Show that x = 2 is a root of the equation`|{:(x, -6, -1),(2,- 3x,x-3),(-3," "2x,x+2):}|` = 0 and solve it completely.

Answer

To show that \( x = 2 \) is a root of the equation represented by the determinant

\[ \begin{vmatrix} x & -6 & -1 \\ 2 & -3x & x-3 \\ -3 & 2x & x+2 \end{vmatrix} = 0 \]

and to solve it completely, we will follow these steps:

### Step 1: Substitute \( x = 2 \) into the determinant
We will first substitute \( x = 2 \) into the determinant to check if it equals zero.

\[ \begin{vmatrix} 2 & -6 & -1 \\ 2 & -3(2) & 2-3 \\ -3 & 2(2) & 2+2 \end{vmatrix} = \begin{vmatrix} 2 & -6 & -1 \\ 2 & -6 & -1 \\ -3 & 4 & 4 \end{vmatrix} \]

### Step 2: Calculate the determinant
Now, we will calculate the determinant using the formula for a 3x3 matrix:

\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]

Where \( a, b, c \) are the elements of the first row, and \( d, e, f, g, h, i \) are the elements of the remaining rows.

Substituting the values:

\[ D = 2((-6)(4) - (-1)(-1)) - (-6)(2(4) - (-1)(-3)) + (-1)(2(4) - (-6)(-3)) \]

Calculating each term:

1. First term: \( 2((-24) - 1) = 2(-25) = -50 \)
2. Second term: \( -6(8 - 3) = -6(5) = -30 \) (note the negative sign)
3. Third term: \( -1(8 - 18) = -1(-10) = 10 \)

Now, combine these:

\[ D = -50 + 30 + 10 = -10 \]

### Step 3: Check if \( D = 0 \)
Since \( D \neq 0 \), we need to calculate the determinant without substituting \( x = 2 \) directly.

### Step 4: Expand the determinant
We will expand the determinant without substituting \( x = 2 \):

\[ D = x \begin{vmatrix} -3x & x-3 \\ 2x & x+2 \end{vmatrix} - (-6) \begin{vmatrix} 2 & x-3 \\ -3 & x+2 \end{vmatrix} - 1 \begin{vmatrix} 2 & -3x \\ -3 & 2x \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. First determinant:
\[ \begin{vmatrix} -3x & x-3 \\ 2x & x+2 \end{vmatrix} = (-3x)(x+2) - (x-3)(2x) = -3x^2 - 6x - 2x^2 + 6x = -5x^2 \]

2. Second determinant:
\[ \begin{vmatrix} 2 & x-3 \\ -3 & x+2 \end{vmatrix} = 2(x+2) - (-3)(x-3) = 2x + 4 + 3x - 9 = 5x - 5 \]

3. Third determinant:
\[ \begin{vmatrix} 2 & -3x \\ -3 & 2x \end{vmatrix} = 2(2x) - (-3)(-3x) = 4x - 9x = -5x \]

### Step 5: Combine the results
Now substituting back into the original determinant:

\[ D = x(-5x^2) + 6(5x - 5) - (-5x) = -5x^3 + 30x - 30 + 5x = -5x^3 + 35x - 30 \]

### Step 6: Set the determinant to zero
Now we set the equation to zero:

\[ -5x^3 + 35x - 30 = 0 \]

### Step 7: Factor out the equation
We can factor out -5:

\[ -5(x^3 - 7x + 6) = 0 \]

### Step 8: Factor the cubic equation
Now we can factor \( x^3 - 7x + 6 \). We already know \( x = 2 \) is a root, so we can divide \( x^3 - 7x + 6 \) by \( x - 2 \):

Using synthetic division:

\[ \begin{array}{r|rrr} 2 & 1 & 0 & -7 & 6 \\ & & 2 & 4 & -6 \\ \hline & 1 & 2 & -3 & 0 \\ \end{array} \]

This gives us:

\[ x^3 - 7x + 6 = (x - 2)(x^2 + 2x - 3) \]

### Step 9: Factor the quadratic
Now we can factor \( x^2 + 2x - 3 \):

\[ x^2 + 2x - 3 = (x + 3)(x - 1) \]

### Final Result
Thus, the complete factorization is:

\[ -5(x - 2)(x + 3)(x - 1) = 0 \]

The roots are:

\[ x = 2, \quad x = -3, \quad x = 1 \]