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graydarl

  • 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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  1. graydarl
    • 2 years ago
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    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|

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

  3. helder_edwin
    • 2 years ago
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    because u r said that the matrix is under the canonical bases of both spaces.

  4. helder_edwin
    • 2 years ago
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    (don't forget to write all vectors as columns, that way everything is easier)

  5. helder_edwin
    • 2 years ago
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    the first one is easy: just multiply the matrix by the vector (column)

  6. helder_edwin
    • 2 years ago
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    the second one: say the matrix is called A then solve the system Ax=0.

  7. helder_edwin
    • 2 years ago
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    for the third one solve the system \[ \large Ax=(2,0,0,2)^t \]

  8. helder_edwin
    • 2 years ago
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    and the last one: u CANNOT compute the inverse of a non-square matrix.

  9. graydarl
    • one year ago
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    thanks

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