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Find all the values of c for which the following matrix is singular: $$\begin{bmatrix} 1 & c & c \\ c & c & c \\ 2 & c & 3 \end{bmatrix}$$

Anyone know how to solve this?

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Write down its determinant as a function of $c$. –  Henning Makholm Apr 22 '12 at 15:47
    
You can create a singular matrix by having a zero row or zero column, or by making two rows equal, or by making two columns equal. These aren't the only ways a matrix can be singular, but they can sometimes be found by quick inspection. The determinant will likely be a cubic in c (c in all three rows and columns, no obvious cancellation) so spotting a solution could help. –  Mark Bennet Apr 22 '12 at 16:09

2 Answers 2

up vote 3 down vote accepted

You could also apply Gaussian Elimination to get: $$ \begin{pmatrix} 1 & c & c \\ 0 & c^2 - c & c - c^2 \\ 0 & 0 & -c + 3\\ \end{pmatrix} $$ This matrix is singular if any element on the diagonal is zero, i.e, if: $$ c^2 - c = 0 \text{ or } -c+3 = 0 $$ which is equivalent to computing the determinant..

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Thats an interesting way of doing it. –  Jim_CS Apr 22 '12 at 16:49

The determinant equals $3c-4c^2+c^3$. Compute it by means of Laplace's algorithm, for instance. Otherwise, use the standard trick for $3 \times 3$ matrices. Hence the matrix is singular when $c=0$. Now you can divide the equation $3c-4c^2+c^3=0$ by $c$, and find the other solutions $c=1$ and $c=3$.

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