Find the inverse of the invertible matrix.
-$$\left[ \begin{array} { l l }
2 & - 1 \
0 & - 3
\end{array} \right]$$
A) $\left[ \begin{array} { c c } \frac { 1 } { 2 } & \frac { 1 } { 6 } \ 0 & - \frac { 1 } { 3 } \end{array} \right]$
B) $\left[ \begin{array} { c c } 0 & - \frac { 1 } { 3 } \ \frac { 1 } { 2 } & - \frac { 1 } { 6 } \end{array} \right]$
C) $\left[ \begin{array} { r r } - \frac { 1 } { 3 } & - \frac { 1 } { 6 } \ 0 & \frac { 1 } { 2 } \end{array} \right]$
D) $\left[ \begin{array} { l l } \frac { 1 } { 2 } & - \frac { 1 } { 6 } \ 0 & - \frac { 1 } { 3 } \end{array} \right]$

Question

Find the inverse of the invertible matrix.
-$$\left[ \begin{array} { l l } 
2 & - 1 \
0 & - 3
\end{array} \right]$$
A) $\left[ \begin{array} { c c } \frac { 1 } { 2 } & \frac { 1 } { 6 } \ 0 & - \frac { 1 } { 3 } \end{array} \right]$
B) $\left[ \begin{array} { c c } 0 & - \frac { 1 } { 3 } \ \frac { 1 } { 2 } & - \frac { 1 } { 6 } \end{array} \right]$
C) $\left[ \begin{array} { r r } - \frac { 1 } { 3 } & - \frac { 1 } { 6 } \ 0 & \frac { 1 } { 2 } \end{array} \right]$
D) $\left[ \begin{array} { l l } \frac { 1 } { 2 } & - \frac { 1 } { 6 } \ 0 & - \frac { 1 } { 3 } \end{array} \right]$

Quizplus · Accepted Answer

The inverse of a 2x2 matrix $\left[ \begin{array} { l l } a & b \ c & d \end{array} \right]$ is $\frac{1}{ad-bc}\left[ \begin{array} { l l } d & -b \ -c & a \end{array} \right]$. For the given matrix, $a=2$, $b=-1$, $c=0$, $d=-3$, so the determinant $ad-bc = -6$. The inverse is $\frac{1}{-6}\left[ \begin{array} { l l } -3 & 1 \ 0 & 2 \end{array} \right] = \left[ \begin{array} { l l } \frac{1}{2} & -\frac{1}{6} \ 0 & -\frac{1}{3} \end{array} \right]$.

Find the Inverse of the Invertible Matrix [2−10−3]\left[ \begin{array} { l l } 2 & - 1 \\0 & - 3\end{array} \right][20​−1−3​]

Find the Inverse of the Invertible Matrix $\left[ \begin{array} { l l } 2 & - 1 \\0 & - 3\end{array} \right]$