Let A=\begin{bmatrix}4 & 0 &5 \\-1 & 3 & 2

CMIIh 2021-09-15 Answered

Let \(A=\begin{bmatrix}4 & 0 &5 \\-1 & 3 & 2 \end{bmatrix}\),
\(B=\begin{bmatrix}1 & 1 &1 \\3 & 5 & 7 \end{bmatrix}\),
\(C=\begin{bmatrix}2 & -3 \\0 & 1 \end{bmatrix}\)
Find: \(\displaystyle{3}{A}-{B}\), and \(\displaystyle{C}\times{B}+{A}\)

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Expert Answer

Sadie Eaton
Answered 2021-09-16 Author has 18652 answers

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

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