The coordinate matrix of x relative to the standard basis can be found by taking a linear combination of the basis vectors in B, using the coordinates of x in B as the coefficients.

The basis vectors in B are (2, -1) and (0, 1). The coordinates of x in B are 5 and 1.

So, we take 5 times the first basis vector and add it to 1 times the second basis vector:

5*(2, -1) + 1*(0, 1) = (10, -5) + (0, 1) = (10, -4)

So, the coordinate matrix of x relative to the standard basis S is (10, -4).

Question

The coordinate matrix of x relative to the standard basis can be found by taking a linear combination of the basis vectors in B, using the coordinates of x in B as the coefficients.

The basis vectors in B are (2, -1) and (0, 1). The coordinates of x in B are 5 and 1.

So, we take 5 times the first basis vector and add it to 1 times the second basis vector:

5*(2, -1) + 1*(0, 1) = (10, -5) + (0, 1) = (10, -4)

So, the coordinate matrix of x relative to the standard basis S is (10, -4).

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The coordinate matrix of x relative to the standard basis can be found by taking a linear combination of the basis vectors in B, using the coordinates of x in B as the coefficients.

The basis vectors in B are (2, -1) and (0, 1). The coordinates of x in B are 5 and 1.

So, we take 5 times the first basis vector and add it to 1 times the second basis vector:

5*(2, -1) + 1*(0, 1) = (10, -5) + (0, 1) = (10, -4)

So, the coordinate matrix of x relative to the standard basis S is (10, -4).

Given the coordinate matrix of x relative to a (nonstandard) basis B for Rn, find the coordinate matrix of x relative to the standard basis.B = {(2, −1), (0, 1)},[x]B = 5 1[x]S =

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