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Model Applied Situations with Matrix Operations
The \perp Shape in the Figure Below Is Shown Using 9 Pixels

Question 10

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Model Applied Situations with Matrix Operations
The \perp shape in the figure below is shown using 9 pixels in a 3×33 \times 3 grid. The color levels are given to the right of the figure. Use the matrix [131131333]\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 leaving the black alone and changing the light grey to dark grey. Use matrix addition to accomplish this.   A)     \left[\begin{array}{lll}1 & 3 & 1 \\ 1 & 3 & 1 \\ 3 & 3 & 3\end{array}\right]+\left[\begin{array}{lll}1 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 0 & 0\end{array}\right]=\left[\begin{array}{lll}2 & 3 & 2 \\ 2 & 3 & 2 \\ 3 & 3 & 3\end{array}\right]   B)    \left[\begin{array}{lll} 1 & 3 & 1 \\ 1 & 3 & 1 \\ 3 & 3 & 3 \end{array}\right]+\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right]=\left[\begin{array}{lll} 2 & 3 & 2 \\ 2 & 3 & 2 \\ 3 & 3 & 3 \end{array}\right]   C)    \left[\begin{array}{lll} 1 & 3 & 1 \\ 1 & 3 & 1 \\ 3 & 3 & 3 \end{array}\right]+\left[\begin{array}{rrr} 0 & -1 & 0 \\ 0 & -1 & 0 \\ -1 & -1 & -1 \end{array}\right]=\left[\begin{array}{lll} 1 & 2 & 1 \\ 1 & 2 & 1 \\ 3 & 2 & 2 \end{array}\right]   D)    \left[\begin{array}{lll} 1 & 3 & 1 \\ 1 & 3 & 1 \\ 3 & 3 & 3 \end{array}\right]+\left[\begin{array}{rrr} 1 & -1 & 1 \\ 1 & -1 & 1 \\ -1 & -1 & -1 \end{array}\right]=\left[\begin{array}{lll} 1 & 2 & 1 \\ 1 & 2 & 1 \\ 3 & 2 & 2 \end{array}\right]
-Adjust the contrast by leaving the black alone and changing the light grey to dark grey. Use matrix addition to accomplish this.


A)
[131131333]+[101101000]=[232232333] \left[\begin{array}{lll}1 & 3 & 1 \\ 1 & 3 & 1 \\ 3 & 3 & 3\end{array}\right]+\left[\begin{array}{lll}1 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 0 & 0\end{array}\right]=\left[\begin{array}{lll}2 & 3 & 2 \\ 2 & 3 & 2 \\ 3 & 3 & 3\end{array}\right]
B)
[131131333]+[111111111]=[232232333]\left[\begin{array}{lll}1 & 3 & 1 \\1 & 3 & 1 \\3 & 3 & 3\end{array}\right]+\left[\begin{array}{lll}1 & 1 & 1 \\1 & 1 & 1 \\1 & 1 & 1\end{array}\right]=\left[\begin{array}{lll}2 & 3 & 2 \\2 & 3 & 2 \\3 & 3 & 3\end{array}\right]
C)
[131131333]+[010010111]=[121121322]\left[\begin{array}{lll}1 & 3 & 1 \\1 & 3 & 1 \\3 & 3 & 3\end{array}\right]+\left[\begin{array}{rrr}0 & -1 & 0 \\0 & -1 & 0 \\-1 & -1 & -1\end{array}\right]=\left[\begin{array}{lll}1 & 2 & 1 \\1 & 2 & 1 \\3 & 2 & 2\end{array}\right]
D)
[131131333]+[111111111]=[121121322]\left[\begin{array}{lll}1 & 3 & 1 \\1 & 3 & 1 \\3 & 3 & 3\end{array}\right]+\left[\begin{array}{rrr}1 & -1 & 1 \\1 & -1 & 1 \\-1 & -1 & -1\end{array}\right]=\left[\begin{array}{lll}1 & 2 & 1 \\1 & 2 & 1 \\3 & 2 & 2\end{array}\right]

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