Question

If \[ \begin{vmatrix} x & 2 & -1 \\ 1 & x & 5 \\ 3 & 2 & x \end{vmatrix} = 0, \quad \text{then the real value of} \quad x \quad \text{is:} \]

• A. 4
• B. -3
• C. 2
• D. -1
• E. -4

Hint:

When calculating the determinant of a 3x3 matrix, use cofactor expansion. You can then simplify the resulting expression to find the value of \( x \).

Answer 1

Answer: (E)

The given expression is a determinant:

\[ \begin{vmatrix} x & 2 & -1 \\ 1 & x & 5 \\ 3 & 2 & x \end{vmatrix} = 0. \] We will calculate the determinant of the 3x3 matrix. Using cofactor expansion along the first row:

\[ \text{Determinant} = x \begin{vmatrix} x & 5 \\ 2 & x \end{vmatrix} - 2 \begin{vmatrix} 1 & 5 \\ 3 & x \end{vmatrix} + (-1) \begin{vmatrix} 1 & x \\ 3 & 2 \end{vmatrix}. \] Step 1: Now, calculate each of the 2x2 determinants:

\[ \begin{vmatrix} x & 5 \\ 2 & x \end{vmatrix} = x \cdot x - 5 \cdot 2 = x^2 - 10, \] \[ \begin{vmatrix} 1 & 5 \\ 3 & x \end{vmatrix} = 1 \cdot x - 5 \cdot 3 = x - 15, \] \[ \begin{vmatrix} 1 & x \\ 3 & 2 \end{vmatrix} = 1 \cdot 2 - x \cdot 3 = 2 - 3x. \] Step 2: Substitute these values into the cofactor expansion:

\[ \text{Determinant} = x(x^2 - 10) - 2(x - 15) - (2 - 3x). \] Simplifying the expression:

\[ \text{Determinant} = x^3 - 10x - 2x + 30 - 2 + 3x, \] \[ \text{Determinant} = x^3 - 9x + 28. \] Step 3: Set the determinant equal to zero:

\[ x^3 - 9x + 28 = 0. \] Step 4: To find the roots of the cubic equation, we can use the Rational Root Theorem. The possible rational roots are the factors of 28 divided by the factors of 1, i.e., \( \pm 1, \pm 2, \pm 4, \pm 7, \pm 14, \pm 28 \). Testing \( x = -4 \):

\[ (-4)^3 - 9(-4) + 28 = -64 + 36 + 28 = 0. \] Thus, \( x = -4 \) is a solution.

Thus, the correct answer is option (E).