Mathematics: Convert 14/3 into a mixed number

Echelon form of augmented matrix Step-by-step explanation:We have, linear systems Find the echelon form of the augmented matrix.Firstly we find the augmented matrix (A) of the given linear system Now find the echelon form From above echelon form of augmented matrix system is consistent .So, system has always a solution exist for every condition of Yes, If we take point then system of linear equation has exist a solution. yes, if pick randomly then always exist a solution of given system...

Convert 14/3 into a mixed number - id14311416184, PompousCoyote

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tavidallaz5988

11.11.2022

Mathematics

Echelon form of augmented matrix $\left[\begin{matrix} 2&3&-5&b_1\\0&\frac{47}{2} &\frac{101}{2} &b_2-\frac{17}{2}b_1 \\0&0&\frac{-10}{3} &b_3-2\frac{b_1}{6} \end{matrix}\right]$

Step-by-step explanation:

We have, linear systems

2x_1+3x_2-5x_3=b_1

17x_1+2x_2+8x_3=b_2

-x_1+x_2-5x_3=b_3

(a) Find the echelon form of the augmented matrix.

Firstly we find the augmented matrix (A) of the given linear system

$A=\left[\begin{matrix} 2&3&-5&b_1\\17&2&8&b_2\\-1&1&-5&b_3\end{matrix}\right]$

Now find the echelon form

$=\left[\begin{matrix} 2&3&-5&b_1\\0&\frac{47}{2} &\frac{101}{2} &b_2-\frac{17}{2}b_1 \\0&\frac{5}{2} &\frac{-15}{2} &b_3+\frac{b_1}{2} \end{matrix}\right]^{R_2\rightarrow \frac{17}{2}R_1, R_3\rightarrow R_3+\frac{1}{2}R_1 }$

$=\left[\begin{matrix} 2&3&-5&b_1\\0&\frac{47}{2} &\frac{101}{2} &b_2-\frac{17}{2}b_1 \\0&0&\frac{-10}{3} &b_3-2\frac{b_1}{6} \end{matrix}\right]^{R_2\rightarrow R_3-\frac{5}{6}R_1 }$

(b) From above echelon form of augmented matrix system is consistent .

So, system has always a solution exist for every condition of b_1,b_2, b_3

(d) yes, if pick randomly b_1,b_2, b_3 then always exist a solution of given system of linear equation.

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Convert 14/3 into a mixed number

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