Find the inverse of the matrix, if it exists
-$$A = \left[ \begin{array} { r r }
5 & 0 \
- 4 & - 6
\end{array} \right]$$

A)
$$\left[ \begin{array} { c c }
- \frac { 1 } { 6 } & 0 \
- \frac { 2 } { 15 } & \frac { 1 } { 5 }
\end{array} \right]$$

B) $\mathrm { A }$ is not invertible

C)
$$\left[ \begin{array} { c c }
\frac { 1 } { 5 } & 0 \
\frac { 2 } { 15 } & - \frac { 1 } { 6 }
\end{array} \right]$$
D)
$$\left[ \begin{array} { c c }
\frac { 1 } { 5 } & 0 \
- \frac { 2 } { 15 } & - \frac { 1 } { 6 }
\end{array} \right]$$

Question

Find the inverse of the matrix, if it exists
-$$A = \left[ \begin{array} { r r } 
5 & 0 \
- 4 & - 6
\end{array} \right]$$

A)
$$\left[ \begin{array} { c c } 
- \frac { 1 } { 6 } & 0 \
- \frac { 2 } { 15 } & \frac { 1 } { 5 }
\end{array} \right]$$

B) $\mathrm { A }$ is not invertible

C)
$$\left[ \begin{array} { c c } 
\frac { 1 } { 5 } & 0 \
\frac { 2 } { 15 } & - \frac { 1 } { 6 }
\end{array} \right]$$
D)
$$\left[ \begin{array} { c c } 
\frac { 1 } { 5 } & 0 \
- \frac { 2 } { 15 } & - \frac { 1 } { 6 }
\end{array} \right]$$

Quizplus · Accepted Answer

The inverse of a 2x2 matrix $\left[ \begin{array} { c c } a & b \ c & d \end{array} \right]$ is $\frac{1}{ad-bc}\left[ \begin{array} { c c } d & -b \ -c & a \end{array} \right]$, given $ad-bc \neq 0$. For matrix A, $ad-bc = (5)(-6) - (0)(-4) = -30$, so the inverse exists and is $\frac{1}{-30}\left[ \begin{array} { c c } -6 & 0 \ 4 & 5 \end{array} \right] = \left[ \begin{array} { c c } \frac{1}{5} & 0 \ -\frac{2}{15} & -\frac{1}{6} \end{array} \right]$.

Find the Inverse of the Matrix, If It Exists - A=[50−4−6]A = \left[ \begin{array} { r r } 5 & 0 \\- 4 & - 6\end{array} \right]A=[5−4​0−6​]

Find the Inverse of the Matrix, If It Exists
- $A = \left[ \begin{array} { r r } 5 & 0 \\- 4 & - 6\end{array} \right]$