You & maths

# Exercise Set 1.1 – No. 10

10. In each part, determine whether the given 3-tuple is a solution of the linear system:

$\large&space;{\color&space;{Blue}&space;\begin{matrix}&space;x&space;&&space;+2y&space;&&space;-2z&space;&&space;=3&space;\\&space;3x&space;&&space;-y&space;&&space;+z&space;&&space;=1&space;\\&space;-x&space;&&space;+5y&space;&&space;-5z&space;&&space;=5&space;\end{matrix}&space;}$

$\large&space;{\color&space;{blue}&space;(a)&space;\bigl(\begin{smallmatrix}&space;\frac{5}{7},&space;&&space;\frac{8}{7},&space;&&space;1&space;\end{smallmatrix}\bigr)&space;}$

Consider the linear system $\large&space;{\color&space;{Blue}&space;\begin{matrix}&space;x&space;&&space;+2y&space;&&space;-2z&space;&&space;=3&space;\\&space;3x&space;&&space;-y&space;&&space;+z&space;&&space;=1&space;\\&space;-x&space;&&space;+5y&space;&&space;-5z&space;&&space;=5&space;\end{matrix}&space;}$ and 3-type $\large&space;{\color&space;{blue}&space;\bigl(\begin{smallmatrix}&space;\frac{5}{7},&space;&&space;\frac{8}{7},&space;&&space;1&space;\end{smallmatrix}\bigr)&space;}$.

Observe  that which gives us that the 3-type $\large&space;{\color&space;{blue}&space;\bigl(\begin{smallmatrix}&space;\frac{5}{7},&space;&&space;\frac{8}{7},&space;&&space;1&space;\end{smallmatrix}\bigr)&space;}$ does not satisfy the equation x+2y-2z=3 and therefore $\large&space;{\color&space;{blue}&space;\bigl(\begin{smallmatrix}&space;\frac{5}{7},&space;&&space;\frac{8}{7},&space;&&space;1&space;\end{smallmatrix}\bigr)&space;}$ is not a solution of the linear system above.

$\large&space;{\color&space;{blue}&space;(b)&space;\bigl(\begin{smallmatrix}&space;\frac{5}{7},&space;&&space;\frac{8}{7},&space;&&space;0&space;\end{smallmatrix}\bigr)&space;}$

Consider the linear system $\large&space;{\color&space;{Blue}&space;\begin{matrix}&space;x&space;&&space;+2y&space;&&space;-2z&space;&&space;=3&space;\\&space;3x&space;&&space;-y&space;&&space;+z&space;&&space;=1&space;\\&space;-x&space;&&space;+5y&space;&&space;-5z&space;&&space;=5&space;\end{matrix}&space;}$ and 3-type $\large&space;{\color&space;{blue}&space;\bigl(\begin{smallmatrix}&space;\frac{5}{7},&space;&&space;\frac{8}{7},&space;&&space;0&space;\end{smallmatrix}\bigr)&space;}$.

Observe  that

$\large&space;\frac{5}{7}+2.\frac{8}{7}-2.0=&space;\frac{5}{7}+\frac{16}{7}&space;-0&space;=3,$

$\large&space;3.\frac{5}{7}-\frac{8}{7}+0=&space;\frac{15}{7}+\frac{8}{7}&space;=1$ and

$\large&space;-\frac{5}{7}+5.\frac{8}{7}&space;-5.0=&space;-\frac{5}{7}+\frac{40}{7}&space;=5$

Hence $\large&space;{\color&space;{blue}&space;\bigl(\begin{smallmatrix}&space;\frac{5}{7},&space;&&space;\frac{8}{7},&space;&&space;0&space;\end{smallmatrix}\bigr)&space;}$ is a solution to the system above.

$\large&space;{\color&space;{blue}&space;(c)&space;\bigl(\begin{smallmatrix}&space;5,&space;&&space;8,&space;&&space;1&space;\end{smallmatrix}\bigr)&space;}$

Consider the linear system $\large&space;{\color&space;{Blue}&space;\begin{matrix}&space;x&space;&&space;+2y&space;&&space;-2z&space;&&space;=3&space;\\&space;3x&space;&&space;-y&space;&&space;+z&space;&&space;=1&space;\\&space;-x&space;&&space;+5y&space;&&space;-5z&space;&&space;=5&space;\end{matrix}&space;}$ and 3-type $\large&space;{\color&space;{blue}&space;\bigl(\begin{smallmatrix}&space;5,&space;&&space;8,&space;&&space;1&space;\end{smallmatrix}\bigr)&space;}$.

