Guided Example #1 - Inverse
Find the inverse, 
A−1, of 
A=|4253|
a. 
|32−1−522|
b. 
|−4−2−5−3|
c. 
|3−2−54|
d. None of the above
The correct answer is A. To see why, view each of the steps below.
Change signs of the other 2 elements:
|3−2−54|
Find the determinant of A
-5%255Cleft(2%255Cright)%253D12-10%253D2 "\begin{vmatrix}4&2\\5&3\end{vmatrix}=4\left(3\right)-5\left(2\right)=12-10=2")
|4253|=4(3)−5(2)=12−10=2
Multiply result of [2] by 
1detA
![LaTeX: A^{-1}=\frac{1}{detA}[\left(3,-2\right),\left(-5,4\right)]=\frac{1}{2}\begin{vmatrix}3&-2\\-5&4\end{vmatrix}=\begin{vmatrix}\frac{3}{2}&-1\\\frac{-5}{2}&2\end{vmatrix}](https://uciunex.instructure.com/equation_images/A%255E%257B-1%257D%253D%255Cfrac%257B1%257D%257BdetA%257D%255B%255Cleft(3%252C-2%255Cright)%252C%255Cleft(-5%252C4%255Cright)%255D%253D%255Cfrac%257B1%257D%257B2%257D%255Cbegin%257Bvmatrix%257D3%2526-2%255C%255C-5%25264%255Cend%257Bvmatrix%257D%253D%255Cbegin%257Bvmatrix%257D%255Cfrac%257B3%257D%257B2%257D%2526-1%255C%255C%255Cfrac%257B-5%257D%257B2%257D%25262%255Cend%257Bvmatrix%257D "A^{-1}=\frac{1}{detA}[\left(3,-2\right),\left(-5,4\right)]=\frac{1}{2}\begin{vmatrix}3&-2\\-5&4\end{vmatrix}=\begin{vmatrix}\frac{3}{2}&-1\\\frac{-5}{2}&2\end{vmatrix}")
A−1=1detA[(3,−2),(−5,4)]=12|3−2−54|=|32−1−522|.
Change signs of the other 2 elements:
|3−2−54|
Find the determinant of A
|4253|=4(3)−5(2)=12−10=2
Multiply result of [2] by 1detA
A−1=1detA[(3,−2),(−5,4)]=12|3−2−54|=|32−1−522|.