1.3. Inverses of $2 \times 2$ matrices

Problem V-9. For the general $2 \times 2$ matrix $A = \left[\begin{array}{cc}a&b\ c&d\end{array}\right]$, show that $AB = ($det$A) I$, where $B = \left[\begin{array}{rr}d&-b\ -c&a\end{array}\right]$.

This gives a handy formula for the inverse of a $2 \times 2$ matrix, so learn it:

$\left[\begin{array}{cc}a&b\ c&d\end{array}\right]^{-1} = {{\frac{\displaysty... ...{\displaystyle \Delta}}} \left[\begin{array}{rr}d&-b\ -c& a\end{array}\right]$, where $\Delta = ad-bc$, if $\Delta \neq 0$.

Notice that the inverse is a scaled version of the $2 \times 2$ matrix obtained by switching the diagonal entries of the original matrix and and negating the off-diagonal entries (while not switching them).

Problem V-10. Re-do Problem V- using this formula.