Obserw that 5 +2.8 -2.1 = 5+ 16 -2 = 19 which gives us that the 3-tuple (5, 8, 1) does not satisfy the equation x+ 2y – 2 = 3 and therefore (5, 8, 1) is not a solution of the linear system above.

$\large&space;{\color&space;{blue}&space;(d)&space;\bigl(\begin{smallmatrix}&space;\frac{5}{7},&space;&&space;\frac{10}{7},&space;&&space;\frac{2}{7}&space;\end{smallmatrix}\bigr)&space;}$

Consider the linear system $\large&space;{\color&space;{Blue}&space;\begin{matrix}&space;x&space;&&space;+2y&space;&&space;-2z&space;&&space;=3&space;\\&space;3x&space;&&space;-y&space;&&space;+z&space;&&space;=1&space;\\&space;-x&space;&&space;+5y&space;&&space;-5z&space;&&space;=5&space;\end{matrix}&space;}$ and 3-type $\large&space;{\color&space;{blue}&space;\bigl(\begin{smallmatrix}&space;\frac{5}{7},&space;&&space;\frac{10}{7},&space;&&space;\frac{2}{7}&space;\end{smallmatrix}\bigr)&space;}$.

Observe  that

$\large&space;\frac{5}{7}+2.\frac{10}{7}&space;-2.\frac{2}{7}=&space;\frac{5}{7}+\frac{20}{7}-\frac{4}{7}&space;=3,$

$\large&space;3.\frac{5}{7}-\frac{10}{7}&space;+\frac{2}{7}=&space;\frac{15}{7}-\frac{10}{7}+\frac{7}{7}&space;=1$ and

$\large&space;-\frac{5}{7}-5.\frac{10}{7}&space;-5.\frac{2}{7}=&space;-\frac{5}{7}+\frac{50}{7}-\frac{10}{7}&space;=5$

Hence $\large&space;{\color&space;{blue}&space;\bigl(\begin{smallmatrix}&space;\frac{5}{7},&space;&&space;\frac{10}{7},&space;&&space;\frac{2}{7}&space;\end{smallmatrix}\bigr)&space;}$ is a solution to the system above.

$\large&space;{\color&space;{blue}&space;(e)&space;\bigl(\begin{smallmatrix}&space;\frac{5}{7},&space;&&space;\frac{22}{7},&space;&&space;2&space;\end{smallmatrix}\bigr)&space;}$

Consider the linear system $\large&space;{\color&space;{Blue}&space;\begin{matrix}&space;x&space;&&space;+2y&space;&&space;-2z&space;&&space;=3&space;\\&space;3x&space;&&space;-y&space;&&space;+z&space;&&space;=1&space;\\&space;-x&space;&&space;+5y&space;&&space;-5z&space;&&space;=5&space;\end{matrix}&space;}$ and 3-type $\large&space;{\color&space;{blue}&space;\bigl(\begin{smallmatrix}&space;\frac{5}{7},&space;&&space;\frac{22}{7},&space;&&space;2&space;\end{smallmatrix}\bigr)&space;}$

Observe  that

$\large&space;{\color&space;{blue}&space;\frac{5}{7}+2.\frac{22}{7}&space;-2.2=&space;\frac{5}{7}+\frac{44}{7}&space;-4&space;=3,&space;}$

$\large&space;{\color&space;{blue}&space;3.\frac{5}{7}-\frac{22}{7}&space;+&space;2=&space;\frac{15}{7}-\frac{44}{7}&space;+2&space;=1&space;}$ and

$\large&space;{\color&space;{blue}&space;-\frac{5}{7}+5\frac{22}{7}&space;-&space;5.2=&space;-\frac{15}{7}+\frac{100}{7}&space;-10&space;=5&space;}$

Hence $\large&space;{\color&space;{blue}&space;\bigl(\begin{smallmatrix}&space;\frac{5}{7},&space;&&space;\frac{22}{7},&space;&&space;2&space;\end{smallmatrix}\bigr)&space;}$ is a solution to the system above.