The solutions of the equation `|(x,2,-1),(2,5,x),(-1,2,x)|=0` are

Options

A. `3,-1`
B. `-3,1`
C. `3,1`
D. `-3,-1`

Answer

To solve the equation \( |(x, 2, -1), (2, 5, x), (-1, 2, x)| = 0 \), we need to evaluate the determinant and set it equal to zero.

### Step 1: Write the determinant
The determinant can be expressed as follows:

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

### Step 2: Calculate the determinant using the rule of Sarrus or cofactor expansion
We can calculate the determinant \( D \) using cofactor expansion along the first row:

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

### Step 3: Calculate the 2x2 determinants
1. For the first 2x2 determinant:
\[ \begin{vmatrix} 5 & x \\ 2 & x \end{vmatrix} = (5)(x) - (2)(x) = 5x - 2x = 3x \]

2. For the second 2x2 determinant:
\[ \begin{vmatrix} 2 & x \\ -1 & x \end{vmatrix} = (2)(x) - (-1)(x) = 2x + x = 3x \]

3. For the third 2x2 determinant:
\[ \begin{vmatrix} 2 & 5 \\ -1 & 2 \end{vmatrix} = (2)(2) - (5)(-1) = 4 + 5 = 9 \]

### Step 4: Substitute back into the determinant equation
Now substituting these values back into the determinant expression:

\[ D = x(3x) - 2(3x) - 1(9) \]
\[ D = 3x^2 - 6x - 9 \]

### Step 5: Set the determinant equal to zero
Now we set the determinant equal to zero:

\[ 3x^2 - 6x - 9 = 0 \]

### Step 6: Simplify the equation
Divide the entire equation by 3:

\[ x^2 - 2x - 3 = 0 \]

### Step 7: Factor the quadratic equation
Now we can factor the quadratic:

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

### Step 8: Solve for x
Setting each factor to zero gives us:

1. \( x - 3 = 0 \) → \( x = 3 \)
2. \( x + 1 = 0 \) → \( x = -1 \)

### Final Answer
The solutions of the equation are \( x = 3 \) and \( x = -1 \).

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