Proving Invertibility of a Matrix: Ax=e1

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SUMMARY

The discussion centers on proving the invertibility of a matrix A through the equation Ax=e1, where e1 represents the first column of the identity matrix. It is established that if A is invertible, then A^{-1}Ax equals e1, leading to the conclusion that A^{-1}e1 yields the first column of the inverse of A. Participants are encouraged to verify this by multiplying various matrices by e1 to observe the outcome, reinforcing the concept that the product of a matrix and e1 results in the first column of that matrix.

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  • Basic concepts of linear algebra
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Students and professionals in mathematics, particularly those studying linear algebra, as well as educators seeking to clarify concepts related to matrix invertibility and its applications.

Ali Asadullah
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Let A be an invertible matrix.
Then Ax=e1 will give us the first column of the inverse of A.
Where e1 is the first column of the identity matrix.

How can we prove this fact??:confused:
 
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From [itex]Ax= e_1[/itex] and the fact that A is invertible, we have [itex]A^{-1}Ax= x= A^{-1}e_1[/itex]. Now, can you convince yourself that any matrix times [itex]e_1[/itex] is the first column of the matrix? Try multiplying a few matrices times [itex]e_1[/itex] and see what happens:
What is
[tex]\begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{bmatrix}\begin{bmatrix}1 \\ 0\end{bmatrix}[/tex]

What is
[tex]\begin{bmatrix}a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{bmatrix}\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}[/tex]
 
OMG that was too simple thank u HallsofIvy :)
 

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