Solution

Elementary Linear algebra

Chapter 1.1

Exercise 10

(a)

(b)

To determine whether the given $3$-tuple is a solution, we will substitute it in place of variables $x,\ y$ and $z$ and see it satisfies each of the three equations.

Substitute $\left(\dfrac{5}{7}, \ \dfrac{8}{7},\ 0\right)$ into the first equation:

$$\dfrac{5}{7}+2\cdot \dfrac{8}{7}-2\cdot 0 = \dfrac{5}{7}+\dfrac{16}{7} = \dfrac{21}{7} = 3$$

Substitute $\left(\dfrac{5}{7}, \ \dfrac{8}{7},\ 0\right)$ into the second equation:

$$3\cdot \dfrac{5}{7} - \dfrac{8}{7}+0 = \dfrac{15}{7}-\dfrac{8}{7} = \dfrac{7}{7} = 1$$

Substitute $\left(\dfrac{5}{7}, \ \dfrac{8}{7},\ 0\right)$ into the third equation:

$$-\dfrac{5}{7}+5\cdot \dfrac{8}{7} - 5\cdot 0 = -\dfrac{5}{7} +\dfrac{40}{7} = \dfrac{35}{7} = 5$$

All three equations are satisfied, hence $\left(\dfrac{5}{7}, \ \dfrac{8}{7},\ 0\right)$ is a solution to the given system.

(c)

To determine whether the given $3$-tuple is a solution, we will substitute it in place of variables $x,\ y$ and $z$ and see it satisfies each of the three equations.

Substitute $(5,\ 8, \ 1)$ into the first equation:

$$5+2\cdot 8-2\cdot 1 = 5+16-2 =19 \neq 3$$

The given 3-tuple does not satisfy the first equation hence it is not the solution to the system.

(d)

To determine whether the given $3$-tuple is a solution, we will substitute it in place of variables $x,\ y$ and $z$ and see it satisfies each of the three equations.

Substitute $\left(\dfrac{5}{7}, \ \dfrac{10}{7},\ \dfrac{2}{7}\right)$ into the first equation:

$$\dfrac{5}{7}+2\cdot \dfrac{10}{7}-2\cdot \dfrac{2}{7} = \dfrac{5}{7}+\dfrac{20}{7}-\dfrac{4}{7} = \dfrac{21}{7} = 3$$

Substitute $\left(\dfrac{5}{7}, \ \dfrac{10}{7},\ \dfrac{2}{7}\right)$ into the second equation:

$$3\cdot \dfrac{5}{7} - \dfrac{10}{7}+\dfrac{2}{7} = \dfrac{15}{7}-\dfrac{10}{7}+\dfrac{2}{7}= \dfrac{7}{7} = 1$$

Substitute $\left(\dfrac{5}{7}, \ \dfrac{10}{7},\ \dfrac{2}{7}\right)$ into the third equation:

$$-\dfrac{5}{7}+5\cdot \dfrac{10}{7} - 5\cdot \dfrac{2}{7} = -\dfrac{5}{7} +\dfrac{50}{7} - \dfrac{10}{7} = \dfrac{35}{7} = 5$$

All three equations are satisfied, hence $\left(\dfrac{5}{7}, \ \dfrac{8}{7},\ \dfrac{2}{7}\right)$ is a solution to the given system.

(e)

To determine whether the given $3$-tuple is a solution, we will substitute it in place of variables $x,\ y$ and $z$ and see it satisfies each of the three equations.

Substitute $\left(\dfrac{5}{7}, \ \dfrac{22}{7},\ 2\right)$ into the first equation:

$$\dfrac{5}{7}+2\cdot \dfrac{22}{7}-2\cdot 2 = \dfrac{5}{7}+\dfrac{44}{7}-2 = \dfrac{49}{7} - 2 = 3$$

Substitute $\left(\dfrac{5}{7}, \ \dfrac{22}{7},\ 2\right)$ into the second equation:

$$3\cdot \dfrac{5}{7} - \dfrac{22}{7}+2 = \dfrac{15}{7}-\dfrac{22}{7}+2 = -1 +2 = 1$$

Substitute $\left(\dfrac{5}{7}, \ \dfrac{22}{7},\ 2\right)$ into the third equation:

$$-\dfrac{5}{7}+5\cdot \dfrac{22}{7} - 5\cdot 2 = -\dfrac{5}{7} +\dfrac{110}{7} -10 = \dfrac{105}{7} - 10 = 5$$

All three equations are satisfied, hence $\left(\dfrac{5}{7}, \ \dfrac{22}{7},\ 2\right)$ is a solution to the given system.