1. Understanding the Problem and Key Concepts

The problem asks for the sum of the real roots of a given equation involving a $3 \times 3$ determinant set to zero. To solve this, we need to convert the determinant equation into a standard polynomial equation and then use a powerful tool called Vieta's formulas.

• Determinant Expansion: For a $3 \times 3$ matrix, say $A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$, its determinant can be expanded using the cofactor expansion method along any row or column. For example, expanding along the first row ($R_1$): \det(A) = a \cdot \left| {\matrix{ e & f \\ h & i \cr } } \right| - b \cdot \left| {\matrix{ d & f \\ g & i \cr } } \right| + c \cdot \left| {\matrix{ d & e \\ g & h \cr } } \right| Each $2 \times 2$ determinant \left| {\matrix{ p & q \\ r & s \cr } } \right| is evaluated as $ps - qr$. Remember the alternating signs for the cofactors: $(+ - +)$ for $R_1$, $(- + -)$ for $R_2$, and $(+ - +)$ for $R_3$.

• Vieta's Formulas: These formulas relate the coefficients of a polynomial equation to the sums and products of its roots. For a general cubic equation of the form $ax^3 + bx^2 + cx + d = 0$, if $x_1, x_2, x_3$ are its roots (real or complex), then:

  • Sum of the roots: $x_1 + x_2 + x_3 = -\frac{b}{a}$
  • Sum of the roots taken two at a time: $x_1x_2 + x_2x_3 + x_3x_1 = \frac{c}{a}$
  • Product of the roots: $x_1x_2x_3 = -\frac{d}{a}$ Since a cubic equation with real coefficients always has at least one real root, and any complex roots appear in conjugate pairs, the sum of all roots (given by Vieta's formula) will be the sum of the real roots if all roots are real, or the sum of the one real root and the two complex conjugates if there are complex roots. In this problem, we are specifically asked for the sum of real roots. Once we find the polynomial, we can check if its roots are all real.

2. Step-by-Step Solution

Step 2.1: Write Down the Determinant Equation The given equation is: \left| {\matrix{ x & { - 6} & { - 1} \cr 2 & { - 3x} & {x - 3} \cr { - 3} & {2x} & {x + 2} \cr } } \right| = 0

Step 2.2: Expand the Determinant We will expand the determinant along the first row ($R_1$) because it's a straightforward approach and typically involves less complexity for the first element. The expansion follows the pattern $a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}$, where $C_{ij}$ is the cofactor. x \cdot \left| {\matrix{ { - 3x} & {x - 3} \cr {2x} & {x + 2} \cr } } \right| - (-6) \cdot \left| {\matrix{ 2 & {x - 3} \cr { - 3} & {x + 2} \cr } } \right| + (-1) \cdot \left| {\matrix{ 2 & { - 3x} \cr { - 3} & {2x} \cr } } \right| = 0

• Explanation for signs:
  • The first term is $x$ (element $a_{11}$) times its minor, with a positive sign (since $(-1)^{1+1} = +1$).
  • The second term is $-6$ (element $a_{12}$) times its minor, with a negative sign (since $(-1)^{1+2} = -1$). Hence, we have $-(-6)$, which simplifies to $+6$.
  • The third term is $-1$ (element $a_{13}$) times its minor, with a positive sign (since $(-1)^{1+3} = +1$). Hence, we have $+(-1)$, which simplifies to $-1$.

Step 2.3: Evaluate the $2 \times 2$ Minors Now, we calculate the value of each $2 \times 2$ determinant:

• First Minor ($M_{11}$): M_{11} = \left| {\matrix{ { - 3x} & {x - 3} \cr {2x} & {x + 2} \cr } } \right| $= (-3x)(x + 2) - (x - 3)(2x)$ $= (-3x^2 - 6x) - (2x^2 - 6x)$ $= -3x^2 - 6x - 2x^2 + 6x$ $= (-3x^2 - 2x^2) + (-6x + 6x)$ $= -5x^2$

• Second Minor ($M_{12}$): M_{12} = \left| {\matrix{ 2 & {x - 3} \cr { - 3} & {x + 2} \cr } } \right| $= (2)(x + 2) - (x - 3)(-3)$ $= (2x + 4) - (-3x + 9)$ $= 2x + 4 + 3x - 9$ $= 5x - 5$

• Third Minor ($M_{13}$): M_{13} = \left| {\matrix{ 2 & { - 3x} \cr { - 3} & {2x} \cr } } \right| $= (2)(2x) - (-3x)(-3)$ $= 4x - (9x)$ $= 4x - 9x$ $= -5x$

Step 2.4: Form the Cubic Equation Substitute the evaluated minors back into the expanded determinant equation from Step 2.2: $x \cdot (-5x^2) + 6 \cdot (5x - 5) - 1 \cdot (-5x) = 0$ Now, distribute and combine like terms to form the polynomial equation: $-5x^3 + (30x - 30) + 5x = 0$ $-5x^3 + (30x + 5x) - 30 = 0$ $-5x^3 + 35x - 30 = 0$ To simplify, we can divide the entire equation by $-5$: $x^3 - 7x + 6 = 0$ This is a cubic equation of the form $ax^3 + bx^2 + cx + d = 0$. Comparing coefficients, we have:

• $a = 1$
• $b = 0$ (since there is no $x^2$ term)
• $c = -7$
• $d = 6$

Step 2.5: Apply Vieta's Formulas The sum of the roots of a cubic equation $ax^3 + bx^2 + cx + d = 0$ is given by $-\frac{b}{a}$. In our equation, $x^3 - 7x + 6 = 0$: Sum of roots $= -\frac{b}{a} = -\frac{0}{1} = 0$.

To confirm that these are real roots, we can factor the polynomial: By inspection, $x=1$ is a root ($1^3 - 7(1) + 6 = 1 - 7 + 6 = 0$). Using polynomial division or synthetic division, we can divide $(x^3 - 7x + 6)$ by $(x-1)$: $(x-1)(x^2 + x - 6) = 0$ Further factoring the quadratic: $(x-1)(x+3)(x-2) = 0$ The roots are $x=1, x=-3, x=2$. All these roots are real. The sum of these real roots is $1 + (-3) + 2 = 0$.

3. Tips for Success and Common Mistakes

• Sign Errors: The most common mistake in determinant expansion is getting the signs wrong for the cofactors. Remember the checkerboard pattern of signs: $\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}$.
• Algebraic Errors: Be extremely careful when expanding products and combining like terms, especially with negative signs. Distribute thoroughly.
• Vieta's Formulas: Ensure you remember the formulas correctly, especially the sign for the sum of roots ($-\frac{b}{a}$).
• Double-Checking: If time permits, expanding the determinant using a different row or column can serve as a good double-check for your polynomial equation.

4. Conclusion By expanding the given $3 \times 3$ determinant, we obtained the cubic equation $x^3 - 7x + 6 = 0$. Using Vieta's formulas, the sum of the roots is given by $-\frac{b}{a}$, where $b=0$ (the coefficient of $x^2$) and $a=1$ (the coefficient of $x^3$). Therefore, the sum of the real roots is $0$.

The final answer is $\boxed{\text{0}}$.