Question

Question: Write each of the vectors (8.1)as a linear combination of the vectors (9,0,7)and (0,-9,13). Hint: To get the right xcomponent in (1,4,-5), you have to use (1/9)(9,0,7). How do you get the right ycomponent? Is the zcomponent now, correct?

Short answer

The vectors given in the linear combinations of u, and v. The set of Vectors given in (8.1) are $(1,4,-5),\text{(5,2,1)},\text{(2, - 1,3)},\text{(3, - 6,11)}$, then the linear combinations are:

$\text{(1,4,}-\text{5)}=\text{1u}-\text{4v,(5,2,1)}=\text{5u}-\text{2v,(2,}-\text{1,3)}=\text{2u}+\text{1v,(3,}-\text{6,11)}=\text{3u}+\text{6v}$

Step-by-step solution

Step-1 Definition of Linear Combination

Given vectors${\text{v}}_1{\text{,v}}_2\text{,}.....{\text{,v}}_i$ in ${\text{R}}^{\text{n}}$, and real number${\text{c}}_{\text{1}}{\text{,c}}_{\text{2}}\text{,}.....{\text{c}}_{\text{j}}$ the vector obtained by$\text{w}={\text{c}}_{\text{1}}{\text{v}}_{\text{1}}{\text{+c}}_{\text{2}}{\text{v}}_{\text{2}}\text{+}......{\text{+c}}_{\text{i}}{\text{v}}_{\text{i}}$ is called a linear combination of ${\text{v}}_1{\text{,v}}_2\text{,}.....{\text{,v}}_i$.

Given Parameters

The given vectors are $\text{u}=\frac{\text{1}}{\text{9}}\left(9,0,7\right)$and $\text{v}=\frac{\text{1}}{\text{9}}\left(\text{0,}-\text{9,13}\right)$.

Write the vectors as a linear combination of u and v by keeping the right x component as $(1,4,-5),\text{(5,2,1)},\text{(2, - 1,3)},\text{(3, - 6,11)}$, and equate it to the linear combination of u, and v.

Step-3 Equating the given parameter

Substitute the value of u and v in the equation $\text{(1,4,}-\text{5)}={\text{k}}_{\text{1}}\text{u}+{\text{k}}_{\text{2}}\text{v}$.

$\text{(1,4,}-\text{5)}=\frac{k_1}{9}\text{(9,0,7)}+\frac{k_2}{9}(0,-9,13)$

Equatelocalid="1664199613583" ${\text{k}}_{\text{1}}$to first term and${\text{k}}_2$to second term of the right-hand side.

$$\begin{gathered} {\text{k}}_1=1 \\ {k}_2=-4 \end{gathered}$$

The equation becomes $\text{(1,4,}-\text{5)}=\text{1u}-\text{4v}$.

Find the value of by substituting the value of${\text{k}}_{\text{1}}$and ${\text{k}}_2$.

$\text{z}=\frac{1}{9}(9,0,7)+\frac{-4}{9}\left(0,-9,13\right)$

Solve the equation.

$$\begin{gathered} \text{z}=\text{(1,0,}\frac{\text{7}}{\text{9}}\text{)}+\left(\text{0,}-\text{4,}\frac{\text{13}}{\text{9}}\right) \\ \text{z}=\frac{\text{7}}{\text{9}}-\frac{\text{4}\times \text{13}}{\text{9}} \\ \text{z}=\frac{\text{7}}{\text{9}}-\frac{\text{52}}{\text{9}} \\ \text{z}=\frac{-\text{45}}{\text{9}} \end{gathered}$$

Further, solve.

$z=-5$

Similarly, solve other three.

$$\begin{gathered} \text{(5,2,1)}={\text{k}}_{\text{1}}\text{u}+{\text{k}}_{\text{2}}\text{v} \\ \text{(5,2,1)}=\frac{{\text{k}}_{\text{1}}}{\text{9}}\text{(9,0,7)}+\frac{{\text{k}}_{\text{2}}}{\text{9}}\text{(0,}-\text{9,13)} \end{gathered}$$

On equating${\text{k}}_{\text{1}}$is 5 and${\text{k}}_2$is -2.

$\text{(5,2,1)}=\text{5u}-\text{2v}$

So,

$$\begin{gathered} \text{z}=\frac{\text{5}\times \text{7}}{\text{9}}-\frac{\text{2}\times \text{13}}{\text{9}} \\ \text{z}=\frac{\text{35}-\text{26}}{\text{9}} \\ \text{z}=\frac{\text{9}}{\text{9}} \\ \text{z}=\text{1} \end{gathered}$$

$\text{(2,}-\text{1,3)}={\text{k}}_{\text{1}}\text{u}+{\text{k}}_{\text{2}}\text{v}$

Substitute the value of u and v.

$\text{(2,}-\text{1,3)}=\frac{{\text{k}}_{\text{1}}}{\text{9}}\text{(9,0,7)}+\frac{{\text{k}}_{\text{2}}}{\text{9}}\text{(0,}-\text{9,13)}$

So, from this${\text{k}}_{\text{1}}$is 2 and ${\text{k}}_2\text{=1}$.

$\Rightarrow \text{(2,}-\text{1,3)}=\text{2u}+\text{1v}$

Solve for z.

$$\begin{gathered} \text{z}=\frac{\text{2}\times \text{7}}{\text{9}}+\frac{\text{13}}{\text{9}} \\ \text{z}=\frac{\text{27}}{\text{9}} \\ \text{z}=\text{3} \end{gathered}$$

$\text{(3,}-\text{6,11)}={\text{k}}_{\text{1}}\text{u}+{\text{k}}_{\text{2}}\text{v}$

Substitute the value of u and v.

$\text{(3,}-\text{6,11)}=\frac{{\text{k}}_{\text{1}}}{\text{9}}\text{(9,0,7)}+\frac{{\text{k}}_{\text{2}}}{\text{9}}\text{(0,}-\text{9,13)}$

On equating${\text{k}}_{\text{1}}$is 3 and ${\text{k}}_2\text{=6}$.

Above equation becomes $\text{(3,}-\text{6,11)}=\text{3u}+\text{6v}$.

Therefore,

$$\begin{gathered} \text{z}=\frac{\text{3}\times \text{7}}{\text{9}}+\frac{\text{6}\times \text{13}}{\text{9}} \\ \text{z}=\frac{\text{21}+\text{78}}{\text{9}} \\ \text{z}=\text{11} \end{gathered}$$