How do you find the inverse of A=((1, 1, 1, 0), (1, 1, 0, -1), (0, 1, 0, 1), (0, 1, 1, 0))?

Mar 2, 2016

 A^(-1) = ( (1, 0, 0, -1), (-1/2, 1/2, 1/2, 1/2), (1/2, -1/2, -1/2, 1/2), (1/2, -1/2, 1/2, -1/2) )

Explanation:

You can compute the inverse by writing the identity matrix $I$ at the right side of your matrix $A$, like this:

( (1, 1, 1, 0, |, 1, 0, 0, 0), (1, 1, 0, -1, |, 0, 1, 0, 0), (0, 1, 0, 1, |, 0, 0, 1, 0), (0, 1, 1, 0, |, 0, 0, 0, 1) )

Afterwards you need to perform transformations (adding rows, multiplying rows with a scalar number or swapping two rows) to transform the matrix at the left side to the identity matrix.
If you perform exactly the same operations at the right side, you will gain the inverse matrix ${A}^{- 1}$ at the right side:

$\left(A | I\right) \to \left(I | {A}^{- 1}\right)$

Let me show you how to do it in this case:

( (1, 1, 1, 0, |, 1, 0, 0, 0), (1, 1, 0, -1, |, 0, 1, 0, 0), (0, 1, 0, 1, |, 0, 0, 1, 0), (0, 1, 1, 0, |, 0, 0, 0, 1) )

It is already nice that you have a $1$ in the first row, first column, and zeroes in the first column, third and fourth rows.

Eliminate the $1$ in the first column, second row by performing
$\left(- 1\right) \cdot I + I I \to \textcolor{b l u e}{I I}$:

=> ( (1, 1, 1, 0, |, 1, 0, 0, 0), (color(blue)(0), color(blue)(0), color(blue)(-1), color(blue)(-1), |, color(blue)(-1), color(blue)(1), color(blue)(0), color(blue)(0)), (0, 1, 0, 1, |, 0, 0, 1, 0), (0, 1, 1, 0, |, 0, 0, 0, 1) )

The first column is already finished. Let's concentrate on the second:

Swap the rows $I I$ and $I I I$ in order to have a $1$ in the second column, second row:

=> ( (1, 1, 1, 0, |, 1, 0, 0, 0), (0, 1, 0, 1, |, 0, 0, 1, 0), (0, 0, -1, -1, |, -1, 1, 0, 0), (0, 1, 1, 0, |, 0, 0, 0, 1) )

Eliminate the $1$ in the second column, fourth row by computing $\left(- 1\right) \cdot I I + I V \to \textcolor{\mathmr{and} a n \ge}{I V}$:

=> ( (1, 1, 1, 0, |, 1, 0, 0, 0), (0, 1, 0, 1, |, 0, 0, 1, 0), (0, 0, -1, -1, |, -1, 1, 0, 0), (color(orange)(0),color(orange)(0), color(orange)(1), color(orange)(-1), |, color(orange)(0), color(orange)(0), color(orange)(-1), color(orange)(1)) )

Swap the rows $I I I$ and $I V$ in order to have a $1$ in the third column, third row:

=> ( (1, 1, 1, 0, |, 1, 0, 0, 0), (0, 1, 0, 1, |, 0, 0, 1, 0), (0, 0, 1, -1, |, 0, 0, -1, 1), (0, 0, -1, -1, |, -1, 1, 0, 0) )

Eliminate the $- 1$ in the third column, fourth row by computing $I I I + I V \to I V$:

=> ( (1, 1, 1, 0, |, 1, 0, 0, 0), (0, 1, 0, 1, |, 0, 0, 1, 0), (0, 0, 1, -1, |, 0, 0, -1, 1), (0, 0, 0, -2, |, -1, 1, -1, 1) )

Generate a $1$ in the fourth column, fourth row by computing $\left(- \frac{1}{2}\right) \cdot I V \to \textcolor{red}{I V}$:

=> ( (1, 1, 1, 0, |, 1, 0, 0, 0), (0, 1, 0, 1, |, 0, 0, 1, 0), (0, 0, 1, -1, |, 0, 0, -1, 1), (color(red)(0), color(red)(0), color(red)(0), color(red)(1), |, color(red)(1/2), color(red)(-1/2), color(red)(1/2), color(red)(-1/2)) )

Now, the diagonal of the left matrix and all the elements below the diagonal are ready.
The next step would be eliminating all the non-zero values above the diagonal.

Eliminate the $- 1$ in the third row, fourth column by computing $I V + I I I \to \textcolor{v i o \le t}{I I I}$:

=> ( (1, 1, 1, 0, |, 1, 0, 0, 0), (0, 1, 0, 1, |, 0, 0, 1, 0), (color(violet)(0), color(violet)(0), color(violet)(1), color(violet)(0), |, color(violet)(1/2), color(violet)(-1/2), color(violet)(-1/2), color(violet)(1/2)), (0, 0, 0, 1, |, 1/2, -1/2, 1/2, -1/2) )

Eliminate the $1$ in the second row, fourth column by computing $\left(- 1\right) \cdot I V + I I \to \textcolor{g r e e n}{I I}$:

=> ( (1, 1, 1, 0, |, 1, 0, 0, 0), (color(green)(0), color(green)(1), color(green)(0), color(green)(0), |, color(green)(-1/2), color(green)(1/2), color(green)(1/2), color(green)(1/2)), (0, 0, 1, 0, |, 1/2, -1/2, -1/2, 1/2), (0, 0, 0, 1, |, 1/2, -1/2, 1/2, -1/2) )

Eliminate the $1$ in the first row, third column by computing $\left(- 1\right) \cdot I I I + I \to \textcolor{t u r q u o i s e}{I}$:

=> ( (color(turquoise)(1), color(turquoise)(1), color(turquoise)(0), color(turquoise)(0), |, color(turquoise)(1/2), color(turquoise)(1/2), color(turquoise)(1/2), color(turquoise)(-1/2)), (0, 1, 0, 0, |, -1/2, 1/2, 1/2, 1/2), (0, 0, 1, 0, |, 1/2, -1/2, -1/2, 1/2), (0, 0, 0, 1, |, 1/2, -1/2, 1/2, -1/2) )

And the last step: eliminate the $1$ in the first row, second column by computing $\left(- 1\right) \cdot I I + I \to \textcolor{b r o w n}{I}$:

=> ( (color(brown)(1), color(brown)(0), color(brown)(0), color(brown)(0), |, color(brown)(1), color(brown)(0), color(brown)(0), color(brown)(-1)), (0, 1, 0, 0, |, -1/2, 1/2, 1/2, 1/2), (0, 0, 1, 0, |, 1/2, -1/2, -1/2, 1/2), (0, 0, 0, 1, |, 1/2, -1/2, 1/2, -1/2) )

 A^(-1) = ( (1, 0, 0, -1), (-1/2, 1/2, 1/2, 1/2), (1/2, -1/2, -1/2, 1/2), (1/2, -1/2, 1/2, -1/2) )
You can check if the computation was correct by computing $A \cdot {A}^{- 1}$ or ${A}^{- 1} \cdot A$. In both cases, the identity matrix $I$ should be the result.