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

May 21, 2016

${A}^{- 1} = \left(\begin{matrix}- 1 & - 1 & 4 \\ 1 & 2 & - 5 \\ 0 & - 1 & 2\end{matrix}\right)$

#### Explanation:

• Create an augmented matrix by adding three more columns to the right of $A$ in the form of an identity matrix.

$\left(\begin{matrix}1 & 2 & 3 \\ 2 & 2 & 1 \\ 1 & 1 & 1\end{matrix}\right) \to \left(\begin{matrix}1 & 2 & 3 & 1 & 0 & 0 \\ 2 & 2 & 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1\end{matrix}\right)$

• Perform row operations to make the left hand side of the augmented matrix into an identity matrix.

Add/subtract multiples of rows to/from other rows to make the left hand side into an upper triangular, then a diagonal matrix. Finally multiply rows by suitable factors to make the diagonal of the left hand side all $1$'s...

$\left(\begin{matrix}1 & 2 & 3 & 1 & 0 & 0 \\ 2 & 2 & 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1\end{matrix}\right)$

$\to \left(\begin{matrix}1 & 2 & 3 & 1 & 0 & 0 \\ 2 & 2 & 1 & 0 & 1 & 0 \\ 0 & - 1 & - 2 & - 1 & 0 & 1\end{matrix}\right)$

$\to \left(\begin{matrix}1 & 2 & 3 & 1 & 0 & 0 \\ 0 & - 2 & - 5 & - 2 & 1 & 0 \\ 0 & - 1 & - 2 & - 1 & 0 & 1\end{matrix}\right)$

$\to \left(\begin{matrix}1 & 2 & 3 & 1 & 0 & 0 \\ 0 & - 2 & - 5 & - 2 & 1 & 0 \\ 0 & 0 & \frac{1}{2} & 0 & - \frac{1}{2} & 1\end{matrix}\right)$

$\to \left(\begin{matrix}1 & 0 & - 2 & - 1 & 1 & 0 \\ 0 & - 2 & - 5 & - 2 & 1 & 0 \\ 0 & 0 & \frac{1}{2} & 0 & - \frac{1}{2} & 1\end{matrix}\right)$

$\to \left(\begin{matrix}1 & 0 & 0 & - 1 & - 1 & 4 \\ 0 & - 2 & - 5 & - 2 & 1 & 0 \\ 0 & 0 & \frac{1}{2} & 0 & - \frac{1}{2} & 1\end{matrix}\right)$

$\to \left(\begin{matrix}1 & 0 & 0 & - 1 & - 1 & 4 \\ 0 & - 2 & 0 & - 2 & - 4 & 10 \\ 0 & 0 & \frac{1}{2} & 0 & - \frac{1}{2} & 1\end{matrix}\right)$

$\to \left(\begin{matrix}1 & 0 & 0 & - 1 & - 1 & 4 \\ 0 & 1 & 0 & 1 & 2 & - 5 \\ 0 & 0 & 1 & 0 & - 1 & 2\end{matrix}\right)$

• Discard the first three columns to leave you with the inverse matrix.

$\to \left(\begin{matrix}- 1 & - 1 & 4 \\ 1 & 2 & - 5 \\ 0 & - 1 & 2\end{matrix}\right)$

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Note that this method will work for square matrices of arbitrary size.