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If a system of N linear equations in N unknowns is represented by the product of a coefficient matrix and a column vector , and each row in that coefficient matrix is linearly independent, does that imply the system has a unique solution.

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3 Answers 3

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Hint: (Taken from Wikipedia)

The following are equivalent:

1) $A$ is invertible

2) $A^t$ is invertible

3) The columns of $A$ are linearly independent

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I don't understand is it the columns or the rows? which has to be linearly independent both? or one of them? –  Ethan Nov 28 '12 at 7:52
    
$A$ is invertible if and only if $A^t$ is invertible. This means that te columns of $A$ are linearly independent if and only if the rows of $A$ are linearly independent. If eany of these conditions are met, then $A$ is invertible, and hence has a unique solution $x = A^{-1} b$. –  JavaMan Nov 28 '12 at 20:52

yes it has a unique solution

for the problem :$Ax=0$ where $A$ = coefficient matrix $x$ = column vector has only $1$ solution which is $x=0$ (as the rows of $A$ are linearly independent $\implies A $ can be reduced to $I_{n\times n}$,hence finding the solution of $Ax=0$ is same as finding the solution for $I_{n\times n}x=0$).

for the problem :$Ax=y$ there will be only one solution $x=A^{-1} \cdot y$

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It implies that the coefficient matrix $A$ is invertible.
Therefore the system $Ax=b$ has the unique solution $x=A^{-1}b.$

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