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

Jan 7, 2017

${A}^{-} 1 = \left(\begin{matrix}0 & - \frac{1}{2} & 1 \\ - 1 & \frac{3}{2} & - 1 \\ 1 & - \frac{1}{2} & 0\end{matrix}\right)$

#### Explanation:

A matrix, $\setminus A$, is invertible if and only if its determinant $| A | \ne 0$. There are several ways to invert a matrix inclining finding the adjoint, row reduction or even a calculator. I will use the adjoint method, which has several steps;

• Calculating the Matrix of Minors,
• Form the Matrix of Cofactors, $c o f \left(A\right)$
• Form the adjoint matrix, $a \mathrm{dj} \left(A\right)$
• Multiply $a \mathrm{dj} \left(A\right)$ by $\frac{1}{|} A |$ to form the inverse ${A}^{-} 1$

At some point we need to calculate $| A |$, or $\det \left(A\right)$, and this can also be used to test if the matrix is actually invertible so I prefer to do this first;

$A = \left(\begin{matrix}1 & 1 & 2 \\ 2 & 2 & 2 \\ 2 & 1 & 1\end{matrix}\right)$,

If we expand about the first row and "strike out" the row and column to form a smaller determinant and alternate signs we get;

$| A | = + \left(1\right) | \left(2 , 2\right) , \left(1 , 1\right) | - \left(1\right) | \left(2 , 2\right) , \left(2 , 1\right) | + \left(2\right) | \left(2 , 2\right) , \left(2 , 1\right) |$
$\setminus \setminus \setminus \setminus \setminus = \left\{\left(2\right) \left(1\right) - \left(1\right) \left(2\right)\right\} - \left\{\left(2\right) \left(1\right) - \left(2\right) \left(2\right)\right\} + 2 \left\{\left(2\right) \left(1\right) - \left(2\right) \left(2\right)\right\}$
$\setminus \setminus \setminus \setminus \setminus = 0 - \left(- 2\right) + 2 \left(- 2\right)$
$\setminus \setminus \setminus \setminus \setminus = - 2$

As $A \ne 0 \implies A$ is invertible, so we now calculate the matrix of minors by systematically working through each element in the matrix and "strike out" that row and columns and form the determinant of the remaining elements, as follows:

M=( ( |(2, 2), (1, 1)|, |(2, 2), (2, 1)|, |(2, 2), (2, 1)| ), ( |(1, 2), (1, 1)|, |(1, 2), (2, 1)|, |(1, 1), (2, 1)| ), ( |(1, 2), (2, 2)|, |(1, 2), (2, 2)|, |(1, 1), (2, 2)| ) )
$\setminus \setminus \setminus \setminus = \left(\begin{matrix}0 & - 2 & - 2 \\ - 1 & - 3 & - 1 \\ - 2 & - 2 & 0\end{matrix}\right)$

We now form the matrix of cofactors, $c o f \left(A\right)$, by taking the above matrix of minors and applying the alternate sign matrix as in

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

Where we change the sign of those elements with the minus sign to get;

$c o f \left(A\right) = \left(\begin{matrix}0 & 2 & - 2 \\ 1 & - 3 & 1 \\ - 2 & 2 & 0\end{matrix}\right)$

Then we form the adjoint matrix by transposing the matrix of cofactors, $c o f \left(A\right)$, so;

$a \mathrm{dj} \left(A\right) = c o f {\left(A\right)}^{T}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {\left(\begin{matrix}0 & 2 & - 2 \\ 1 & - 3 & 1 \\ - 2 & 2 & 0\end{matrix}\right)}^{T}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(\begin{matrix}0 & 1 & - 2 \\ 2 & - 3 & 2 \\ - 2 & 1 & 0\end{matrix}\right)$

And then finally we multiply by the reciprocal of the determinant to get:

${A}^{-} 1 = \frac{1}{|} A | a \mathrm{dj} \left(A\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(- \frac{1}{2}\right) \left(\begin{matrix}0 & 1 & - 2 \\ 2 & - 3 & 2 \\ - 2 & 1 & 0\end{matrix}\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(\begin{matrix}0 & - \frac{1}{2} & 1 \\ - 1 & \frac{3}{2} & - 1 \\ 1 & - \frac{1}{2} & 0\end{matrix}\right)$

We can easily check if this is the correct answer as $A {A}^{-} 1 = I$, So
$A {A}^{-} 1 = \left(\begin{matrix}1 & 1 & 2 \\ 2 & 2 & 2 \\ 2 & 1 & 1\end{matrix}\right) \left(\begin{matrix}0 & - \frac{1}{2} & 1 \\ - 1 & \frac{3}{2} & - 1 \\ 1 & - \frac{1}{2} & 0\end{matrix}\right)$
 \ \ \ \ \ \ \ \ \ \ = ( (0-1+2\ \ \ , -1/2+3/2-1\ \ \ , 1-1+0), (-1+3-2,-1+3-1, -2+3-1 ), (1-1+0, 1-1+0, 2-1+0))
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = I$ (or sometimes we write ${I}_{3}$).

Jan 7, 2017

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

#### Explanation:

Here's the row reduction way, less to remember.....

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

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

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

switch R2 and R3

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

$R 1 \to R 1 + 2 R 3 , R 2 \to R 2 - 3 R 3$

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

$R 1 \to R 1 + R 2 , R 2 \to - R 2 , R 3 \to - R$#

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

Jan 7, 2017

${A}^{- 1} = \left(\begin{matrix}0 & - \frac{1}{2} & 1 \\ - 1 & \frac{3}{2} & - 1 \\ 1 & - \frac{1}{2} & 0\end{matrix}\right)$

#### Explanation:

Here's one method:

Given:

$A = \left(\begin{matrix}1 & 1 & 2 \\ 2 & 2 & 2 \\ 2 & 1 & 1\end{matrix}\right)$

Make an augmented matrix by adding three columns containing the entries of a $3 \times 3$ identity matrix:

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

Perform a series of row operations to make the left half of this augmented matrix into an identity matrix.

Subtract row 3 from row 2 to get:

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

Add row 2 to row 3 to get:

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

Divide row 3 by $2$ to get:

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

Subtract row 1 from row 3 to get:

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

Subtract row 2 from row 1 to get:

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

Add row 3 to row 1 and row 2 ro get:

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

Multiply row 3 by $- 1$ to get:

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

Now we can read ${A}^{- 1}$ from the right hand $3$ columns:

${A}^{- 1} = \left(\begin{matrix}0 & - \frac{1}{2} & 1 \\ - 1 & \frac{3}{2} & - 1 \\ 1 & - \frac{1}{2} & 0\end{matrix}\right)$

One of the advantages of this method is that it works for square matrices of any size.