Question

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

Answer 1

`A^(-1) = ((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:

Create an augmented matrix by adding `4` columns like a `4xx4` identity 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))`

Then perform a sequence of row operations to make the left hand `4xx4` matrix into an identity matrix:

Divide row 1 by `2` to get:

`((1,0,0,0,1/2,0,0,0),(1,2,0,0,0,1,0,0),(0,1,2,0,0,0,1,0),(0,0,1,2,0,0,0,1))`

Subtract row 1 from row 2 to get:

`((1,0,0,0,1/2,0,0,0),(0,2,0,0,-1/2,1,0,0),(0,1,2,0,0,0,1,0),(0,0,1,2,0,0,0,1))`

Divide row 2 by `2` to get:

`((1,0,0,0,1/2,0,0,0),(0,1,0,0,-1/4,1/2,0,0),(0,1,2,0,0,0,1,0),(0,0,1,2,0,0,0,1))`

Subtract row 2 from row 3 to get:

`((1,0,0,0,1/2,0,0,0),(0,1,0,0,-1/4,1/2,0,0),(0,0,2,0,1/4,-1/2,1,0),(0,0,1,2,0,0,0,1))`

Divide row 3 by `2` to get:

`((1,0,0,0,1/2,0,0,0),(0,1,0,0,-1/4,1/2,0,0),(0,0,1,0,1/8,-1/4,1/2,0),(0,0,1,2,0,0,0,1))`

Subtract row 3 from row 4 to get:

`((1,0,0,0,1/2,0,0,0),(0,1,0,0,-1/4,1/2,0,0),(0,0,1,0,1/8,-1/4,1/2,0),(0,0,0,2,-1/8,1/4,-1/2,1))`

Divide row 4 by `2` to get:

`((1,0,0,0,1/2,0,0,0),(0,1,0,0,-1/4,1/2,0,0),(0,0,1,0,1/8,-1/4,1/2,0),(0,0,0,1,-1/16,1/8,-1/4,1/2))`

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

`((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))`

This method works for square matrices of any size.

Answer 2

#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 = ((0,0,0,0),(1,0,0,0),(0,1,0,0),(0,0,1,0))` we have

`A = 2I_4+L_1` where `I_4` is the `4`-order identity matrix.

Now the `A` inverse is a matrix

`B=((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}))`

such that

`AcdotB = I_4` or

`2B+L_1cdotB=I_4`

calling now

`i_0 = 2B+L_1cdotB-I_4`
`i_1 = L_1cdot i_0`
`i_2 = L_1cdot i_1`
`i_3 = L_1cdot i_2`

As we can verify

`L_1cdotB= ((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}))`

then

`i_3 =((0,0,0,0),(0,0,0,0),(0,0,0,0),(2b_{11}-1,2b_{12},2b_{13},2b_{14}))=0_4`

and

`b_{11}= 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} = -1/4, b_{22} = 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) )#