Question

Find the elementary row operation that transforms the first matrix into the second, and then find the reverse row operation that transforms the second matrix into the first.

30.\(\left[ {\begin{array}{*{20}{c}}1&3&{ - 4}\\0&{ - 2}&6\\0&{ - 5}&9\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}1&3&{ - 4}\\0&1&{ - 3}\\0&{ - 5}&9\end{array}} \right]\)

Short answer

The row operation to transform the first matrix into the second is the multiplication of row two by \( - \frac{1}{2}\), i.e., \({R_2} \to - \frac{1}{2}{R_2}\) in the first matrix. And the reverse row operation is the multiplication of row two by \( - 2\), i.e., \({R_2} \to - 2{R_2}\) in the second matrix.

Step-by-step solution

Rewrite the given matrices

The given matrices are:

\(\left[ {\begin{array}{*{20}{c}}1&3&{ - 4}\\0&{ - 2}&6\\0&{ - 5}&9\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}1&3&{ - 4}\\0&1&{ - 3}\\0&{ - 5}&9\end{array}} \right]\)

Apply the elementary row operation

A basic principle states that row operations do not affect the solution set of a linear system.

Perform an elementaryrow operation.

Multiply row two by \( - \frac{1}{2}\) in the first matrix, \(\left[ {\begin{array}{*{20}{c}}1&3&{ - 4}\\0&{ - 2}&6\\0&{ - 5}&9\end{array}} \right]\), and transform it to the second matrix, \(\left[ {\begin{array}{*{20}{c}}1&3&{ - 4}\\0&1&{ - 3}\\0&{ - 5}&9\end{array}} \right]\).

Apply the elementary row operation

A basic principle states that row operations do not affect the solution set of a linear system.

Perform an elementary row operation.

Multiply row two by \( - 2\) in the second matrix, \(\left[ {\begin{array}{*{20}{c}}1&3&{ - 4}\\0&1&{ - 3}\\0&{ - 5}&9\end{array}} \right]\), and transform it to the first matrix, \(\left[ {\begin{array}{*{20}{c}}1&3&{ - 4}\\0&{ - 2}&6\\0&{ - 5}&9\end{array}} \right]\).

Thus, the row operation is the multiplication of row two by \( - \frac{1}{2}\) in the first matrix. And the reverse row operation is the multiplication of row two by \( - 2\) in the second matrix.