Can you factorise the determinant of the following using elementary row column operations:
\begin{bmatrix} 1 & 1 & 1 \\
1 & \sin t & \cos t \\
1 & \sin^2 t & \cos^2 t \end{bmatrix}
I can get many different determinants but can't get them to simplify down to anything nice
I got to the determinant of:
\begin{bmatrix} 0 & 1 & \sin t \\
1 & \sin t – \cos t & \cos t \\
0 & 1-\sin t -\cos t & \cos^2 t – \cos t \end{bmatrix}
Couldn't get any further with the operations so tried to just expand but didn't lead to the correct answer of:
$(1-\sin t)(1- \cos t)(\cos t – \sin t)$
Best Answer
Here's two ways to think about it: the first is just go ahead and expand the determinant right away and then factor properly. This method gives you $$ 1\cdot (\cos^2 t\sin t-\cos t\sin^2 t)-1\cdot (\cos^2 t-\sin^2 t)+1\cdot (\cos t-\sin t). $$ Writing $\cos^2 t\sin t-\cos t\sin^2 t=(\cos t-\sin t)\cos t\sin t$, then $\cos^2 t-\sin^2 t=(\cos t-\sin t)(\cos + \sin t)$, you can pull out this factor and the resulting quadratic is easy to factor, which will give you the answer.
Another way is you know that the determinant is a polynomial of the entries with degree at most $2$ in $\sin t$ and $\cos t$ and degree 3 total in $\sin t$ and $\cos t$ (this is actually a Vandermonde matrix). This can be made more rigorous, but you know when two columns of the matrix are equal, the determinant is zero (this is sufficient but not necessary). Here, this happens when $\sin t=1$, $\cos t=1$, or $\sin t=\cos t$. This suggests that the determinant factors as $(1-\sin t)(1-\cos t)(\cos t-\sin t)$, as whenever one of these factors is zero, the matrix has zero determinant.