Question

6. Find a basis for the subspace of R3 spanned by S (42,30,54), (14,10, 18),(7,5,6)). 7. Given that [xlg [4,5,3]', the coordinate matrix of x relative to a (nonstandard) basis B((,1,0(1,0,1),(0,0,0)). Find the coordinate vector of x relative to the standard basis in R3 8. Find the coordinate matrix of x=(-3,28,6) in Rs relative to the basis B=((3,8,0),(5,0,11),( 1,5,7), 9. Find the transition matrix from B ((1,7),(-2, -2))to B'- ((-28,0),(-4,4)) 10 Perform a rotation of axes to eliminate the xy-term, and sketch the graph of the conic defined by the function 13x2 + 13y210xy-36 0.

Answer 1

6) Given set is : S = {(42,30,54),(14,10,18),(7,5,6)}

Here, the first vector (42,30,54) is a linear combination of the second vector (14,10,18).

Then, the set of linearly independent vectors = {(14,10,18),(7,5,6)}.

R³ is a vector space of dimension 3. The standard basis for R³ is {(1,0,0),(0,1,0),(0,0,1)}.

Now, (7,5,6) = 7*(1,0,0)+5*(0,1,0)+6*(0,0,1)

Since the coefficient of (1,0,0) in the representation of (7,5,6) is non-zero, by replacement theorem (7,5,6) can replace (1,0,0) in the basis {(1,0,0),(0,1,0),(0,0,1)} and {(7,5,6),(0,1,0),(0,0,1)} can be a new basis for R³.

Again, (14,10,18) = 2*(7,5,6)+0*(0,1,0)+6*(0,0,1)

Since the coefficient of (0,0,1) in the representation of (14,10,18) is non-zero, by replacement theorem (14,10,18) can replace (0,0,1) in the basis {(7,5,6),(0,1,0),(0,0,1)} and {(7,5,6),(0,1,0),(14,10,18)} can be a new basis for R³.

Therefore, required basis is = {(7,5,6),(0,1,0),(14,10,18)}.

7) Given $[x]_B=[4,5,3]^T$ where B = {(1,1,0),(1,0,1),(0,0,0)}.

Then, x = 4*(1,1,0)+5*(1,0,1)+3*(0,0,0)

i.e., x = (9,4,5)

Now, (9,4,5) = 9*(1,0,0)+4*(0,1,0)+5*(0,0,1)

Therefore, the coordinate vector of x relative to the standard basis of R³ ,i.e., {(1,0,0),(0,1,0),(0,0,1)} is = $[9,4,5]^T$.

8) Let us consider a relation a(3,8,0)+b(5,0,11)+c(1,5,7) = (-3,28,6) where a, b, c are real numbers.

Then, 3a+5b+15c = -3

8a+5c = 28

11b+7c = 6

Solving we get, a = 113/25, b = 198/125, c = -204/125

Therefore, the required coordinate matrix is = $\begin{bmatrix} 113/25\\ 198/125\\ -204/125 \end{bmatrix}$ .

9) Here, (1,7) = (-2/7)*(-28,0)+(7/4)*(-4,4)

and, (-2,-2) = (1/7)*(-28,0)+(-1/2)*(-4,4)

Therefore, the required transition matrix is = $\begin{bmatrix} -2/7 & 1/7\\ 7/4 & -1/2\\ \end{bmatrix}$ .