Question

Consider the basis B of R2 consisting of vectors

5
-6
and
-2
-2

Find x in R2 whose coordinate vector relative to the basis B is

[x]B = [-6]
[ 2 ]

Answer 1

Answer: The required vector x is

`x=\begin{bmatrix}-(8)/(11)\\ (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=-(8)/(11)`

and from equation (i), we get

`5*\left(-(8)/(11)\right)-2b=-6\\\\\\\Rightarrow 2b=-(40)/(11)+6\\\\\\\Rightarrow 2b=(26)/(11)\\\\\\\Rightarrow b=(13)/(11).`

Thus, the required vector x is

`x=\begin{bmatrix}-(8)/(11)\\ (13)/(11)\end{bmatrix}`.

Answer 2

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)