Exercise Set 1.1 – No. 10

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

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

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

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

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

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

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

Observe  that 

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

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

\large -\frac{5}{7}+5.\frac{8}{7} -5.0= -\frac{5}{7}+\frac{40}{7} =5

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

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

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

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

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

Observe  that 

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

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

\large -\frac{5}{7}-5.\frac{10}{7} -5.\frac{2}{7}= -\frac{5}{7}+\frac{50}{7}-\frac{10}{7} =5

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

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

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

Observe  that 

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

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

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

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