Show that the matrices A = [1 2 -1 4] and B = [2/3 -1/3 1/6 1/6] are multiplicative inverses in M_2, 2.

Proof:Let's Multiply them![tex]AB = \left[\begin{array}{ccc}1&2\\-1&4\\\end{array}\right]\left[\begin{array}{ccc}2/3&-1/3\\1/6&1/6\\\end{array}\right]=\left[\begin{array}{ccc} 1(2/3) + 2(1/6) & 1(-1/3) + 2(1/6)\\-1(2/3) + 4(1/6)&-1(-1/3) + 4(1/6)\\\end{array}\right][/tex]If you don't remember how to multiply matrices, don't worry. In order to get AB we focus on the rows of A, and the columns of B. The first row of A is [1 2] and the first column of B is [2/3 1/6], if we do de dot multiplication we get 1(2/3) + 2(1/6) = 2/3 + 1/3 = 1.The first row of A is [1 2] and the second column of B is [-1/3 1/6]. Here we get 1(-1/3) + 2(1/6) = -1/3 +  1/3 = 0.The second row of A is [-1 4] and the first column of B is  [2/3 1/6]. Here we get -1(2/3) + 4(1/6) = -2/3 + 2/3 = 0.The second row of A is [-1 4] and the second column of B is [-1/3 1/6]. Here we get -1(-1/3) + 4(1/6) = 1/3 + 2/3 = 1.Then  [tex]AB = \left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right][/tex]We also have to see what happens with [tex]BA[/tex]. [tex]BA = \left[\begin{array}{ccc}2/3&-1/3\\1/6&1/6\\\end{array}\right]\left[\begin{array}{ccc}1&2\\-1&4\\\end{array}\right] = \left[\begin{array}{ccc} (2/3)(1) + (-1/3)(-1) & (2/3)(2) + (-1/3)(4)\\(1/6)(1) + (1/6)(-1)&(1/6)(2) + (1/6)(4)\\\end{array}\right]= \left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right][/tex]With this, we can conclude that [tex] AB = I = BA[/tex]where [tex]I[/tex] is the identity matrix. An by definition, that means that [tex]A[/tex] and  [tex]B[/tex] are multiplicative inverses.