Answer:
C. T is not one-to-one because the standard matrix A has a free variable.
Step-by-step explanation:
Given
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]](https://tex.z-dn.net/?f=T%28x_1%2Cx_2%2Cx_3%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-5%264%5C%5C0%261%26-6%5C%5C0%260%260%5Cend%7Barray%7D%5Cright%5D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx_1%26x_2%26x_3%5Cend%7Barray%7D%5Cright%5D)
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
A rational number is 5.3 . It's rational because it can be written as a fraction a/b
- that is 5 3/10 or 53/10.
An irrational number between the 2 values would be √29
Enter this into ur calculator and you'll get an answer like 5.385164807 but this number carries on beyond bounds and can't be written as a fraction a/b It is irrational.
The second one because the second one equals to 12 while all the other equal to 6
