Answer:
[tex]x_{1} = 0\\ x_{2} = 5 \\x_{3} =-4[/tex]
Step-by-step explanation:
[tex]\left[\begin{array}{cccc}1&-1&-1&-1\\2&3&5&-5\\1&-2&3&-22\end{array}\right][/tex]
new [tex]R_{2}[/tex]=[tex]R_{2} -2R_{3}[/tex]
= [tex]\left[\begin{array}{cccc}1&-1&-1&-1\\0&7&-1&39\\1&-2&3&-22\end{array}\right][/tex]
(first )new [tex]R_{3} = R_{3} -R_{1}[/tex]
=[tex]\left[\begin{array}{cccc}1&-1&-1&-1\\0&7&-1&39\\0&1&-4&21\end{array}\right][/tex]
to make [tex]R_{32}[/tex] = 0 from the new matrix;
(second ) new [tex]R_{3}[/tex] = new [tex]R_{2}[/tex] - (first) new [tex]R_{3}[/tex]
= [tex]\left[\begin{array}{cccc}1&-1&-1&-1\\0&7&-1&39\\0&0&27&-108\end{array}\right][/tex] ------------------------- eqn k
therefore [tex]27x_{3}= -108\\x_{3} = -4[/tex]
[tex]7x_{2}-x_{3}=39[/tex]
[tex]x_{2} = 5[/tex]
And [tex]x_{1}- x_{2}- x_{3} =-1[/tex] from first line of equation k
[tex]x_{1} = 0[/tex]
Hence
[tex]x_{1} = 0\\ x_{2} = 5 \\x_{3} =-4[/tex]
