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Alja [10]
3 years ago
6

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.
Mathematics
1 answer:
xz_007 [3.2K]3 years ago
8 0

Answer:

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

Step-by-step explanation:

Given

T(x_1,x_2,x_3) = (x_1-5x_2+4x_3,x_2 - 6x_3)

Required

Determine if it is linear or onto

Represent the above as a matrix.

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]

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

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