# How do you find the inverse of A=((1, 2, 1), (2, 5, 4), (1, 4, 9)) ?

Mar 2, 2017

The inverse is $= \left(\begin{matrix}\frac{29}{4} & - \frac{7}{2} & \frac{3}{4} \\ - \frac{7}{2} & 2 & - \frac{1}{2} \\ \frac{3}{4} & - \frac{1}{2} & \frac{1}{4}\end{matrix}\right)$

#### Explanation:

For a matrix $A$ to be invertible, the determinant $\det A \ne 0$

Let's calculate

$\det A = | \left(1 , 2 , 1\right) , \left(2 , 5 , 4\right) , \left(1 , 4 , 9\right) |$

$= 1 \cdot | \left(5 , 4\right) , \left(4 , 9\right) | - 2 \cdot | \left(2 , 4\right) , \left(1 , 9\right) | + 1 \cdot | \left(2 , 5\right) , \left(1 , 4\right) |$

$= 1 \cdot \left(45 - 16\right) - 2 \cdot \left(18 - 4\right) + 1 \cdot \left(8 - 5\right)$

$= 29 - 28 + 3$

$= 4$

As $\det A \ne 0$, matrix $A$ is invertible

We calculate the matrix of cofactors

$C = \left(\begin{matrix}| \left(5 4\right) & \left(4 9\right) | & - | \left(2 4\right) & \left(1 9\right) | & | \left(2 5\right) & \left(1 4\right) | \\ - | \left(2 1\right) & \left(4 9\right) | & | \left(1 1\right) & \left(1 9\right) | & - | \left(1 2\right) & \left(1 4\right) | \\ | \left(2 1\right) & \left(5 4\right) | & - | \left(1 1\right) & \left(2 4\right) | & | \left(1 2\right) & \left(2 5\right) |\end{matrix}\right)$

$= \left(\begin{matrix}29 & - 14 & 3 \\ - 14 & 8 & - 2 \\ 3 & - 2 & 1\end{matrix}\right)$

The transpose of $C$ is

${C}^{T} = \left(\begin{matrix}29 & - 14 & 3 \\ - 14 & 8 & - 2 \\ 3 & - 7 & 1\end{matrix}\right)$

The inverse is

${A}^{-} 1 = \frac{1}{\det} A \cdot {C}^{T}$

$= \frac{1}{4} \left(\begin{matrix}29 & - 14 & 3 \\ - 14 & 8 & - 2 \\ 3 & - 2 & 1\end{matrix}\right)$

$= \left(\begin{matrix}\frac{29}{4} & - \frac{14}{4} & \frac{3}{4} \\ - \frac{14}{4} & \frac{8}{4} & - \frac{2}{4} \\ \frac{3}{4} & - \frac{2}{4} & \frac{1}{4}\end{matrix}\right)$

Verification

$A \cdot {A}^{-} 1 = \left(\begin{matrix}1 & 2 & 1 \\ 2 & 5 & 4 \\ 1 & 4 & 9\end{matrix}\right) \cdot \left(\begin{matrix}\frac{29}{4} & - \frac{14}{4} & \frac{3}{4} \\ - \frac{14}{4} & \frac{8}{4} & - \frac{2}{4} \\ \frac{3}{4} & - \frac{2}{4} & \frac{1}{4}\end{matrix}\right)$

$= \left(\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}\right)$

$= I$