Question

As in Problem $\mathbf{}\mathbf{6}$,write$\mathbf{}\mathbf{V}\mathbf{=}\mathbf{4}\mathbf{i}\mathbf{-}\mathbf{5}\mathbf{j}$ in terms of the basis vectors$\mathbf{}\mathbf{}\mathbf{i}\mathbf{-}\mathbf{4}\mathbf{j}\mathbf{}$and$\mathbf{5}\mathbf{i}\mathbf{+}\mathbf{2}\mathbf{j}$.

Short answer

The vector in terms of basis vectors is$\mathrm{V}=\frac{3}{2}(\mathrm{i}-4\mathrm{j})+\frac{1}{2}(5\mathrm{i}+2\mathrm{j})$ .

Step-by-step solution

Use of Cramer’s Rule

Cramer’s rule is a strategy for solving systems of linear equations with the same number of unknowns as equations. The approach entails solving a series of equations using determinants and ratios to acquire a distinct set of solutions for a linear system.

Given Parameters

The given vector equations are $\mathrm{V}=4\mathrm{i}-5\mathrm{j},\mathrm{A}=\mathrm{i}-4\mathrm{j}$and$\mathrm{B}=5\mathrm{i}+2\mathrm{j}$

Write $V=4i-5j$in terms of the basis vectors $A=i-4j$and$\mathrm{B}=5\mathrm{i}+2\mathrm{j}$ .

Finding the vector V in terms of  A and B

Follow Cramer’s rule.

$a=\frac{\left|\begin{array}{cc}{B}_x& B_y\\ V_x& V_y\end{array}\right|}{\left|\begin{array}{cc}{B}_x& B_y\\ A_x& A_y\end{array}\right|}$

role="math" localid="1659076477717" $a=\frac{\left|\begin{array}{cc}5& 2\\ 4& -5\end{array}\right|}{\left|\begin{array}{cc}5& 2\\ 1& -4\end{array}\right|}$

Simplify for a .

$\begin{array}{l}\mathrm{a}=\frac{-25-8}{-20-2}\\ \mathrm{a}=\frac{3}{2}\end{array}$

Find the value of b.

$$\begin{gathered} \mathrm{b}=\frac{\left|\begin{array}{cc}{\mathrm{A}}_{\mathrm{x}}& {\mathrm{A}}_{\mathrm{y}}\\ {\mathrm{V}}_{\mathrm{x}}& {\mathrm{V}}_{\mathrm{y}}\end{array}\right|}{\left|\begin{array}{cc}{\mathrm{A}}_{\mathrm{x}}& {\mathrm{A}}_{\mathrm{y}}\\ {\mathrm{B}}_{\mathrm{x}}& {\mathrm{B}}_{\mathrm{y}}\end{array}\right|} \\ \mathrm{b}=\frac{\left|\begin{array}{cc}1& -4\\ 4& -5\end{array}\right|}{\left|\begin{array}{cc}1& -4\\ 5& 2\end{array}\right|} \end{gathered}$$

Simplify further for b .

$\begin{array}{l}\mathrm{b}=\frac{-5+16}{2+20}\\ \mathrm{b}=\frac{1}{2}\end{array}$

The vector V in terms of basis vectors is $\mathrm{V}=\frac{3}{2}A+\frac{1}{2}B$.

Substitute the values ofandin the equation $\mathrm{V}=\frac{3}{2}A+\frac{1}{2}B$.

Therefore, the vector in terms of basis vectors is $\mathrm{V}=\frac{3}{2}(\mathrm{i}-4\mathrm{j})+\frac{1}{2}(5\mathrm{i}+2\mathrm{j})$.