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$A=\left(\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right)$

I don't know how to start. Will be grateful for a clue.

Edit: Matrix ranks and Det have not yet been presented in the material.

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    Perhaps you learned how to do Gaussian reduction to determine whether a matrix is invertible? Maybe called "echelon form" or something? Mark is right ... we don't know what you have done so far. So ... as the instructor for help!2012-11-05

6 Answers 6

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Row operations or column operationas don't change the rank of the matrix. So it is invertible iff anything row/column equivalent to it is invertible.

Subtract 1 times the second column from the third, 1 times the first column from the second column; this gives $\begin{pmatrix} 1&1&1\\ 4&1&1 \\7&1&1\end{pmatrix}.$ This is clearly singular since two columns are equal.

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Notice that $A=\left(\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right)\left(\begin{array}{c}1\\-2\\1\end{array}\right)=\left(\begin{array}{c}0\\0\\0\end{array}\right)$ So, if $A$ were invertible, this would imply $\left(\begin{array}{c}1\\-2\\1\end{array}\right)=\left(\begin{array}{c}0\\0\\0\end{array}\right).$ This is again a variation on Ihf's answer.

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Note that $L_3-L_2=L_2-L_1$. What does that imply about the rank of $A$?

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If you really have to do this with only elementary tools, you can set up the equation $\pmatrix{1 & 2 & 3 \cr 4 & 5 & 6 \cr 7 & 8 & 9}\pmatrix{a & b & c \cr d & e & f \cr g & h & i}=\pmatrix{1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1}$ This gives you $9$ equations, but they can be divided nicely into groups of three by considering each column separately. For instance, we get $a+2d+3g=1$ $4a+5d+6g=0$ $7a+8d+9g=0$ This system is easily shown to be inconsistent, for example by subtracting the second equation from the third equation, and by subtracting the first equation from the second equation. You will then obtain the two equations $3a+3d+3g=0$ $3a+3d+3g=-1$ which cannot be true at the same time.

(This argument is really the same as the one given by lhf.)

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Best way to check for inverse existence would be to calculate the Determinant. If it's non zero inverse will exist.

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    The only argument which seems to be sensible in a case where determinant is not known (understood) is showing that two rows/ columns are linearly dependent.2012-11-05
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Hint: What is the rank of $A$ ? what does this tells you ?