Let A=[405−132] B=[111357] C=[2−301] Find 3A–B;C⋅B+A.

shadsiei

shadsiei

Answered question

2021-10-15

Let A=[405132]
B=[111357]
C=[2301]
Find 3AB;CB+A.

Answer & Explanation

Bentley Leach

Bentley Leach

Skilled2021-10-16Added 109 answers

A=[405132]
B=[111357]
C=[2301]
1) (3AB)=3×[405132][111357]=
=[12015396][111357]=
=[11114641]
2) (C)=[2301]=[2031]
(CB)=[2031]×[111357]=
=[222024]
(CB+A)=[222024]+[405132]=
=[627136]

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