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

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.

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

Follow Cramer’s rule.

$$\begin{gathered} a=\frac{\left|\begin{array}{cc}{B}_x& B_y\\ V_x& V_y\end{array}\right|}{\left|\begin{array}{cc}{B}_x& B_y\\ A_x& A_y\end{array}\right|} \\ a=\frac{\left|\begin{array}{cc}5& 2\\ 4& -5\end{array}\right|}{\left|\begin{array}{cc}5& 2\\ 1& -4\end{array}\right|} \end{gathered}$$

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.

role="math" localid="1659078238504" $$\begin{gathered} b=\frac{\left|\begin{array}{cc}{A}_x& A_y\\ V_x& V_y\end{array}\right|}{\left|\begin{array}{cc}{A}_x& A_y\\ B_x& B_y\end{array}\right|} \\ b=\frac{\left|\begin{array}{cc}1& -4\\ 4& -5\end{array}\right|}{\left|\begin{array}{cc}1& -4\\ 5& 2\end{array}\right|} \end{gathered}$$

Simplify further for b .

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

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

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

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