Solve for x given the following equation involving a determinant. $$\left| \begin{array} { c c }
x - 6 & - 8 \
- 1 & x - 4
\end{array} \right| = 0$$
A) $x = 2,8$
B) $x = - 2,8$
C) $x = 4,16$
D) $x = 2 , - 8$
E) $x = - 2 , - 8$

Question

Solve for x given the following equation involving a determinant. $$\left| \begin{array} { c c } 
x - 6 & - 8 \
- 1 & x - 4
\end{array} \right| = 0$$
A) $x = 2,8$
B) $x = - 2,8$
C) $x = 4,16$
D) $x = 2 , - 8$
E) $x = - 2 , - 8$

Quizplus · Accepted Answer

To solve for $x$, we need to calculate the determinant of the given matrix and set it equal to zero. The determinant of a 2x2 matrix $\left| \begin{array} { c c } a & b \ c & d \end{array} \right|$ is calculated as $ad - bc$. Applying this to the given matrix, we get $(x - 6)(x - 4) - (-8)(-1) = x^2 - 10x + 24 - 8 = x^2 - 10x + 16 = 0$. Factoring this quadratic equation, we find $(x - 2)(x - 8) = 0$, which gives us the solutions $x = 2$ and $x = 8$.

Solve for X Given the Following Equation Involving a Determinant ∣x−6−8−1x−4∣=0\left| \begin{array} { c c } x - 6 & - 8 \\- 1 & x - 4\end{array} \right| = 0​x−6−1​−8x−4​​=0

Solve for X Given the Following Equation Involving a Determinant $\left| \begin{array} { c c } x - 6 & - 8 \\- 1 & x - 4\end{array} \right| = 0$