# How do you find the inverse of A=((2, 0, 0, 0), (1, 2, 0, 0), (0, 1, 2, 0), (0, 0, 1, 2))?

Jul 16, 2016

${A}^{- 1} = \left(\begin{matrix}\frac{1}{2} & 0 & 0 & 0 \\ - \frac{1}{4} & \frac{1}{2} & 0 & 0 \\ \frac{1}{8} & - \frac{1}{4} & \frac{1}{2} & 0 \\ - \frac{1}{16} & \frac{1}{8} & - \frac{1}{4} & \frac{1}{2}\end{matrix}\right)$

#### Explanation:

Create an augmented matrix by adding $4$ columns like a $4 \times 4$ identity matrix:

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

Then perform a sequence of row operations to make the left hand $4 \times 4$ matrix into an identity matrix:

Divide row 1 by $2$ to get:

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

Subtract row 1 from row 2 to get:

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

Divide row 2 by $2$ to get:

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

Subtract row 2 from row 3 to get:

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

Divide row 3 by $2$ to get:

$\left(\begin{matrix}1 & 0 & 0 & 0 & \frac{1}{2} & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & - \frac{1}{4} & \frac{1}{2} & 0 & 0 \\ 0 & 0 & 1 & 0 & \frac{1}{8} & - \frac{1}{4} & \frac{1}{2} & 0 \\ 0 & 0 & 1 & 2 & 0 & 0 & 0 & 1\end{matrix}\right)$

Subtract row 3 from row 4 to get:

$\left(\begin{matrix}1 & 0 & 0 & 0 & \frac{1}{2} & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & - \frac{1}{4} & \frac{1}{2} & 0 & 0 \\ 0 & 0 & 1 & 0 & \frac{1}{8} & - \frac{1}{4} & \frac{1}{2} & 0 \\ 0 & 0 & 0 & 2 & - \frac{1}{8} & \frac{1}{4} & - \frac{1}{2} & 1\end{matrix}\right)$

Divide row 4 by $2$ to get:

$\left(\begin{matrix}1 & 0 & 0 & 0 & \frac{1}{2} & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & - \frac{1}{4} & \frac{1}{2} & 0 & 0 \\ 0 & 0 & 1 & 0 & \frac{1}{8} & - \frac{1}{4} & \frac{1}{2} & 0 \\ 0 & 0 & 0 & 1 & - \frac{1}{16} & \frac{1}{8} & - \frac{1}{4} & \frac{1}{2}\end{matrix}\right)$

We can then read off our inverse matrix from the last four columns:

$\left(\begin{matrix}\frac{1}{2} & 0 & 0 & 0 \\ - \frac{1}{4} & \frac{1}{2} & 0 & 0 \\ \frac{1}{8} & - \frac{1}{4} & \frac{1}{2} & 0 \\ - \frac{1}{16} & \frac{1}{8} & - \frac{1}{4} & \frac{1}{2}\end{matrix}\right)$

This method works for square matrices of any size.

Jul 16, 2016

B = ( (1/2, 0, 0, 0), (-1/4, 1/2, 0, 0), (1/8, -1/4, 1/2, 0), (-1/16, 1/8, -1/4, 1/2) )

#### Explanation:

Calling
${L}_{1} = \left(\begin{matrix}0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0\end{matrix}\right)$ we have

$A = 2 {I}_{4} + {L}_{1}$ where ${I}_{4}$ is the $4$-order identity matrix.

Now the $A$ inverse is a matrix

$B = \left(\begin{matrix}{b}_{12} & {b}_{13} & {b}_{14} & {b}_{15} \\ {b}_{21} & {b}_{22} & {b}_{23} & {b}_{24} \\ {b}_{31} & {b}_{32} & {b}_{33} & {b}_{34} \\ {b}_{41} & {b}_{42} & {b}_{43} & {b}_{44}\end{matrix}\right)$

such that

$A \cdot B = {I}_{4}$ or

$2 B + {L}_{1} \cdot B = {I}_{4}$

calling now

${i}_{0} = 2 B + {L}_{1} \cdot B - {I}_{4}$
${i}_{1} = {L}_{1} \cdot {i}_{0}$
${i}_{2} = {L}_{1} \cdot {i}_{1}$
${i}_{3} = {L}_{1} \cdot {i}_{2}$

As we can verify

${L}_{1} \cdot B = \left(\begin{matrix}0 & 0 & 0 & 0 \\ {b}_{12} & {b}_{13} & {b}_{14} & {b}_{15} \\ {b}_{21} & {b}_{22} & {b}_{23} & {b}_{24} \\ {b}_{31} & {b}_{32} & {b}_{33} & {b}_{34}\end{matrix}\right)$

then

${i}_{3} = \left(\begin{matrix}0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 2 {b}_{11} - 1 & 2 {b}_{12} & 2 {b}_{13} & 2 {b}_{14}\end{matrix}\right) = {0}_{4}$

and

${b}_{11} = \frac{1}{2} , {b}_{12} = 0 , {b}_{13} = 0 , {b}_{14} = 0$

substituting those values onto

${i}_{2} = {0}_{4}$ and solving we obtain

${b}_{21} = - \frac{1}{4} , {b}_{22} = \frac{1}{2} , {b}_{23} = 0 , {b}_{24} = 0$

following with this procedure for ${i}_{1}$ and ${i}_{0}$ we obtain

B = ( (1/2, 0, 0, 0), (-1/4, 1/2, 0, 0), (1/8, -1/4, 1/2, 0), (-1/16, 1/8, -1/4, 1/2) )