion along the first row gives:
$
x \bigl((-3x)(x + 2) - (2x)(x - 3)\bigr)
\;+\;(-6)\bigl(2(x + 2) - (-3)(x - 3)\bigr)
\;+\;(-1)\bigl(\dots\bigr).
$
We must carefully calculate each minor and cofactor.
Step 3: Calculate the cofactors
Cofactor for the (1,1) entry:
$
(-3x)(x + 2) - (2x)(x - 3)
= -3x(x + 2) - 2x(x - 3).
$
Simplify:
$
-3x(x+2) = -3x^2 - 6x,
\quad
-\,2x(x - 3) = -2x^2 + 6x.
$
So together:
$
-3x^2 - 6x - 2x^2 + 6x = -5x^2.
$
Cofactor for the (1,2) entry:
This entry is –6, and the minor’s determinant is
$
2(x+2) - (-3)(x-3).
$
Simplify:
$
2(x+2) = 2x + 4,
\quad
-3(x-3) = -3x + 9.
$
So:
$
2(x+2) - (-3)(x-3) = (2x + 4) + (3x - 9) = 5x - 5.
$
Because the entry is in position (1,2), we include a factor of $(-1)^{1+2} = -1$ in the cofactor. Hence the term from this element is:
$
\,(-6)\cdot\bigl[-\bigl(5x - 5\bigr)\bigr] = 6(5x - 5) = 30x - 30.
$
Cofactor for the (1,3) entry:
This entry is –1. The minor’s determinant for this position is
$
\begin{vmatrix}
2 & -3x \\
-3 & 2x
\end{vmatrix}
= 2(2x) - (-3)(-3x).
$
Simplify:
$
4x - 9x = -5x.
$
The cofactor for position (1,3) has factor $(-1)^{1+3} = (-1)^4 = 1.$ Thus the term from this element is:
$
(-1)\cdot(-5x) = 5x.
$
Step 4: Combine all terms to form the polynomial
Combining the parts from each cofactor expansion:
$
\text{Determinant}
= x \cdot (-5x^2) + 30x - 30 + 5x
= -5x^3 + 30x - 30 + 5x.
$
Simplify to get:
$
-5x^3 + 35x - 30 = 0.
$
Divide throughout by -5 to simplify:
$
x^3 - 7x + 6 = 0.
$
Step 5: Find the sum of the roots
From the simplified cubic polynomial
$
x^3 - 7x + 6 = 0,
$
notice there is no $x^2$ term. The sum of the roots of a monic polynomial
$
x^3 + ax^2 + bx + c = 0
$
is
$
-a.
$
Here $a = 0,$ so the sum of the roots is 0.
Answer
The sum of the real roots of the given equation is 0.
