Question

In Exercises 7–10, the augmented matrix of a linear system has been reduced by row operations to the form shown. In each case, continue the appropriate row operations and describe the solution set of the original system.

8. \(\left( {\begin{aligned}{*{20}{c}}1&{ - 4}&9&0\\0&1&7&0\\0&0&2&0\end{aligned}} \right)\)

Short answer

The solution set of the linear system contains one solution; i.e., \(\left( {\begin{aligned}{*{20}{c}}0\\0\\0\end{aligned}} \right)\).

Step-by-step solution

Rewrite the given augmented matrix of a linear system

The augmented matrix of a linear system is given as

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 4}&9&0\\0&1&7&0\\0&0&2&0\end{aligned}} \right)\)

Perform elementary row operations

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

To obtain \(z\) in the third equation, perform an elementary row operationon the matrix \(\left( {\begin{aligned}{*{20}{c}}1&{ - 4}&9&0\\0&1&7&0\\0&0&2&0\end{aligned}} \right)\) as shown below.

Divide the third row by 2; i.e., \({R_3} \to \frac{1}{2}{R_3}\)to get

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 4}&9&0\\0&1&7&0\\{\frac{0}{2}}&{\frac{0}{2}}&{\frac{2}{2}}&{\frac{0}{2}}\end{aligned}} \right)\)

After the row operation, the matrix becomes

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 4}&9&0\\0&1&7&0\\0&0&1&0\end{aligned}} \right)\)

To eliminate the\(7z\)term in the second equation, perform an elementary row operationon the matrix \(\left( {\begin{aligned}{*{20}{c}}1&{ - 4}&9&0\\0&1&7&0\\0&0&1&0\end{aligned}} \right)\) as shown below.

Subtract 7 times the third row from the second row 2; i.e., \({R_2} \to {R_2} - 7{R_3}\).

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 4}&9&0\\{0 - 7\left( 0 \right)}&{1 - 7\left( 0 \right)}&{7 - 7\left( 1 \right)}&{0 - 7\left( 0 \right)}\\0&0&1&0\end{aligned}} \right)\)

After the row operation, the matrix becomes

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 4}&9&0\\0&1&0&0\\0&0&1&0\end{aligned}} \right)\)

Eliminate the second element of the first row

To eliminate the\( - 4y\)term in the first equation,perform an elementary row operationon the matrix \(\left( {\begin{aligned}{*{20}{c}}1&{ - 4}&9&0\\0&1&0&0\\0&0&1&0\end{aligned}} \right)\) as shown below.

Add 4 times the second row to the first row;i.e., \({R_1} \to {R_1} + 4{R_2}\).

\(\left( {\begin{aligned}{*{20}{c}}{1 + 4\left( 0 \right)}&{ - 4 + 4\left( 1 \right)}&{9 + 4\left( 0 \right)}&{0 + 4\left( 0 \right)}\\0&1&0&0\\0&0&1&0\end{aligned}} \right)\)

After the row operation, the matrix becomes

\(\left( {\begin{aligned}{*{20}{c}}1&0&9&0\\0&1&0&0\\0&0&1&0\end{aligned}} \right)\)

Eliminate the third element of the first row

To eliminate the\(9z\)term in the first equation,perform an elementary row operationon the matrix \(\left( {\begin{aligned}{*{20}{c}}1&0&9&0\\0&1&0&0\\0&0&1&0\end{aligned}} \right)\) as shown below.

Subtract 9 times the third row from the first ro;i.e., \({R_1} \to {R_1} - 9{R_3}\).

\(\left( {\begin{aligned}{*{20}{c}}{1 - 9\left( 0 \right)}&{0 - 9\left( 0 \right)}&{9 - 9\left( 1 \right)}&{0 - 9\left( 0 \right)}\\0&1&0&0\\0&0&1&0\end{aligned}} \right)\)

After the row operation, the matrix becomes

\(\left( {\begin{aligned}{*{20}{c}}1&0&0&0\\0&1&0&0\\0&0&1&0\end{aligned}} \right)\)

Convert the augmented matrix into a system of linear equations to get\(x = 0,\,\,y = 0,\,\,\)and\(z = 0.\)

Hence, the required solution is \(\left( {\begin{aligned}{*{20}{c}}0\\0\\0\end{aligned}} \right)\).