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The givenaugmented matrix of a consistent system is as follows:

\(\left[ {\begin{array}{*{20}{c}}1&{ - 4}&7&g\\0&3&{ - 5}&h\\{ - 2}&5&{ - 9}&k\end{array}} \right]\)

To eliminate the first term of the third row, perform an elementaryrow operationon the matrix \(\left[ {\begin{array}{*{20}{c}}1&{ - 4}&7&g\\0&3&{ - 5}&h\\{ - 2}&5&{ - 9}&k\end{array}} \right]\), as shown below.

Add two times of row one to row three; i.e., \({R_3} \to {R_3} + 2{R_1}\).

\(\left[ {\begin{array}{*{20}{c}}1&{ - 4}&7&g\\0&3&{ - 5}&h\\{ - 2 + 2\left( 1 \right)}&{5 + 2\left( { - 4} \right)}&{ - 9 + 2\left( 7 \right)}&{k + 2\left( g \right)}\end{array}} \right]\)

After the row operation, the matrix becomes:

\(\left[ {\begin{array}{*{20}{c}}1&{ - 4}&7&g\\0&3&{ - 5}&h\\0&{ - 3}&5&{k + 2g}\end{array}} \right]\)

To eliminate the second term of the third row, add the second row to the third row; i.e., \({R_3} \to {R_3} - 3{R_1}\).

\(\left[ {\begin{array}{*{20}{c}}1&{ - 4}&7&g\\0&3&{ - 5}&h\\0&0&0&{k + 2g + h}\end{array}} \right]\)

For the system to beconsistent, it should have uniqueorinfinitely many solutions.

Let c denote the number \(k + 2g + h\). Then the third equation, as represented by the augmented matrix above, is \(0 = b\). This equation is possible if and only if b is equal to zero. It means the original system has a solution if and only if \(k + 2g + h = 0\).

Hence, the required equation is \(2g + h + k = 0\).