Question

If $\begin{vmatrix} x+1 & 1 & 3 \\ 2 & x+2 & 1 \\ 2 & 1 & x+4 \end{vmatrix}=0$, then x will be equal to:

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

Answer 1

None of the given options are correct. The equation simplifies to x^3 + 7x^2 + 5x - 5 = 0, which does not have -1, -4, 1, or 4 as roots.

Answer 2

To solve the determinant equation:

$\begin{vmatrix} x+1 & 1 & 3 \\ 2 & x+2 & 1 \\ 2 & 1 & x+4 \end{vmatrix}=0$

Expanding the determinant along the first row ($R_1$):

$(x+1) \begin{vmatrix} x+2 & 1 \\ 1 & x+4 \end{vmatrix} - 1 \begin{vmatrix} 2 & 1 \\ 2 & x+4 \end{vmatrix} + 3 \begin{vmatrix} 2 & x+2 \\ 2 & 1 \end{vmatrix} = 0$

Calculating the 2x2 determinants:

• $\begin{vmatrix} x+2 & 1 \\ 1 & x+4 \end{vmatrix} = (x+2)(x+4) - (1)(1) = x^2 + 6x + 8 - 1 = x^2 + 6x + 7$
• $\begin{vmatrix} 2 & 1 \\ 2 & x+4 \end{vmatrix} = (2)(x+4) - (1)(2) = 2x + 8 - 2 = 2x + 6$
• $\begin{vmatrix} 2 & x+2 \\ 2 & 1 \end{vmatrix} = (2)(1) - (x+2)(2) = 2 - 2x - 4 = -2x - 2$

Substituting back into the equation:

$(x+1)(x^2 + 6x + 7) - (2x + 6) + 3(-2x - 2) = 0$

Expanding:

$x^3 + 6x^2 + 7x + x^2 + 6x + 7 - 2x - 6 - 6x - 6 = 0$

Combining like terms:

$x^3 + 7x^2 + 5x - 5 = 0$

Checking the given options to see if any are roots of this cubic equation:

• For $x = -1$: $(-1)^3 + 7(-1)^2 + 5(-1) - 5 = -1 + 7 - 5 - 5 = -4 \neq 0$
• For $x = -4$: $(-4)^3 + 7(-4)^2 + 5(-4) - 5 = -64 + 112 - 20 - 5 = 23 \neq 0$
• For $x = 1$: $(1)^3 + 7(1)^2 + 5(1) - 5 = 1 + 7 + 5 - 5 = 8 \neq 0$
• For $x = 4$: $(4)^3 + 7(4)^2 + 5(4) - 5 = 64 + 112 + 20 - 5 = 191 \neq 0$

Since none of the options satisfy the derived equation, the question or options are likely flawed. The correct solution involves solving the cubic equation, which does not have the given integers as roots.