# How do you find the inverse of A=((3, 5), (2, 4))?

Feb 22, 2016

${\left[\begin{matrix}3 & 5 \\ 2 & 4\end{matrix}\right]}^{- 1} = \left[\begin{matrix}2 & - \frac{5}{2} \\ - 1 & \frac{3}{2}\end{matrix}\right]$

#### Explanation:

Find its determinant first.

$\det \left(A\right) = 3 \times 4 - 5 \times 2 = 2$

From here, there are many "different" ways of finding inverse matrix, here are 3 of them:

1) Shortcut for 2x2 matrix

For $A = \left[\begin{matrix}a & b \\ c & d\end{matrix}\right]$, the inverse can be found using the formula:

${A}^{- 1} = \frac{1}{\textcolor{g r e e n}{\det \left(A\right)}} \left[\begin{matrix}d & - b \\ - c & a\end{matrix}\right] = \frac{1}{\textcolor{g r e e n}{a d - b c}} \left[\begin{matrix}d & - b \\ - c & a\end{matrix}\right]$

So for $A = \left[\begin{matrix}3 & 5 \\ 2 & 4\end{matrix}\right]$,

${A}^{- 1} = \frac{1}{2} \left[\begin{matrix}4 & - 5 \\ - 2 & 3\end{matrix}\right] = \left[\begin{matrix}2 & - \frac{5}{2} \\ - 1 & \frac{3}{2}\end{matrix}\right]$

2) Augmented matrix method

Use Gauss-Jordan elimination to transform $\left[A | I\right]$ to $\left[I | {A}^{- 1}\right]$.

The following steps result in ${\left[\begin{matrix}3 & 5 \\ 2 & 4\end{matrix}\right]}^{- 1}$.

$\left[\begin{matrix}3 & 5 | 1 & 0 \\ 2 & 4 | 0 & 1\end{matrix}\right] \to \left[\begin{matrix}6 & 10 | 2 & 0 \\ 6 & 12 | 0 & 3\end{matrix}\right]$

$\to \left[\begin{matrix}6 & 10 | 2 & 0 \\ 0 & 2 | - 2 & 3\end{matrix}\right]$

$\to \left[\begin{matrix}6 & 0 | 12 & - 15 \\ 0 & 2 | - 2 & 3\end{matrix}\right]$

$\to \left[\begin{matrix}1 & 0 | 2 & - \frac{5}{2} \\ 0 & 1 | - 1 & \frac{3}{2}\end{matrix}\right]$

So we see that ${\left[\begin{matrix}3 & 5 \\ 2 & 4\end{matrix}\right]}^{- 1} = \left[\begin{matrix}2 & - \frac{5}{2} \\ - 1 & \frac{3}{2}\end{matrix}\right]$.

${A}^{- 1} = \frac{1}{\det \left(A\right)} \text{adj} \left(A\right)$

The adjoint of $n \times n$ matrix $A$ is another $n \times n$ matrix in which the $\left(i , j\right)$ entry is the $\left(j , i\right)$ cofactor of $A$.

The $\left(i , j\right)$ cofactor of A is the determinant of the $\left(n - 1\right) \times \left(n - 1\right)$ matrix that results from deleting row $i$ and column $j$ of $A$, multiplied by ${\left(- 1\right)}^{\left(i + j\right)}$. This means that you multiply the determinant by $- 1$ if $\left(i + j\right)$ is odd and you don't do anything if $\left(i + j\right)$ is even.

So for a $2 \times 2$ matrix, let $A = \left[\begin{matrix}a & b \\ c & d\end{matrix}\right]$, then the adjoint is given by

$\text{adj} \left(A\right) = {\left[\begin{matrix}\det \left(\left[d\right]\right) & - \det \left(\left[c\right]\right) \\ - \det \left(\left[b\right]\right) & \det \left(\left[a\right]\right)\end{matrix}\right]}^{T}$

$= {\left[\begin{matrix}d & - c \\ - b & a\end{matrix}\right]}^{T}$

$= \left[\begin{matrix}d & - b \\ - c & a\end{matrix}\right]$

So,

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

This is the "shortcut" we used in method 1).

$\left[\begin{matrix}3 & 5 \\ 2 & 4\end{matrix}\right] \left[\begin{matrix}2 & - \frac{5}{2} \\ - 1 & \frac{3}{2}\end{matrix}\right]$ and $\left[\begin{matrix}2 & - \frac{5}{2} \\ - 1 & \frac{3}{2}\end{matrix}\right] \left[\begin{matrix}3 & 5 \\ 2 & 4\end{matrix}\right]$
Both should give $\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]$.