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Let

A=⎡⎣⎢10−211−1−215⎤⎦⎥ and b=⎡⎣⎢3−1−7⎤⎦⎥.
Define the linear transformation T:R2→R2
as T(x)=Ax
. Find a vector x
whose image under T
is b
.
x=
Is the vector x
unique? (enter YES or NO)

User Deorst
by
7.9k points

2 Answers

2 votes

Final answer:

To find the vector x whose image under the linear transformation T is b, we need to solve the equation Ax=b. The vector x is not unique as there can be multiple solutions to the equation Ax=b.

Step-by-step explanation:

To find the vector x whose image under the linear transformation T is b, we need to solve the equation Ax=b.

Given A=⎡⎣⎢10−211−1−215⎤⎦⎥ and b=⎡⎣⎢3−1−7⎤⎦⎥, we have:

⎡⎣⎢10−211−1−215⎤⎦⎥ ⎡⎣⎢x_1x_2⎤⎦⎥ = ⎡⎣⎢3−1−7⎤⎦⎥

Multiplying both sides by the inverse of A:

⎡⎣⎢10−211−1−215⎤⎦⎥-1 ⎡⎣⎢10−211−1−215⎤⎦⎥ ⎡⎣⎢x_1x_2⎤⎦⎥ = ⎡⎣⎢10−211−1−215⎤⎦⎥-1 ⎡⎣⎢3−1−7⎤⎦⎥

Finally, we can determine the values of x by evaluating the resulting expression.

The vector x is not unique as there can be multiple solutions to the equation Ax=b.

User Alisabzevari
by
8.2k points
6 votes

Final answer:

To find the vector x such that the linear transformation T represented by matrix A results in vector b, we solve the matrix equation Ax = b. The solution vector x will be unique if matrix A is invertible, signified by its determinant being non-zero.

Step-by-step explanation:

The student is asked to find a vector x such that when transformed by a linear transformation, represented by matrix A, results in vector b. This is effectively solving the matrix equation Ax = b for x. To find x, we must solve the system of linear equations that arises from this matrix equation.

We set up the equation as:

  1. Substitute the given values of A and b to get a system of linear equations.
  2. Solve these equations simultaneously for the vector x.

If matrix A is invertible (which means that its determinant is not zero), there will be a unique solution vector x for which Ax = b. To check if the vector x is unique, we can calculate the determinant of A. If it's non-zero, we have a unique solution, otherwise, we might have infinitely many solutions or none at all.

User Jonathan Cremin
by
8.6k points
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