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
Answer:
8
Step-by-step explanation:
Two different approaches:
<u>Method 1</u>
Apply radical rule √(ab) = √a√b to simplify the radicals:
√98 = √(49 x 2) = √49√2 = 7√2
√50 = √(25 x 2) = √25√2 = 5√2
Therefore, (√98 - √50)² = (7√2 - 5√2)²
= (2√2)²
= 4 x 2
= 8
<u>Method 2</u>
Use the perfect square formula: (a - b)² = a² - 2ab + b²
where a = √98 and b = √50
So (√98 - √50)² = (√98)² - 2√98√50 + (√50)²
= 98 - 2√98√50 + 50
= 148 - 2√98√50
Apply radical rule √(ab) = √a√b to simplify radicals:
√98 = √(49 x 2) = √49√2 = 7√2
√50 = √(25 x 2) = √25√2 = 5√2
Therefore, 148 - 2√98√50 = 148 - (2 × 7√2 × 5√2)
= 148 - 140
= 8
Answer:
all students on the track-and-field team
Step-by-step explanation:
if the coach wants to find out if they prefer track or field he/she has to ask the people who are experiencing it right now and tell him or her if they prefer this or not
