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

A)

\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]

\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]

\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]

\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]

Solve the Problem A=[−12]A = \left[ \begin{array} { l l } - 1 & 2 \end{array} \right]A=[−1​2​]

Multiple Choice

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Solve the Problem $A = \left[ \begin{array} { l l } - 1 & 2 \end{array} \right]$

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