How do you solve #((8, 7), (1, 1))x=((3, -6), (-2, 9))#?

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Answer 1 · 2018-08-14T22:27:35

#x = ((17,-69),(-19,78))#

Let's try multiplying both sides by #((1, -7), (-1, 8))# and see what happens:

#((1, -7),(-1, 8))((8, 7),(1,1))x=((1,0),(0,1))x = x#

whereas:

#((1, -7),(-1,8))((3,-6),(-2,9)) = ((17,-69),(-19,78))#

So:

#x = ((1, -7),(-1, 8))((8, 7),(1,1))x = ((1, -7),(-1,8))((3,-6),(-2,9)) = ((17,-69),(-19,78))#

In general, the multiplicative inverse of #((a,b),(c,d))# is:

#1/abs((a,b),(c,d)) ((d, -b),(-c, a))#

and in our example:

#abs((8, 7),(1, 1)) = 8 * 1 - 7 * 1 = 8 - 7 = 1#