The idea is to find a linear combination a_1(5, -6) + a_2(-2, -2) = (-6, 2) It boils down to a system of equations: Take the augmented matrix:
[5−6−2−2−62] Reduced form: [1001−8111311] -(8/11)*(5, -6) + (13/11)*(-2, -2) = (-6, 2)Answer: The required vector x is
[tex]x=\begin{bmatrix}-\dfrac{8}{11}\\ \dfrac{13}{11}\end{bmatrix}[/tex].
Step-by-step explanation: Given that a basis B of R² consists of vectors (5, -6) and (-2, -2).
We are to find the vector x in R² whose co-ordinate vector relative to the basis B is [tex]\begin{bmatrix}-6\\ 2\end{bmatrix}[/tex].
Let us consider that a, b are scalars such that
[tex]a(5,-6)+b(-2,-2)=(-6,2)\\\\\Rightarrow (5a-2b,-6a-2b)=(-6,2)\\\\\Rightarrow 5a-2b=-6~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\\\\-6a-2b=2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)[/tex]
Subtracting equation (ii) from equation (i), we get
[tex](5a-2b)-(-6a-2b)=-6-2\\\\\Rightarrow 11a=-8\\\\\Rightarrow a=-\dfrac{8}{11}[/tex]
and from equation (i), we get
[tex]5\times\left(-\dfrac{8}{11}\right)-2b=-6\\\\\\\Rightarrow 2b=-\dfrac{40}{11}+6\\\\\\\Rightarrow 2b=\dfrac{26}{11}\\\\\\\Rightarrow b=\dfrac{13}{11}.[/tex]
Thus, the required vector x is
[tex]x=\begin{bmatrix}-\dfrac{8}{11}\\ \dfrac{13}{11}\end{bmatrix}[/tex].
