Model Applied Situations with Matrix Operations
The

\perp

shape in the figure below is shown using 9 pixels in a

3 \times 3

grid. The color levels are given to the right of the figure. Use the matrix

\left[ \begin{array} { l l l } 1 & 3 & 1 \\ 1 & 3 & 1 \\ 3 & 3 & 3 \end{array} \right]

that represents a digital photograph of the

\perp

shape to solve the problem. $Model Applied Situations with Matrix Operations The \perp shape in the figure below is shown using 9 pixels in a 3 \times 3 grid. The color levels are given to the right of the figure. Use the matrix \left[ \begin{array} { l l l } 1 & 3 & 1 \\ 1 & 3 & 1 \\ 3 & 3 & 3 \end{array} \right] that represents a digital photograph of the \perp shape to solve the problem. -Adjust the contrast by changing the black to white and the light grey to dark grey. Use matrix addition to accomplish this. A) \left[ \begin{array} { l l l } 1 & 3 & 1 \\ 1 & 3 & 1 \\ 3 & 3 & 3 \end{array} \right] + \left[ \begin{array} { r r r } 1 & - 3 & 1 \\ 1 & - 3 & 1 \\ - 3 & - 3 & - 3 \end{array} \right] = \left[ \begin{array} { l l l } 2 & 0 & 2 \\ 2 & 0 & 2 \\ 0 & 0 & 0 \end{array} \right] B) \left[ \begin{array} { l l l } 1 & 3 & 1 \\ 1 & 3 & 1 \\ 3 & 3 & 3 \end{array} \right] + \left[ \begin{array} { r r r } - 1 & 3 & - 1 \\ - 1 & 3 & - 1 \\ 3 & 3 & 3 \end{array} \right] = \left[ \begin{array} { l l l } 2 & 0 & 2 \\ 2 & 0 & 2 \\ 0 & 0 & 0 \end{array} \right] C) \left[ \begin{array} { l l l } 1 & 3 & 1 \\ 1 & 3 & 1 \\ 3 & 3 & 3 \end{array} \right] + \left[ \begin{array} { l l l } - 1 & - 1 & - 1 \\ - 1 & - 1 & - 1 \\ - 1 & - 1 & - 1 \end{array} \right] = \left[ \begin{array} { l l l } 0 & 2 & 0 \\ 0 & 2 & 0 \\ 2 & 2 & 2 \end{array} \right] D) \left[ \begin{array} { l l l } 1 & 3 & 1 \\ 1 & 3 & 1 \\ 3 & 3 & 3 \end{array} \right] + \left[ \begin{array} { r r r } 0 & - 3 & 0 \\ 0 & - 3 & 0 \\ - 3 & - 3 & - 3 \end{array} \right] = \left[ \begin{array} { l l l } 1 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 0 & 0 \end{array} \right]$
-Adjust the contrast by changing the black to white and the light grey to dark grey. Use matrix addition to accomplish this.
A)

\left[ \begin{array} { l l l } 1 & 3 & 1 \\ 1 & 3 & 1 \\ 3 & 3 & 3 \end{array} \right] + \left[ \begin{array} { r r r } 1 & - 3 & 1 \\ 1 & - 3 & 1 \\ - 3 & - 3 & - 3 \end{array} \right] = \left[ \begin{array} { l l l } 2 & 0 & 2 \\ 2 & 0 & 2 \\ 0 & 0 & 0 \end{array} \right]

\left[ \begin{array} { l l l } 1 & 3 & 1 \\ 1 & 3 & 1 \\ 3 & 3 & 3 \end{array} \right] + \left[ \begin{array} { r r r } - 1 & 3 & - 1 \\ - 1 & 3 & - 1 \\ 3 & 3 & 3 \end{array} \right] = \left[ \begin{array} { l l l } 2 & 0 & 2 \\ 2 & 0 & 2 \\ 0 & 0 & 0 \end{array} \right]

\left[ \begin{array} { l l l } 1 & 3 & 1 \\1 & 3 & 1 \\3 & 3 & 3\end{array} \right] + \left[ \begin{array} { l l l } - 1 & - 1 & - 1 \\- 1 & - 1 & - 1 \\- 1 & - 1 & - 1\end{array} \right] = \left[ \begin{array} { l l l } 0 & 2 & 0 \\0 & 2 & 0 \\2 & 2 & 2\end{array} \right]

\left[ \begin{array} { l l l } 1 & 3 & 1 \\1 & 3 & 1 \\3 & 3 & 3\end{array} \right] + \left[ \begin{array} { r r r } 0 & - 3 & 0 \\0 & - 3 & 0 \\- 3 & - 3 & - 3\end{array} \right] = \left[ \begin{array} { l l l } 1 & 0 & 1 \\1 & 0 & 1 \\0 & 0 & 0\end{array} \right]

Understand What Is Meant by Equal Matrices Find Values for the Variables

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Understand What Is Meant by Equal Matrices
Find Values for the Variables

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