Use the inverse formula $A ^ { - 1 } = \frac { 1 } { a d - b c } \left[ \begin{array} { c c }
d & - b \
- c & a
\end{array} \right]$ to find the inverse of the 2×2 matrix (if it exists).
$\left[ \begin{array} { l l }
2 & 3 \
1 & 5
\end{array} \right]$
A) $\frac { 1 } { 7 } \left[ \begin{array} { l l }
5 & 3 \
1 & 2
\end{array} \right]$
B) $\frac { 1 } { 7 } \left[ \begin{array} { c c }
- 5 & - 3 \
1 & 2
\end{array} \right]$
C) $\frac { 1 } { 7 } \left[ \begin{array} { c c }
- 5 & 3 \
1 & - 2
\end{array} \right]$
D) $\frac { 1 } { 7 } \left[ \begin{array} { c c }
5 & - 3 \
- 1 & 2
\end{array} \right]$
E) $\frac { 1 } { 7 } \left[ \begin{array} { l l }
- 5 & - 3 \
- 1 & - 2
\end{array} \right]$

Question

Use the inverse formula $A ^ { - 1 } = \frac { 1 } { a d - b c } \left[ \begin{array} { c c } 
d & - b \
- c & a
\end{array} \right]$ to find the inverse of the 2×2 matrix (if it exists).  
 $\left[ \begin{array} { l l } 
2 & 3 \
1 & 5
\end{array} \right]$  
A)  $\frac { 1 } { 7 } \left[ \begin{array} { l l } 
5 & 3 \
1 & 2
\end{array} \right]$
B) $\frac { 1 } { 7 } \left[ \begin{array} { c c } 
- 5 & - 3 \
1 & 2
\end{array} \right]$
C)  $\frac { 1 } { 7 } \left[ \begin{array} { c c } 
- 5 & 3 \
1 & - 2
\end{array} \right]$
D)  $\frac { 1 } { 7 } \left[ \begin{array} { c c } 
5 & - 3 \
- 1 & 2
\end{array} \right]$
E)  $\frac { 1 } { 7 } \left[ \begin{array} { l l } 
- 5 & - 3 \
- 1 & - 2
\end{array} \right]$

Quizplus · Accepted Answer

The determinant of the matrix is $ad - bc = (2)(5) - (3)(1) = 10 - 3 = 7$. Using the inverse formula, we swap $a$ and $d$, change the signs of $b$ and $c$, and divide by the determinant, resulting in $\frac{1}{7}\left[ \begin{array}{cc} 5 & -3 \ -1 & 2 \end{array} \right]$.

Use the Inverse Formula A−1=1ad−bc[d−b−ca]A ^ { - 1 } = \frac { 1 } { a d - b c } \left[ \begin{array} { c c } d & - b \\- c & a\end{array} \right]A−1=ad−bc1​[d−c​−ba​]

Use the Inverse Formula $A ^ { - 1 } = \frac { 1 } { a d - b c } \left[ \begin{array} { c c } d & - b \\- c & a\end{array} \right]$