Use the graphing calculator to solve if possibleA=begin{bmatrix}1 & 0&5 1 & -5&70&3&-4 end{bmatrix}B=begin{bmatrix}3 & -5&3 2&3&14&1&-3end{bmatrix}C=begin{bmatrix}5 & 2&3 2& -1&0 end{bmatrix}D=begin{bmatrix}5 -34 end{bmatrix}Find the value in row 2 column 3 of AB-3B

naivlingr 2021-03-09 Answered

Use the graphing calculator to solve if possible
\(A=\begin{bmatrix}1 & 0&5 \\1 & -5&7\\0&3&-4 \end{bmatrix}\\ B=\begin{bmatrix}3 & -5&3 \\2&3&1\\4&1&-3\end{bmatrix}\\ C=\begin{bmatrix}5 & 2&3 \\2& -1&0 \end{bmatrix}\\ D=\begin{bmatrix}5 \\-3\\4 \end{bmatrix}\)
Find the value in row 2 column 3 of \(AB-3B\)

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Expert Answer

sweererlirumeX
Answered 2021-03-10 Author has 12380 answers
Step 1
Given:
The given matrices are
\(A=\begin{bmatrix}1 & 0&5 \\1 & -5&7\\0&3&-4 \end{bmatrix}\)
\(B=\begin{bmatrix}3 & -5&3 \\2&3&1\\4&1&-3\end{bmatrix}\)
\(C=\begin{bmatrix}5 & 2&3 \\2& -1&0 \end{bmatrix}\)
\(D=\begin{bmatrix}5 \\-3\\4 \end{bmatrix}\)
To find:
The value in row 2 column 3 of AB-3B.
Step 2
The matrices are \(A=\begin{bmatrix}1 & 0&5 \\1 & -5&7\\0&3&-4 \end{bmatrix},B=\begin{bmatrix}3 & -5&3 \\2&3&1\\4&1&-3\end{bmatrix}\)
Now,
\(AB-3B=\begin{bmatrix}1 & 0&5 \\1 & -5&7\\0&3&-4 \end{bmatrix}\begin{bmatrix}3 & -5&3 \\2&3&1\\4&1&-3\end{bmatrix}-3\begin{bmatrix}3 & -5&3 \\2&3&1\\4&1&-3\end{bmatrix}\)
\(AB-3B=\begin{bmatrix}23 & 0&-12 \\21 & -13&-23\\-10&5&15 \end{bmatrix}-\begin{bmatrix}9 & -15&9 \\6&9&3\\12&3&-9\end{bmatrix}\)
\(AB-3B=\begin{bmatrix}14 & 15&-21 \\ 15&-22&-26\\-22&2&24\end{bmatrix}\)
The value in row 2 column 3 of AB-3B is -26.
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