# How do you find the inverse of A=((2, 3, 1), (4, 3, 1), (1, 2, 4))?

Dec 8, 2016

${A}^{- 1} = \frac{1}{20} \left(\begin{matrix}- 10 & 10 & 0 \\ 15 & - 7 & - 2 \\ - 5 & 1 & 6\end{matrix}\right)$

#### Explanation:

In place of each element of $A = \left(\begin{matrix}2 & 3 & 1 \\ 4 & 3 & 1 \\ 1 & 2 & 4\end{matrix}\right)$ write down the determinant of the $2 \times 2$ matrix formed by deleting the row and column that the current element is in:

$\left(\begin{matrix}\left\mid \begin{matrix}3 & 1 \\ 2 & 4\end{matrix} \right\mid & \left\mid \begin{matrix}4 & 1 \\ 1 & 4\end{matrix} \right\mid & \left\mid \begin{matrix}4 & 3 \\ 1 & 2\end{matrix} \right\mid \\ \left\mid \begin{matrix}3 & 1 \\ 2 & 4\end{matrix} \right\mid & \left\mid \begin{matrix}2 & 1 \\ 1 & 4\end{matrix} \right\mid & \left\mid \begin{matrix}2 & 3 \\ 1 & 2\end{matrix} \right\mid \\ \left\mid \begin{matrix}3 & 1 \\ 3 & 1\end{matrix} \right\mid & \left\mid \begin{matrix}2 & 1 \\ 4 & 1\end{matrix} \right\mid & \left\mid \begin{matrix}2 & 3 \\ 4 & 3\end{matrix} \right\mid\end{matrix}\right) = \left(\begin{matrix}10 & 15 & 5 \\ 10 & 7 & 1 \\ 0 & - 2 & - 6\end{matrix}\right)$

Invert the signs of alternate elements in a checkboard pattern:

$\left(\begin{matrix}+ & - & + \\ - & + & - \\ + & - & +\end{matrix}\right)$

to get:

$\left(\begin{matrix}10 & - 15 & 5 \\ - 10 & 7 & - 1 \\ 0 & 2 & - 6\end{matrix}\right)$

Transpose to get:

$\left(\begin{matrix}10 & - 10 & 0 \\ - 15 & 7 & 2 \\ 5 & - 1 & - 6\end{matrix}\right)$

This will be a scalar multiple of the inverse matrix - the multiplier being $\left\mid A \right\mid$.

So we find:

$\left(\begin{matrix}2 & 3 & 1 \\ 4 & 3 & 1 \\ 1 & 2 & 4\end{matrix}\right) \left(\begin{matrix}10 & - 10 & 0 \\ - 15 & 7 & 2 \\ 5 & - 1 & - 6\end{matrix}\right) = \left(\begin{matrix}- 20 & 0 & 0 \\ 0 & - 20 & 0 \\ 0 & 0 & - 20\end{matrix}\right)$

Hence:

${A}^{- 1} = - \frac{1}{20} \left(\begin{matrix}10 & - 10 & 0 \\ - 15 & 7 & 2 \\ 5 & - 1 & - 6\end{matrix}\right) = \frac{1}{20} \left(\begin{matrix}- 10 & 10 & 0 \\ 15 & - 7 & - 2 \\ - 5 & 1 & 6\end{matrix}\right)$