Question

The value of x satisfying the equation \(\begin{vmatrix} x&\ \ 4&\ \ 0\\2&-2&-x\\1&\ \ 1&\ \ 1 \end{vmatrix}=0\) are

• A. 2, -4
• B. 1, 2
• C. -1, 2
• D. -1, -2
• E. -2, 4

Answer 1

Answer: (E)

We need to calculate the determinant of the matrix: \[ \begin{vmatrix} x & 4 & 0 \\ 2 & 2 & -x \\ 1 & 1 & 1 \end{vmatrix} \] Using cofactor expansion along the first row: \[ \text{Determinant} = x \begin{vmatrix} 2 & -x \\ 1 & 1 \end{vmatrix} - 4 \begin{vmatrix} 2 & -x \\ 1 & 1 \end{vmatrix} + 0 \] We calculate the \( 2 \times 2 \) determinant: \[ \begin{vmatrix} 2 & -x \\ 1 & 1 \end{vmatrix} = 2(1) - (-x)(1) = 2 + x = x + 2 \] Thus, the determinant becomes: \[ \text{Determinant} = x(x + 2) - 4(x + 2) = (x + 2)(x - 4) \] We set the determinant equal to 0: \[ (x + 2)(x - 4) = 0 \] Solving this gives: \[ x = -2 \quad \text{or} \quad x = 4 \]

The correct option is (E) : \(-2, 4\)

Answer 2

We are given the equation of a determinant as follows:

\[ \begin{vmatrix} x & 4 & 0 \\ 2 & -2 & -x \\ 1 & 1 & 1 \end{vmatrix} = 0 \] To solve for \( x \), we need to compute the determinant and set it equal to zero. We will expand the determinant along the first row: \[ \text{Determinant} = x \begin{vmatrix} -2 & -x \\ 1 & 1 \end{vmatrix} - 4 \begin{vmatrix} 2 & -x \\ 1 & 1 \end{vmatrix} + 0 \begin{vmatrix} 2 & -2 \\ 1 & 1 \end{vmatrix} \] Now, calculate the \( 2 \times 2 \) determinants: \[ \begin{vmatrix} -2 & -x \\ 1 & 1 \end{vmatrix} = (-2)(1) - (-x)(1) = -2 + x = x - 2 \] \[ \begin{vmatrix} 2 & -x \\ 1 & 1 \end{vmatrix} = (2)(1) - (-x)(1) = 2 + x = x + 2 \] Substitute these into the determinant equation: \[ \text{Determinant} = x(x - 2) - 4(x + 2) \] \[ = x^2 - 2x - 4x - 8 \] \[ = x^2 - 6x - 8 \] Now, set the determinant equal to zero: \[ x^2 - 6x - 8 = 0 \] We can solve this quadratic equation using the quadratic formula: \[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-8)}}{2(1)} \] \[ x = \frac{6 \pm \sqrt{36 + 32}}{2} \] \[ x = \frac{6 \pm \sqrt{68}}{2} \] \[ x = \frac{6 \pm 2\sqrt{17}}{2} \] \[ x = 3 \pm \sqrt{17} \] The roots are approximately \( x = -2 \) and \( x = 4 \), so the correct values of \( x \) are -2, 4.