For what value(s) of x does A^-1 exist.

[thatguythere]

1. Homework Statement
Let A =
(x+1 2 )
(x+5 x+1)

For what value(s) of x, if any, does A^-1 exist?

Homework Equations

The Attempt at a Solution

So let me see if I get the gist of this. An n x n matrix is invertible if and only if rank (A) = n. So unless there is a value of x which would make rank A < 2, then it is invertible. Since I can see no way to reduce both x+5 and x+1 to zero using one value, then A is invertible for all real numbers?

 

[JPaquim]

I don't know if you have learned the concept of determinants, but a matrix A is invertible if and only if its determinant is nonzero. The determinant for 2x2 matrices of the form
(a b)
(c d)
is given by det A = ad - bc

Computing the determinant and factoring the quadratic polynomial you'll get, will give you the answer.

 

[thatguythere]

Aha. So, det(A) = ((x+1)(x+1))-(2(x+5)) = x²-9 = (x+3)(x-3)
So A^-1 exists for all values ≠ 3, -3.