Consider the following sets of matrices: M2(R) is the set of all 2 x 2 real matrices; GL2(R) is the subset of M2(R) with non-zero determinant: SL2(R) is the subset of GL2(R) with determinant 1. We know that multiplication is a binary operation on M2(R); show that it is an induced operation on the other 2 sets (You may freely use known facts from Math 3A for this).

Step-by-step explanation:

REcall the following definition of induced operation.

Let * be a binary operation over a set S and H a subset of S. If for every a,b elements in H it happens that a*b is also in H, then the binary operation that is obtained by restricting * to H is called the induced operation.

So, according to this definition, we must show that given two matrices of the specific subset, the product is also in the subset.

For this problem, recall this property of the determinant. Given A,B matrices in Mn(R) then det(AB) = det(A)*det(B).

Case SL2(R):

Let A,B matrices in SL2(R). Then, det(A) and det(B) is different from zero. So

.

So AB is also in SL2(R).

Case GL2(R):

Let A,B matrices in GL2(R). Then, det(A)= det(B)=1 is different from zero. So

.

So AB is also in GL2(R).

With these, we have proved that the matrix multiplication over SL2(R) and GL2(R) is an induced operation from the matrix multiplication over M2(R).

1.3

step-by-step explanation:

3x2=6   9x+6

bring down the 6 on both sides

6+6=12

12= 9x

divide by 9

12/9 =

1.3 =x