# Prove that $A=\left(\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right)$ is not invertible

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

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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What do you know about matrices so far? And from what perspective? What do you understand a matrix to be (an array of numbers satisfying some properties, the expression of a linear mapping of space)? – Mark Bennet Nov 5 '12 at 12:20
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! – GEdgar Nov 5 '12 at 14:54

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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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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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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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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I suspect that this may be more advance than what the OP may know at this stage – Belgi Nov 5 '12 at 11:49
@Belgi I think the determinant of a matrix is one of the first thing one learns before its invertibility. – user31280 Nov 5 '12 at 12:49
@F'OlaYinka I checked about ten textbooks in linear algebra, and most of them (all but two rather old books) covered inverse matrices before determinants. It's really the natural order, as the main motivation for introducing determinants is to tell whether the matrix is invertible or not. – Per Manne Nov 5 '12 at 14:12
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. – Aseem Dua Nov 5 '12 at 14:44

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

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Hint: What is the rank of $A$ ? what does this tells you ?

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