Question

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

Answer 1

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|

Answer 2

u have
\[ \large f(x)=\begin{pmatrix}
1 & 1 & 2 & 1 & 0 & 2\\ 1 & 2 & 3 & 1 & 0 & 1\\
0 & 1 & 2 & 1 & 1 & 1\\ 1 & 1 & 1 & 1 & 1 & 1
\end{pmatrix}x \]

Answer 3

because u r said that the matrix is under the canonical bases of both spaces.

Answer 4

(don't forget to write all vectors as columns, that way everything is easier)

Answer 5

the first one is easy: just multiply the matrix by the vector (column)

Answer 6

the second one: say the matrix is called A then solve the system Ax=0.

Answer 7

for the third one solve the system
\[ \large Ax=(2,0,0,2)^t \]

Answer 8

and the last one: u CANNOT compute the inverse of a non-square matrix.

Answer 9

thanks