Let \( f: \mathbb A_{2 X 2} \rightarrow \mathbb R^4 \) be linear transform defined by:

\[ f( \begin{vmatrix}{x_{1}} && {x_{2}} \\ {x_{3}} && {x_{4}}\end{vmatrix}) = \left( \begin{array}{cccc} x_{1} - x_{2} + 2 * x_{3} + 4 * x_{4} \\ 2 * x_{1} - x_{2} + 4 * x_{3} + 3 * x_{4} \\ 4 * x_{1} - 3 * x_{2} + 2 * x_{3} + x_{4} \\ 5 * x_{1} - x_{2} + 2 * x_{3} + 2 * x_{4} \\ \end{array} \right) \]

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Let \[ A = \{ \begin{vmatrix}{1} && {2} \\ {3} && {4}\end{vmatrix}, \begin{vmatrix}{-2} && {1} \\ {5} && {3}\end{vmatrix}, \begin{vmatrix}{4} && {-2} \\ {1} && {7}\end{vmatrix}, \begin{vmatrix}{7} && {6} \\ {4} && {3}\end{vmatrix} \} \] be a basis for \( \mathbb A_{2 X 2} \)

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and,

\[ B = \{ \left( \begin{array}{cccc} 2 \\ -1 \\ 4 \\ 3 \\ \end{array} \right), \left( \begin{array}{cccc} 5 \\ 2 \\ 1 \\ 4 \\ \end{array} \right), \left( \begin{array}{cccc} 1 \\ -1 \\ 1 \\ 2 \\ \end{array} \right), \left( \begin{array}{cccc} 2 \\ 0 \\ 1 \\ -1 \\ \end{array} \right) \} \] be a basis for \( \mathbb R^4 \)

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If \( M = [a_{ij}]_{4 \: x \: 4} \) represents the linear transform above find \( \displaystyle S = \sum_{i = 1}^{4} \sum_{j = 1}^{4} a_{i j}.\) \(\)

Express the result to four decimal places.

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If your interested you can write a program(in any language) for the General Case below and use it to find the above result.

\(\) General Case:(For program)

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For this case I wrote the matrix in an unconventional manner. \(\)

Let \( f: \mathbb A_{n X n} \rightarrow \mathbb R^{n^2} \) be linear transform defined by:

\[ f( \begin{vmatrix}{x_{1}} && {x_{2}} && {...} && {x_{n}} \\
{x_{n + 1}} && {x_{n + 2}} && {...} && {x_{2 * n}} \\
{...} && {...} && {...} \\

{x_{(j - 1) * n + 1}} && {x_{(j - 1) * n + 2}} && {...} && {x_{j * n}} \\

{...} && {...} && {...} \\

{x_{(n - 1) * n + 1}} && {x_{(n - 1) * n + 2}} && {...} && {x_{n^2}} \\

\end{vmatrix}) = \left( \begin{array}{cccc}
C_{11} * x_{1} + C_{12} * x_{2} + ... + C_{1 * n^2} * x_{n^2} \\
C_{21} * x_{1} + C_{22} * x_{2} + ... + C_{2 * n^2} * x_{n^2} \\
... \\
C_{j1} * x_{1} + C_{j2} * x_{2} + ... + C_{j * n^2} * x_{n^2} \\
... \\
C_{n^21} * x_{1} + C_{n^22} * x_{2} + ... + C_{n^2n^2} * x_{n^2} \\
\end{array} \right) \]

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Let \[ V_{q} = \begin{vmatrix}{x_{1q}} && {x_{2q}} && {...} && {x_{nq}} \\
{x_{(n + 1)q}} && {x_{(n + 2)q}} && {...} && {x_{(2 * n)q}} \\
{...} && {...} && {...} \\

{x_{((j - 1) * n + 1)q}} && {x_{((j - 1) * n + 2)q}} && {...} && {x_{(j * n)q}} \\

{...} && {...} && {...} \\

{x_{((n - 1) * n + 1)q}} && {x_{((n - 1) * n + 2)q}} && {...} && {x_{(n^2)q}} \\

\end{vmatrix} \]

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and \( A = \{V_{q}|(1 <= q <= n^2)\} \) be a basis for \( \mathbb A_{n X n} \).

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Let \[ W_{q} = \left( \begin{array}{cccccc} a_{1q} \\ a_{2q} \\ ... \\ a_{jq} \\ ... \\ a_{(n^2)q} \\ \end{array} \right) \]

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and \( B = \{W_{q}|(1 <= q <= n^2)\} \) be a basis for \( \mathbb R^{n^2} \).

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Write a program in any language to find the matrix \( M = [a_{ij}]_{n^2 \: x \:n^2} \) representation of the general linear transform above and the sum \( \displaystyle S = \sum_{i = 1}^{n^2} \sum_{j = 1}^{n^2} a_{i j} \). \(\)

Make certain \( A \) and \( B \) are bases for \( \mathbb A_{n X n} \) and \( \mathbb R^{n^2} \). \(\)

You can use the program written to find the matrix \( M = [a_{ij}]_{4 \: x \: 4} \) that represents the linear transform above and output \( \displaystyle S = \sum_{i = 1}^{4} \sum_{j = 1}^{4} a_{i j} \).

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Let \[X = \begin{vmatrix}{3} && {-2} \\ {5} && {7}\end{vmatrix} \]

To check the matrix \( M = [a_{ij}]_{4 \: x \: 4} \) found, first find \([f(X)]_B\) without using the matrix \( M = [a_{ij}]_{4 \: x \: 4} \), then find \([f(X)]_B\) using the matrix \( M = [a_{ij}]_{4 \: x \: 4} \).

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