Question

If $\left| \begin{array}{ccc} -1 & 2 & 4 \\ 1 & x & 1 \\ 0 & 3 & 3x \end{array} \right| = -57$, the product of the possible values of $x$ is:

• A. $-24$
• B. $-16$
• C. $16$
• D. $24$

Hint:

For solving determinants and equations involving matrices, use cofactor expansion and then simplify the resulting equation. In quadratic equations, use the quadratic formula to find the solutions.

Answer 1

The Correct Option is B

We are given the determinant of the following matrix: \[ \left| \begin{array}{ccc} -1 & 2 & 4 \\ 1 & x & 1 \\ 0 & 3 & 3x \end{array} \right| = -57. \] We need to calculate the determinant and solve for $x$. Using cofactor expansion along the first row: \[ \text{det} = (-1) \left| \begin{array}{cc} x & 1 \ 3 &\ 3x \end{array} \right| - 2 \left| \begin{array}{cc} 1 & 1 \\ 0 & 3x \end{array} \right| + 4 \left| \begin{array}{cc} 1 & x \\ 0 & 3 \end{array} \right|. \] Now calculate the 2x2 determinants: \[ \left| \begin{array}{cc} x & 1
3 & 3x \end{array} \right| = x(3x) - (1)(3) = 3x^2 - 3, \] \[ \left| \begin{array}{cc} 1 & 1 \ 0 &\ 3x \end{array} \right| = (1)(3x) - (1)(0) = 3x, \] \[ \left| \begin{array}{cc} 1 & x \ 0 &\ 3 \end{array} \right| = (1)(3) - (x)(0) = 3. \] Substituting these into the determinant formula: \[ \text{det} = (-1)(3x^2 - 3) - 2(3x) + 4(3). \] Simplifying: \[ \text{det} = -(3x^2 - 3) - 6x + 12, \] \[ \text{det} = -3x^2 + 3 - 6x + 12, \] \[ \text{det} = -3x^2 - 6x + 15. \] We are given that this determinant equals $-57$: \[ -3x^2 - 6x + 15 = -57. \] Simplifying the equation: \[ -3x^2 - 6x + 72 = 0, \] \[ x^2 + 2x - 24 = 0. \] Now solve for $x$ using the quadratic formula: \[ x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-24)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 96}}{2} = \frac{-2 \pm \sqrt{100}}{2} = \frac{-2 \pm 10}{2}. \] Thus, the two possible values of $x$ are: \[ x = \frac{-2 + 10}{2} = 4 \quad \text{or} \quad x = \frac{-2 - 10}{2} = -6. \] The product of the possible values of $x$ is: \[ 4 \times (-6) = -24. \] Thus, the product of the possible values of $x$ is $-24$.