If `|(1,-3,4),(-5,x+2,2),(4,1,x-6)|=0` then x equals

Options

A. 17,21
B. 0,19
C. 0,35
D. 21,35

Answer

To solve the determinant equation \( |(1,-3,4),(-5,x+2,2),(4,1,x-6)|=0 \), we will proceed step by step.

### Step 1: Write down the determinant
We start with the determinant of the matrix:

\[ D = \begin{vmatrix} 1 & -3 & 4 \\ -5 & x+2 & 2 \\ 4 & 1 & x-6 \end{vmatrix} \]

### Step 2: Expand the determinant
We will expand the determinant using the first row:

\[ D = 1 \cdot \begin{vmatrix} x+2 & 2 \\ 1 & x-6 \end{vmatrix} - (-3) \cdot \begin{vmatrix} -5 & 2 \\ 4 & x-6 \end{vmatrix} + 4 \cdot \begin{vmatrix} -5 & x+2 \\ 4 & 1 \end{vmatrix} \]

### Step 3: Calculate the 2x2 determinants
Now we calculate each of the 2x2 determinants:

1. For \( \begin{vmatrix} x+2 & 2 \\ 1 & x-6 \end{vmatrix} \):

\[ = (x+2)(x-6) - (2)(1) = x^2 - 6x + 2x - 12 - 2 = x^2 - 4x - 14 \]

2. For \( \begin{vmatrix} -5 & 2 \\ 4 & x-6 \end{vmatrix} \):

\[ = (-5)(x-6) - (2)(4) = -5x + 30 - 8 = -5x + 22 \]

3. For \( \begin{vmatrix} -5 & x+2 \\ 4 & 1 \end{vmatrix} \):

\[ = (-5)(1) - (4)(x+2) = -5 - 4x - 8 = -4x - 13 \]

### Step 4: Substitute back into the determinant
Substituting these back into our determinant expression:

\[ D = 1(x^2 - 4x - 14) + 3(-5x + 22) + 4(-4x - 13) \]

### Step 5: Simplify the expression
Now, simplify \( D \):

\[ D = x^2 - 4x - 14 - 15x + 66 - 16x - 52 \]

Combine like terms:

\[ D = x^2 - 4x - 15x - 16x - 14 + 66 - 52 \]
\[ = x^2 - 35x + 0 \]

### Step 6: Set the determinant to zero
Since we want \( D = 0 \):

\[ x^2 - 35x = 0 \]

### Step 7: Factor the equation
Factoring gives:

\[ x(x - 35) = 0 \]

### Step 8: Solve for \( x \)
Setting each factor to zero gives:

1. \( x = 0 \)
2. \( x - 35 = 0 \) → \( x = 35 \)

Thus, the possible values for \( x \) are \( 0 \) and \( 35 \).

### Final Result
The values of \( x \) are \( 0 \) and \( 35 \).

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