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How do I reduce this matrix to row echelon form and hence find the determinant, or is there a way that I am unaware of that finds the determinant of this matrix without having to reduce it row echelon form given this is all I know and there exists no additional information.

$\left[ \begin{array}{ccc} a & 1 & -1 \\ a & 2a+2 & a \\ a-3 & a-3 & a-3 \\ \end{array} \right]$

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    $\begingroup$ It's not hard to make 0 two of the three entries on the third row. That reduces the problem to find the determinant of a 2x2 matrix. $\endgroup$
    – user121880
    May 20, 2014 at 4:13
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    $\begingroup$ ... and for this purpose it may be simpler to use column operations rather than row operations. $\endgroup$ May 20, 2014 at 4:16

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There are lots of ways to do this. One way is to note that $$ p(a)= \begin{vmatrix} a & 1 & -1 \\ a & 2a+2 & a \\ a-3 & a-3 & a-3 \end{vmatrix} $$ is a degree three polynomial. If we find three zeros, then we can find a formula for $p(a)$.

Note that $$ p(3)= \begin{vmatrix} 3 & 1 & -1 \\ 3 & 8 & 3 \\ 0 & 0 & 0 \end{vmatrix}=0 $$ so $(a-3)$ divides $p(a)$. Also note that $$ p(-1)= \begin{vmatrix} -1 & 1 & -1 \\ -1 & 0 & -1 \\ -4 & -4 & -4 \end{vmatrix}=0 $$ since the first and third columns are identical. This means $(a+1)$ divides $p(a)$.

Putting this together, we have that $$ p(a)=(a-3)(a+1)(a-r) $$ Can you find r?

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You may think of the determinant of this matrix as a polynomial of degree $\leq 3$ in $a$. Now there are some values of $a$ that make two rows or two columns identical or one row as a zero row.

For example: if $a=3$, then the last row is zero. What this means is this polynomial will have a factor $(a-3)$. Now try to see if there are other values of $a$ that can make first and third column equal (hence determinant will be zero). That can give you another factor and so on.

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