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 2 years ago
Linear application f: R^6 > R^4 has towards canonical bases the matrix A=1 1 2 1 0 2/
1 2 3 1 0 1/
0 1 2 1 1 1/
1 1 1 1 1 1
We must find :D
1) f(x), X=(1,1, 1, 0, 1, 0)
2) Find a basis in ker(f)
3) Does x with f(x)= 2, 0, 0, 2
4) Inverse of matrix A
Please help me
 2 years ago
Linear application f: R^6 > R^4 has towards canonical bases the matrix A=1 1 2 1 0 2/ 1 2 3 1 0 1/ 0 1 2 1 1 1/ 1 1 1 1 1 1 We must find :D 1) f(x), X=(1,1, 1, 0, 1, 0) 2) Find a basis in ker(f) 3) Does x with f(x)= 2, 0, 0, 2 4) Inverse of matrix A Please help me

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graydarl
 2 years ago
Best ResponseYou've already chosen the best response.0A= 1 1 2 1 0 2 1 2 3 1 0 1  0 1 2 1 1 1 1 1 1 1 1 1

helder_edwin
 2 years ago
Best ResponseYou've already chosen the best response.1u have \[ \large f(x)=\begin{pmatrix} 1 & 1 & 2 & 1 & 0 & 2\\ 1 & 2 & 3 & 1 & 0 & 1\\ 0 & 1 & 2 & 1 & 1 & 1\\ 1 & 1 & 1 & 1 & 1 & 1 \end{pmatrix}x \]

helder_edwin
 2 years ago
Best ResponseYou've already chosen the best response.1because u r said that the matrix is under the canonical bases of both spaces.

helder_edwin
 2 years ago
Best ResponseYou've already chosen the best response.1(don't forget to write all vectors as columns, that way everything is easier)

helder_edwin
 2 years ago
Best ResponseYou've already chosen the best response.1the first one is easy: just multiply the matrix by the vector (column)

helder_edwin
 2 years ago
Best ResponseYou've already chosen the best response.1the second one: say the matrix is called A then solve the system Ax=0.

helder_edwin
 2 years ago
Best ResponseYou've already chosen the best response.1for the third one solve the system \[ \large Ax=(2,0,0,2)^t \]

helder_edwin
 2 years ago
Best ResponseYou've already chosen the best response.1and the last one: u CANNOT compute the inverse of a nonsquare matrix.
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