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S_A_V [24]
3 years ago
5

Consider the basis B of R2 consisting of vectors

Mathematics
2 answers:
adelina 88 [10]3 years ago
5 0

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:

<span>[<span><span>5<span>−6</span></span><span><span>−2</span><span>−2</span></span><span><span>−6</span>2</span></span>]</span> Reduced form: <span>[<span><span>10</span><span>01</span><span><span>−<span>811</span></span><span>1311</span></span></span>]</span> -(8/11)*(5, -6) + (13/11)*(-2, -2) = (-6, 2)

dmitriy555 [2]3 years ago
4 0

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}.

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