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In Exercises 54 through 58, let V be the plane with equation. x1+2x2+3x3=0inR3In each exercise, find the matrix B of the given transformation T from V to V, with respect to the basis[11-1][5-41]Note that the domain and target space of T are restricted to the plane V, so that B will be a22matrix.

54. The orthogonal projection onto the line spanned by vector[11-1]

Short Answer

Expert verified

The solution isB=1-415

Step by step solution

01

Step 1:Solution for the matrix of the given transformation T

Consider the matrix of the given transformation be T

Let V be the plane with equationx1+2x2-3x3=0in R3.

Also we know that the domain and the target space of T are restricted to the plane V so that B will be a22matrix.

Assume fx=acosx+bsinxbe the function

Hence the basis of the equation be[11-1][5-41]

Then, the two equation be as follows.

b1=a1+a2b=2-4a1+5a2

Therefore the matrix be B as follows

B=1-415

Hence the solution.

02

Step 2:Solution for the isomorphism of the given transformation T

ConsiderB=1-415be the matrix of the given transformation T.

B=detB=ad-bc=5+4=90

Thus the transformation T is invertible as well.So T is an isomorphism.

Hence the solution.

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