Answer:
The correct answer is B. \((-1,0,1)\)
Step-by-step explanation:
The given linear system is
\(-4x+4y+5z=9\)
\(-2x+2y+z=3\)
\(-4x+5y=4\)
The augmented matrix is
\(\left[\begin{array}{cccc}-4&4&5&|9\\-2&2&1&|3\\-4&5&0&|4\end{array}\right]\)
To solve this by Gauss-Jordan elimination means that, we need to reduce to reduced row echelon form.
\(-\frac{1}{4}R_1\rightarrow R_1\)
\(\left[\begin{array}{cccc}1&-1&-\frac{5}{4} &\:\:\:\:\:\:|-\frac{9}{4} \\-2&2&1&|3\\-4&5&0&|4\end{array}\right]\)
\(R_2+2R_1\rightarrow R_2\)
\(R_3+4R_1\rightarrow R_3\)
\(\left[\begin{array}{cccc}1&-1&-\frac{5}{4} &|-\frac{9}{4} \\0&0&-\frac{3}{2} &|-\frac{3}{2} \\0&1&-5&|-5\end{array}\right]\)
\(R_2\leftrightarrow R_3\)
\(\left[\begin{array}{cccc}1&-1&-\frac{5}{4} &|-\frac{9}{4} \\0&1&-5 &|-5 \\0&0&-\frac{3}{2}&|-\frac{3}{2}\end{array}\right]\)
\(R_1+R_2\rightarrow R_1\)
\(\left[\begin{array}{cccc}1&0&-\frac{25}{4} &|-\frac{29}{4} \\0&1&-5 &|-5 \\0&0&-\frac{3}{2}&|-\frac{3}{2}\end{array}\right]\)
\(-\frac{2}{3}R_3\rightarrow R_3\)
\(\left[\begin{array}{cccc}1&0&-\frac{25}{4} &|-\frac{29}{4} \\0&1&-5 &|\:\:-5 \\0&0&1&|\:\:\:\:\:\:\:1\end{array}\right]\)
\(R_2+5R_3\rightarrow R_2\)
\(R_1+\frac{25}{4}R_3\rightarrow R_1\)
\(\left[\begin{array}{cccc}1&0&0&|-1 \\0&1&0&|\:\:\:\:\:\:0\\0&0&1&|\:\:\:\:\:\:1\end{array}\right]\)
The matrix is now in the reduced row echelon form.
This gives the solution to be,
\((-1,0,1)\)
