Find the inverse of the matrix. $$\left[ \begin{array} { c c }
1 & 2 \
- 2 & - 1
\end{array} \right]$$
A) $\left[ \begin{array} { c c } - \frac { 1 } { 3 } & - \frac { 2 } { 3 } \ \frac { 2 } { 3 } & \frac { 1 } { 3 } \end{array} \right]$

B)
$\left[ \begin{array} { c c } - \frac { 1 } { 3 } & \frac { 2 } { 3 } \ \frac { 2 } { 3 } & \frac { 1 } { 3 } \end{array} \right]$

C)
$$\left[ \begin{array} { c c }
\frac { 1 } { 3 } & - \frac { 2 } { 3 } \
\frac { 2 } { 3 } & \frac { 1 } { 3 }
\end{array} \right]$$

D) $\left[ \begin{array} { c c } - \frac { 1 } { 3 } & - \frac { 2 } { 3 } \ \frac { 2 } { 3 } & - \frac { 1 } { 3 } \end{array} \right]$

E) $\left[ \begin{array} { c c } - \frac { 1 } { 3 } & - \frac { 2 } { 3 } \ - \frac { 2 } { 3 } & \frac { 1 } { 3 } \end{array} \right]$

Question

Find the inverse of the matrix. $$\left[ \begin{array} { c c } 
1 & 2 \
- 2 & - 1
\end{array} \right]$$
A) $\left[ \begin{array} { c c } - \frac { 1 } { 3 } & - \frac { 2 } { 3 } \ \frac { 2 } { 3 } & \frac { 1 } { 3 } \end{array} \right]$

B)
$\left[ \begin{array} { c c } - \frac { 1 } { 3 } & \frac { 2 } { 3 } \ \frac { 2 } { 3 } & \frac { 1 } { 3 } \end{array} \right]$

C)
$$\left[ \begin{array} { c c } 
\frac { 1 } { 3 } & - \frac { 2 } { 3 } \
\frac { 2 } { 3 } & \frac { 1 } { 3 }
\end{array} \right]$$

D) $\left[ \begin{array} { c c } - \frac { 1 } { 3 } & - \frac { 2 } { 3 } \ \frac { 2 } { 3 } & - \frac { 1 } { 3 } \end{array} \right]$

E) $\left[ \begin{array} { c c } - \frac { 1 } { 3 } & - \frac { 2 } { 3 } \ - \frac { 2 } { 3 } & \frac { 1 } { 3 } \end{array} \right]$

Quizplus · Accepted Answer

The inverse of a 2x2 matrix $\left[ \begin{array} { c c } a & b \ c & d \end{array} \right]$ is $\frac{1}{ad-bc}\left[ \begin{array} { c c } d & -b \ -c & a \end{array} \right]$. For the given matrix, the determinant is $1(-1) - (-2)(2) = -1 + 4 = 3$. Thus, the inverse is $\frac{1}{3}\left[ \begin{array} { c c } -1 & -2 \ 2 & 1 \end{array} \right]$, which simplifies to $\left[ \begin{array} { c c } -\frac{1}{3} & -\frac{2}{3} \ \frac{2}{3} & \frac{1}{3} \end{array} \right]$.

Find the Inverse of the Matrix [12−2−1]\left[ \begin{array} { c c } 1 & 2 \\- 2 & - 1\end{array} \right][1−2​2−1​]

Find the Inverse of the Matrix $\left[ \begin{array} { c c } 1 & 2 \\- 2 & - 1\end{array} \right]$