Question

The augmented matrix is given for a system of equations

If the system is consistent; find the general solution: Otherwise state that there is no solution:

$\left[\overline{)\begin{array}{ccc}1& 2& -3\\ 0& 1& 4\\ 0& 0& 0\end{array}}\begin{array}{c}-9\\ 5\\ 1\end{array}\right]$

• A. No solution
• B. $$\begin{gathered} x_1=-9-2x_2+3x_3 \\ {x}_2\text{is free} \\ {x}_3\text{is free} \end{gathered}$$
• C. $$\begin{gathered} x_1=-19+11x_3 \\ {x}_2=5-4x_3 \\ {x}_3\text{is free} \end{gathered}$$
• D. $$\begin{gathered} x_1=-19+11x_3 \\ {x}_2=5-4x_3 \\ {x}_3=1 \end{gathered}$$

Answer 1

The correct option is C

$$\begin{gathered} x_1=-19+11x_3 \\ {x}_2=5-4x_3 \\ {x}_3\text{is free} \end{gathered}$$

Explanation for the correct option:

Given:

The augmented matrix

$\left[\overline{)\begin{array}{ccc}1& 2& -3\\ 0& 1& 4\\ 0& 0& 0\end{array}}\begin{array}{c}-9\\ 5\\ 1\end{array}\right]$

To find:

The general solution.

Explanation:

From the given augmented matrix we get,

$\begin{array}{rcl}{x}_1+2x_2-3x_3& =& -9\\ x_2+4x_3& =& 5\\ 0x_3& =& 1\end{array}$

Then from the last equation, we have $0=1\text{(is not possible)}$

Thus, $x_3$ is free.

From the second equation, we get

$\begin{array}{rcl}{x}_2+4x_3& =& 5\\ x_2& =& 5-4x_3\end{array}$

Now from the first equation, we get

$\begin{array}{rcl}{x}_1+2x_2-3x_3& =& -9\\ x_1& =& -9-2x_2+3x_3\end{array}$

Substitute $x_2=5-4x_3$ in the above equation.

$\begin{array}{rcl}{x}_1& =& -9-2\left(5-4x_3\right)+3x_3\\ & =& -9-10+8x_3+3x_3\\ & =& -19+11x_3\end{array}$

Therefore, the general solution for the given augmented matrix is

$$\begin{gathered} x_1=-19+11x_3 \\ {x}_2=5-4x_3 \\ {x}_3\text{is free} \end{gathered}$$

Hence, the correct option is $\left(C\right)$.