Let M be a matrix and u and v vectors: $M =[(a, b),(c, d)], v = [(x), (y)], u =[(w), (z)].$ (a) Propose a definition for $u + v$ . (b) Show that your definition obeys $Mv + Mu = M(u + v)$ ?

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Answer 1 · 2016-07-20T03:50:58

Definition of addition of vectors, multiplication of a matrix by a vector and proof of distributive law are below.

For two vectors $v=[(x),(y)]$ and $u=[(w),(z)]$
we define an operation of addition as $u+v=[(x+w),(y+z)]$

Multiplication of a matrix $M=[(a,b),(c,d)]$ by vector $v=[(x),(y)]$ is defined as $M*v =[(a,b),(c,d)]*[(x),(y)] = [(ax+by),(cx+dy)]$

Analogously, multiplication of a matrix $M=[(a,b),(c,d)]$ by vector $u=[(w),(z)]$ is defined as $M*u =[(a,b),(c,d)]*[(w),(z)] = [(aw+bz),(cw+dz)]$

Let's check the distributive law of such definition:
$M*v+M*u= [(ax+by),(cx+dy)]+[(aw+bz),(cw+dz)]=$

$=[(ax+by+aw+bz),(cx+dy+cw+dz)]=$

$=[(a(x+w)+b(y+z)),(c(x+w)+d(y+z)))]=$

$= [(a,b),(c,d)] * [(x+w),(y+z)] = M*(v+u)$

End of proof.

Let M be a matrix and u and v vectors: M =[(a, b),(c, d)], v = [(x), (y)], u =[(w), (z)].M =[(a, b),(c, d)], v = [(x), (y)], u =[(w), (z)]. (a) Propose a definition for u + vu + v. (b) Show that your definition obeys Mv + Mu = M(u + v)Mv + Mu = M(u + v)?