Answer: The required vector x is
\(x=\begin{bmatrix}-\dfrac{8}{11}\\ \dfrac{13}{11}\end{bmatrix}\).
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 \(\begin{bmatrix}-6\\ 2\end{bmatrix}\).
Let us consider that a, b are scalars such that
\(a(5,-6)+b(-2,-2)=(-6,2)\\\\\Rightarrow (5a-2b,-6a-2b)=(-6,2)\\\\\Rightarrow 5a-2b=-6~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\\\\-6a-2b=2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)\)
Subtracting equation (ii) from equation (i), we get
\((5a-2b)-(-6a-2b)=-6-2\\\\\Rightarrow 11a=-8\\\\\Rightarrow a=-\dfrac{8}{11}\)
and from equation (i), we get
\(5\times\left(-\dfrac{8}{11}\right)-2b=-6\\\\\\\Rightarrow 2b=-\dfrac{40}{11}+6\\\\\\\Rightarrow 2b=\dfrac{26}{11}\\\\\\\Rightarrow b=\dfrac{13}{11}.\)
Thus, the required vector x is
\(x=\begin{bmatrix}-\dfrac{8}{11}\\ \dfrac{13}{11}\end{bmatrix}\).
