Key to "Let A=[405−132], B=[111357], C=[2−301] Find: 3A−B, and C×B+A"

High School
Algebra
Algebra I
Algebra foundations

CMIIh

Answered question

2021-09-15

Let $A = [\begin{matrix} 4 & 0 & 5 \\ - 1 & 3 & 2 \end{matrix}]$ ,
$B = [\begin{matrix} 1 & 1 & 1 \\ 3 & 5 & 7 \end{matrix}]$ ,
$C = [\begin{matrix} 2 & - 3 \\ 0 & 1 \end{matrix}]$
Find: $3 A - B$ , and $C \times B + A$

Answer & Explanation

Sadie Eaton

Skilled2021-09-16Added 104 answers

$A = [\begin{matrix} 4 & 0 & 5 \\ - 1 & 3 & 2 \end{matrix}]$
$B = [\begin{matrix} 1 & 1 & 1 \\ 3 & 5 & 7 \end{matrix}]$
$C = [\begin{matrix} 2 & - 3 \\ 0 & 1 \end{matrix}]$
$(3 A - B) = [\begin{matrix} 4 & 0 & 5 \\ - 1 & 3 & 2 \end{matrix}] - [\begin{matrix} 1 & 1 & 1 \\ 3 & 5 & 7 \end{matrix}]$
$B = [\begin{matrix} 12 & 0 & 15 \\ - 3 & 9 & 6 \end{matrix}] - [\begin{matrix} 1 & 1 & 1 \\ 3 & 5 & 7 \end{matrix}]$
$(3 A - B) = [\begin{matrix} 11 & - 1 & 14 \\ - 6 & 4 & 1 \end{matrix}]$
$C \times = {[\begin{matrix} 2 & - 3 \\ 0 & 1 \end{matrix}]}^{t} = [\begin{matrix} 2 & 0 \\ - 3 & 1 \end{matrix}]$
$C \times B = [\begin{matrix} 2 & 0 \\ - 3 & 1 \end{matrix}] [\begin{matrix} 1 & 1 & 1 \\ 3 & 5 & 7 \end{matrix}]$
$= [\begin{matrix} 2 & 2 & 2 \\ 0 & 2 & 4 \end{matrix}]$
$C \times B + A = [\begin{matrix} 2 & 2 & 2 \\ 0 & 2 & 4 \end{matrix}] + [\begin{matrix} 4 & 0 & 5 \\ - 1 & 3 & 2 \end{matrix}]$
$= [\begin{matrix} 6 & 2 & 7 \\ - 1 & 5 & 6 \end{matrix}]$

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