Determine if the specified linear transformation is (a) one-to-one and (b) onto. Justify your answer. T: , T()(,), T()(,), and T()(,), where , , are the columns of the 33 identity matrix. a. Is the linear transformation one-to-one? A. T is one-to-one because T(x)0 has only the trivial solution. B. T is not one-to-one because the columns of the standard matrix A are linearly independent. C. T is not one-to-one because the standard matrix A has a free variable. D. T is one-to-one because the column vectors are not scalar multiples of each other. b. Is the linear transformation onto? A. T is onto because the columns of the standard matrix A span . B. T is not onto because the columns of the standard matrix A span . C. T is onto because the standard matrix A does not have a pivot position for every row. D. T is not onto because the standard matrix A contains a row of zeros.

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MrRoyal MrRoyal · Answer 1 · 2021-06-25T01:56:03+02:00

Answer:

C. T is not one-to-one because the standard matrix A has a free variable.

Step-by-step explanation:

Given

[tex]T(x_1,x_2,x_3) = (x_1-5x_2+4x_3,x_2 - 6x_3)[/tex]

Required

Determine if it is linear or onto

Represent the above as a matrix.

[tex]T(x_1,x_2,x_3) = \left[\begin{array}{ccc}1&-5&4\\0&1&-6\\0&0&0\end{array}\right] \left[\begin{array}{c}x_1&x_2&x_3\end{array}\right][/tex]

From the above matrix, we observe that the matrix does not have a pivot in every column.

This means that the column are not linearly independent, & it has a free variable and as such T is not one-on-one