Solve the equation for x.
-$$\left| \begin{array} { l l }
x & 1 \
2 & x
\end{array} \right| = - 1$$
A) $\{ - 1 \}$
B) $\{ - 1,1 \}$
C) $\{ 1 \}$
D) $\varnothing$

Question

Solve the equation for x.
-$$\left| \begin{array} { l l } 
x & 1 \
2 & x
\end{array} \right| = - 1$$
A) $\{ - 1 \}$
B) $\{ - 1,1 \}$
C) $\{ 1 \}$
D) $\varnothing$

Quizplus · Accepted Answer

To solve the equation, we need to calculate the determinant of the given 2x2 matrix and set it equal to -1. The determinant of a 2x2 matrix $\left| \begin{array} { l l } a & b \ c & d \end{array} \right|$ is calculated as $ad - bc$. Applying this to the given matrix, we get the determinant as $x*x - 1*2 = x^2 - 2$. Setting this equal to -1 gives us the equation $x^2 - 2 = -1$, which simplifies to $x^2 = 1$. Solving for $x$, we find $x = \pm 1$, which means the solution set is $\{-1, 1\}$.

Solve the Equation for X ∣x12x∣=−1\left| \begin{array} { l l } x & 1 \\2 & x\end{array} \right| = - 1​x2​1x​​=−1

Solve the Equation for X $\left| \begin{array} { l l } x & 1 \\2 & x\end{array} \right| = - 1$