Given matrix $A = \left[ \begin{array} { c c }
7 & 15 \
- 14 & 5
\end{array} \right]$ . Find $A ^ { - 1 }$ the inverse matrix.
A) $A ^ { - 1 } = \left[ \begin{array} { c c }
\frac { 1 } { 7 } & \frac { 1 } { 15 } \\
- \frac { 1 } { 14 } & \frac { 1 } { 5 }
\end{array} \right]$
B) $A ^ { - 1 } = \left[ \begin{array} { c c }
\frac { 1 } { 35 } & - \frac { 3 } { 49 } \\
\frac { 2 } { 35 } & \frac { 1 } { 25 }
\end{array} \right]$
C) $A ^ { - 1 } = \left[ \begin{array} { c c }
\frac { 1 } { 49 } & - \frac { 3 } { 49 } \\
\frac { 2 } { 35 } & \frac { 1 } { 35 }
\end{array} \right]$
D) $A ^ { - 1 } = \left[ \begin{array} { c c }
\frac { 2 } { 49 } & - \frac { 1 } { 49 } \\
\frac { 1 } { 35 } & \frac { 3 } { 35 }
\end{array} \right]$
E) $A ^ { - 1 } = \left[ \begin{array} { c c }
- \frac { 1 } { 49 } & \frac { 3 } { 49 } \\
\frac { 2 } { 35 } & - \frac { 1 } { 35 }
\end{array} \right]$

Question

Given matrix $A = \left[ \begin{array} { c c } 
7 & 15 \
- 14 & 5
\end{array} \right]$ . Find $A ^ { - 1 }$ the inverse matrix.
A) $A ^ { - 1 } = \left[ \begin{array} { c c } 
\frac { 1 } { 7 } & \frac { 1 } { 15 } \\
- \frac { 1 } { 14 } & \frac { 1 } { 5 }
\end{array} \right]$
B) $A ^ { - 1 } = \left[ \begin{array} { c c } 
\frac { 1 } { 35 } & - \frac { 3 } { 49 } \\
\frac { 2 } { 35 } & \frac { 1 } { 25 }
\end{array} \right]$
C) $A ^ { - 1 } = \left[ \begin{array} { c c } 
\frac { 1 } { 49 } & - \frac { 3 } { 49 } \\
\frac { 2 } { 35 } & \frac { 1 } { 35 }
\end{array} \right]$
D) $A ^ { - 1 } = \left[ \begin{array} { c c } 
\frac { 2 } { 49 } & - \frac { 1 } { 49 } \\
\frac { 1 } { 35 } & \frac { 3 } { 35 }
\end{array} \right]$
E) $A ^ { - 1 } = \left[ \begin{array} { c c } 
- \frac { 1 } { 49 } & \frac { 3 } { 49 } \\
\frac { 2 } { 35 } & - \frac { 1 } { 35 }
\end{array} \right]$

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 matrix $A$, the determinant is $7 \times 5 - (-14) \times 15 = 35 + 210 = 245$. Thus, $A^{-1} = \frac{1}{245} \left[ \begin{array} { c c } 5 & -15 \ 14 & 7 \end{array} \right]$, which simplifies to $\left[ \begin{array} { c c } \frac{1}{49} & -\frac{3}{49} \ \frac{2}{35} & \frac{1}{35} \end{array} \right]$.

Given Matrix A=[715−145]A = \left[ \begin{array} { c c } 7 & 15 \\- 14 & 5\end{array} \right]A=[7−14​155​]

Given Matrix $A = \left[ \begin{array} { c c } 7 & 15 \\- 14 & 5\end{array} \right]$