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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In this case, C must be a 3x3 matrix. |
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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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