Find the inverse of the matrix #{[-1,2,1],[0,1,1],[1,0,2]}# using row reduction?

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Answer 1 · 2018-02-01T20:06:05

see below sorry for this long answer!

start by augmenting the matrix with the identity, and row reduce both matrices until the RHS is the identity

disclaimer: this may not be the quickest way to the inverse!

#((-1,2,1),(0,1,1),(1,0,2))((|),(|),(|))((1,0,0),(0,1,0),(0,0,1))#

#R'_1=-R_1#

#((1,-2,-1),(0,1,1),(1,0,2))##((|),(|),(|))# #((-1,0,0),(0,1,0),(0,0,1))#

#R'_2=2R_2#

#((1,-2,-1),(0,2,2),(1,0,2))##((|),(|),(|))# #((-1,0,0),(0,2,0),(0,0,1))#

#R'_1=R_1+R_2#

#((1,0,1),(0,2,2),(1,0,2))##((|),(|),(|))# #((-1,2,0),(0,2,0),(0,0,1))#

#R'_2=1/2R_2#
#((1,0,1),(0,1,1),(1,0,2))##((|),(|),(|))# #((-1,2,0),(0,1,0),(0,0,1))#

#R_3=R_3-R_1#

#((1,0,1),(0,1,1),(0,0,1))##((|),(|),(|))# #((-1,2,0),(0,1,0),(1,-2,1))#

#R'_2=R_2-R_3#

#((1,0,1),(0,1,0),(0,0,1))##((|),(|),(|))# #((-1,2,0),(-1,3,-1),(1,-2,1))#

#R'_1=R_1-R_3#

#((1,0,0),(0,1,0),(0,0,1))##((|),(|),(|))# #((-2,4,-1),(-1,3,-1),(1,-2,1))#

#:. " inverse is"#

#((-2,4,-1),(-1,3,-1),(1,-2,1))#