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).
$\left[ \begin{array} { c c }
- 12 & 3 \
5 & 2
\end{array} \right]$
A) $- \frac { 1 } { 39 } \left[ \begin{array} { c c }
2 & - 3 \
- 5 & - 12
\end{array} \right]$
B) $- \frac { 1 } { 39 } \left[ \begin{array} { c c }
- 2 & - 3 \
5 & 12
\end{array} \right]$
C) $- \frac { 1 } { 39 } \left[ \begin{array} { c c }
- 2 & 3 \
5 & - 12
\end{array} \right]$
D) $- \frac { 1 } { 39 } \left[ \begin{array} { c c }
2 & - 3 \
5 & 12
\end{array} \right]$
E) $- \frac { 1 } { 39 } \left[ \begin{array} { l l }
- 2 & - 3 \
- 5 & - 12
\end{array} \right]$

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).  
  $\left[ \begin{array} { c c } 
- 12 & 3 \
5 & 2
\end{array} \right]$  
A)  $- \frac { 1 } { 39 } \left[ \begin{array} { c c } 
2 & - 3 \
- 5 & - 12
\end{array} \right]$
B)  $- \frac { 1 } { 39 } \left[ \begin{array} { c c } 
- 2 & - 3 \
5 & 12
\end{array} \right]$
C)  $- \frac { 1 } { 39 } \left[ \begin{array} { c c } 
- 2 & 3 \
5 & - 12
\end{array} \right]$
D)  $- \frac { 1 } { 39 } \left[ \begin{array} { c c } 
2 & - 3 \
5 & 12
\end{array} \right]$
E)  $- \frac { 1 } { 39 } \left[ \begin{array} { l l } 
- 2 & - 3 \
- 5 & - 12
\end{array} \right]$

Quizplus · Accepted Answer

The determinant of the matrix is $ad - bc = (-12)(2) - (3)(5) = -24 - 15 = -39$. Using the inverse formula, we swap $a$ and $d$, change the signs of $b$ and $c$, and divide by the determinant, resulting in $- \frac { 1 } { 39 } \left[ \begin{array} { c c } 2 & -3 \ -5 & -12 \end{array} \right]$.

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