How do you find the inverse of #((5, 2), (-1, a))#?

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Answer 1 · 2016-02-14T11:29:27

Use the formula for the inverse of a #2xx2# matrix to find:

#((5,2),(-1,a))^(-1) = ((a/(5a+2),-2/(5a+2)),(1/(5a+2),5/(5a+2)))#

In general the inverse of a #2xx2# matrix is given by the formula:

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

In our example, let's calculate the determinant first:

#abs((5,2),(-1,a)) = 5a+2#

So provided #a != -2/5# the determinant is non-zero and the inverse matrix can be written:

#((5,2),(-1,a))^(-1) = 1/(5a+2) ((a,-2),(1,5))=((a/(5a+2),-2/(5a+2)),(1/(5a+2),5/(5a+2)))#