Answer:
x = -3
y = -2
z = 6
or
(-3, -2, 6)
Step-by-step explanation:
So first, lets construct our matrix:
[tex]\left[\begin{array}{ccc|c}1&3&-3&-27\\2&1&1&-2\\1&-1&3&17\end{array}\right][/tex]
Now lets get it into reduced row echelon form. R₂ ⇒ -2R₁ + R₂
[tex]\left[\begin{array}{ccc|c}1&3&-3&-27\\0&-5&7&52\\1&-1&3&17\end{array}\right][/tex]
R₃ ⇒ -R₁ + R₃
[tex]\left[\begin{array}{ccc|c}1&3&-3&-27\\0&-5&7&52\\0&-4&6&44\end{array}\right][/tex]
R₂ ⇒ -1/5R₂
[tex]\left[\begin{array}{ccc|c}1&3&-3&-27\\0&1&-\frac{7}{5}&-\frac{52}{5}\\0&-4&6&44\end{array}\right][/tex]
R₃ ⇒ 4R₂ + R₃
[tex]\left[\begin{array}{ccc|c}1&3&-3&-27\\0&1&-\frac{7}{5}&-\frac{52}{5}\\0&0&\frac{2}{5}&\frac{12}{5}\end{array}\right][/tex]
R₃ ⇒ 5/2R₃
[tex]\left[\begin{array}{ccc|c}1&3&-3&-27\\0&1&-\frac{7}{5}&-\frac{52}{5}\\0&0&1&6\end{array}\right][/tex]
R₂ ⇒ 7/5R₃ + R₂
[tex]\left[\begin{array}{ccc|c}1&3&-3&-27\\0&1&0&-2\\0&0&1&6\end{array}\right][/tex]
R₁ ⇒ 3R₃ + R₁
[tex]\left[\begin{array}{ccc|c}1&3&0&-9\\0&1&0&-2\\0&0&1&6\end{array}\right][/tex]
R₁ ⇒ -3R₂ + R₁
[tex]\left[\begin{array}{ccc|c}1&0&0&-3\\0&1&0&-2\\0&0&1&6\end{array}\right][/tex]
And now we have our matrix in reduced row echelon form. The solution to the system of equations would be:
x = -3
y = -2
z = 6
(-3, -2, 6)
I hope you find my answer and explanation to be helpful. Happy studying.
