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Am I correct to say that this matrix $C$ cannot be found $$C\times\left(\begin{array}{cc} 9 & 1\\ 4 & 6\\ 3 & 4\end{array}\right) = \left(\begin{array}{cc}9&1\\4 & 6\\ 3&4\end{array}\right)$$

because the columns does not match the rows?

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en.wikipedia.org/wiki/… –  pedja Apr 28 '12 at 5:43
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In this case, C must be a 3x3 matrix.

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how does that work? 3X3 x2X3 != 2X3 –  Xabi Apr 28 '12 at 5:50
    
3x3 X 3x2 = 3x2 –  clookid Apr 28 '12 at 5:53
    
Try $C=I_3$ and multiplying it out for yourself. Not positive it's the only solution, but it should work. –  Mike Apr 28 '12 at 5:58
    
Oops! :/ Should have read that better –  Xabi Apr 28 '12 at 6:00
    
@Fatz: first the number of rows, then the number of columns. Your two matrices, other than $C$, are both $3\times 2$. –  Arturo Magidin Apr 28 '12 at 6:06
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No, you are not. $C=I_{3\times 3}$ where $I_{3\times 3}$ is the identity matrix works.

Note that your equation can be viewed as:

$$ Cv_1=v_1,\qquad Cv_2=v_2 $$ where $v_1=(9,4,3)^T$, $v_2=(1,6,4)^T$. This implies that $v_1$ and $v_2$ are eigenvectors with respect to the eigenvalue $\lambda = 1$ of $C$. (And thus $C$ is not unique.)

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