0
$\begingroup$

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.

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

$\endgroup$
  • 1
    $\begingroup$ 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 '14 at 4:13
  • 2
    $\begingroup$ ... and for this purpose it may be simpler to use column operations rather than row operations. $\endgroup$ – Robert Israel May 20 '14 at 4:16
1
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.