Question

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.

Answer 1

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