Another way to solve square systems is to use the inverse of the coefficient matrix. For a 2 x 2 matrix, the inverse has a fairly simple form. Suppose A = the inverse of A is

a 0. Then Le di A-1 = 1_ 1_ d 61 det(A) (-ca]

Recall that det (A) = ad - bc. The above formula does not work if det (A) = 0, and in fact the matrix does not have an inverse in this case.

(a) Create your own 2 x 2 matrix A with non-zero entries and non-zero determinant. Verify that AA-1 = I and A-1A= I by multiplying the matrices in both orders. This shows that multiplying by the inverse is the matrix version of "dividing" by A. I is the identity matrix, which functions as one for multiplication purposes.

(b) Solve the system in problem 2(a) by using an inverse matrix. Why does the method not apply to parts (b) and (c)?

(c) For which value(s) of k will [2 k -1] 5]

fail to have an inverse?