Question

Consider the basis B of R2 consisting of vectors

Answer 1

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 2

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

$x=\begin{bmatrix}-\dfrac{8}{11}\\ \dfrac{13}{11}\end{bmatrix}$.