For the matrices #A= [(2,1),(3,2)]# and B#[(-2,-2),(3,2)]#, how to calculate A+B, BA and kA?

1 Answer
Jan 20, 2018

See the explanation below

Explanation:

#A=((2,1),(3,2))#

#B=((-2,-2),(3,2))#

#"Matrix addition"#

#((a,b),(c,d))+((e,f),(g,h))=((a+e,b+f),(c+g,d+h))#

#A+B=((2,1),(3,2))+((-2,-2),(3,2))=((0,-1),(6,4))#

#"Matrix multiplication"#

#((a,b),(c,d))xx((e,f),(g,h))=((ae+bg,af+bh),(ce+gd,cf+dh))#

#BxxA=((-2,-2),(3,2))*((-2,-2),(3,2))=((-2,0),(0,-2))=-2I#

#I=((1,0),(0,1))#

#"Multiplication by a constant"#

#k xx((a,b),(c,d))=((ka,kb),(kc,k))#

#kA=k((2,1),(3,2))=((2k,k),(3k,2k))#