Find the inverse of the matrix #[[6,-2,-3],[-1,8,-7],[4,-4,6]]#?

2 Answers
Sep 2, 2017

Answer:

The inverse is #=((5/62,3/31,19/124),(-11/124,6/31,45/248),(-7/62,2/31,23/124))#

Explanation:

The matrix is #A=((6,-2,-3),(-1,8,-7),(4,-4,6))#

We calculate the inverse of the matrix as follows

#((6,-2,-3,:,1,0,0),(-1,8,-7,:,0,1,0),(4,-4,6,:,0,0,1))#

#R3larrR3+4R2#, #=>#, #((6,-2,-3,:,1,0,0),(-1,8,-7,:,0,1,0),(0,28,-22,:,0,4,1))#

#R2larrR1+6R2#, #=>#, #((6,-2,-3,:,1,0,0),(0,46,-45,:,1,6,0),(0,28,-22,:,0,4,1))#

#R1larr(R1)/6#, #=>#, #((1,-1/3,-1/2,:,1/6,0,0),(0,46,-45,:,1,6,0),(0,28,-22,:,0,4,1))#

#R2larr(R2)/46#, #=>#, #((1,-1/3,-1/2,:,1/6,0,0),(0,1,-45/46,:,1/46,3/23,0),(0,28,-22,:,0,4,1))#

#R3larr(R3)/28#, #=>#, #((1,-1/3,-1/2,:,1/6,0,0),(0,1,-45/46,:,1/46,3/23,0),(0,1,-11/14,:,0,1/7,1/28))#

#R3larr(R3-R2)#, #=>#, #((1,-1/3,-1/2,:,1/6,0,0),(0,1,-45/46,:,1/46,3/23,0),(0,0,31/161,:,-1/46,2/161,1/28))#

#R3larr(R3)*(161/31)#, #=>#, #((1,-1/3,-1/2,:,1/6,0,0),(0,1,-45/46,:,1/46,3/23,0),(0,0,1,:,-7/62,2/31,23/124))#

#R2larr(R2)+(45/46)R3#, #=>#, #((1,-1/3,-1/2,:,1/6,0,0),(0,1,0,:,-11/124,6/31,45/248),(0,0,1,:,-7/62,2/31,23/124))#

#R1larr(R1)+(1/3)R2#, #=>#, #((1,0,-1/2,:,17/124,2/31,15/248),(0,1,0,:,-11/124,6/31,45/248),(0,0,1,:,-7/62,2/31,23/124))#

#R1larr(R1)+(1/2)R3#, #=>#, #((1,0,0,:,5/62,3/31,19/124),(0,1,0,:,-11/124,6/31,45/248),(0,0,1,:,-7/62,2/31,23/124))#

Sep 2, 2017

Answer:

#((6, -2, -3),(-1,8,-7),(4,-4,6))^(-1) = 1/248((20,24,38),(-22,48,45),(-28,16,46))#

Explanation:

Given:

#((6, -2, -3),(-1,8,-7),(4,-4,6))#

The nice thing about the augmented matrix approach is that it works for matrices of any size, but in the case of #3xx3# matrices it is practical to calculate the adjunct matrix...

Consider the matrix formed by replacing each entry by the determinant of the #2xx2# matrix formed by the entries of the other rows and columns, reckoned cyclically:

#((abs((8,-7),(-4,6)), abs((-7,-1),(6,4)), abs((-1,8),(4,-4))),(abs((-4, 6),(-2,-3)), abs((6,4),(-3,6)), abs((4,-4),(6,-2))),(abs((-2,-3),(8,-7)), abs((-3,6),(-7,-1)), abs((6,-2),(-1,8))))#

#=((48-28, -28+6, 4-32),(12+12,36+12,-8+24),(14+24,3+42,48-2))#

#=((20,-22,-28),(24,48,16),(38,45,46))#

Transpose this matrix to get:

#((20,24,38),(-22,48,45),(-28,16,46))#

Multiplying by the original we find:

#((6, -2, -3),(-1,8,-7),(4,-4,6))((20,24,38),(-22,48,45),(-28,16,46)) = ((248,0,0),(0,248,0),(0,0,248))#

So:

#((6, -2, -3),(-1,8,-7),(4,-4,6))^(-1) = 1/248((20,24,38),(-22,48,45),(-28,16,46))#