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

2 Answers
Jul 16, 2016

Answer:

#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.

Jul 16, 2016

Answer:

#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) )#