Question

We’ve shown that matrix multiplication is generally not commutative, meaning that as a general rule for two

Answer 1

Answer:

a) Observe that

$G=\left[\begin{array}{ccc}2&6\\4&0\end{array}\right] =2\left[\begin{array}{ccc}1&3\\2&0\end{array}\right] =2F$

Then, $FG=F(2F)=2FF, \text{ and } GF=(2F)F=2FF$

b) The zero matrix satisfies that for every matrix B such that the product is well defined, $0B=0=B0.$

Since the matrix G is the zero matrix then $FG=F0=0F=GF$

c) The identity(Id) matrix satisfies that for  that for every matrix B such that the product is well defined Id*B=B=B*Id. Observe that G is the identity matrix, then FG=F*Id=F=Id*F=GF

d) Observe that

$G=\left[\begin{array}{ccc}3&0&0\\0&3&0\\0&0&3\end{array}\right] =3\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] =3Id$.

Then $GF=3Id*F=3F\\FG=F(3Id)=3F*Id=3F$