Respuesta :
Answer:
Echelon form of augmented matrix [tex]\left[\begin{matrix} 2&3&-5&b_1\\0&\frac{47}{2} &\frac{101}{2} &b_2-\frac{17}{2}b_1 \\0&0&\frac{-10}{3} &b_3-2\frac{b_1}{6} \end{matrix}\right][/tex]
Step-by-step explanation:
We have, linear systems
[tex]2x_1+3x_2-5x_3=b_1[/tex]
[tex]17x_1+2x_2+8x_3=b_2[/tex]
[tex]-x_1+x_2-5x_3=b_3[/tex]
[tex](a)[/tex] Find the echelon form of the augmented matrix.
Firstly we find the augmented matrix (A) of the given linear system
[tex]A=\left[\begin{matrix} 2&3&-5&b_1\\17&2&8&b_2\\-1&1&-5&b_3\end{matrix}\right][/tex]
Now find the echelon form
[tex]=\left[\begin{matrix} 2&3&-5&b_1\\0&\frac{47}{2} &\frac{101}{2} &b_2-\frac{17}{2}b_1 \\0&\frac{5}{2} &\frac{-15}{2} &b_3+\frac{b_1}{2} \end{matrix}\right]^{R_2\rightarrow \frac{17}{2}R_1, R_3\rightarrow R_3+\frac{1}{2}R_1 }[/tex]
[tex]=\left[\begin{matrix} 2&3&-5&b_1\\0&\frac{47}{2} &\frac{101}{2} &b_2-\frac{17}{2}b_1 \\0&0&\frac{-10}{3} &b_3-2\frac{b_1}{6} \end{matrix}\right]^{R_2\rightarrow R_3-\frac{5}{6}R_1 }[/tex]
[tex](b)[/tex] From above echelon form of augmented matrix system is consistent .
So, system has always a solution exist for every condition of [tex]b_1,b_2, b_3[/tex]
[tex](c)[/tex] Yes, If we take point [tex](0,1,1)[/tex] then system of linear equation has exist a solution.
[tex](d)[/tex] yes, if pick randomly [tex]b_1,b_2, b_3[/tex] then always exist a solution of given system of linear equation.
