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frx
Question: For which real constants a and b does the vectors [(1,a,0),(a,1,1),(-b,-1,-1)] lie in the same plane? So know that the lines are in the same plane when the determinant is 0 so I've calulated it like this: \[A=\left[\begin{matrix}1 & a & 0 \\ a & 1 & 1 \\-b &-1&-1 \end{matrix}\right]\] \[\det(A)=1\left[\begin{matrix}1 & 1 \\ -1 & -1\end{matrix}\right]-a \left[\begin{matrix}a & 1\\ -b & -1\end{matrix}\right]+0= -a(-a+b)=a( a-b)\] \[a(a-b)=0 \] \[a=0, b=a\] It's right but the student solution guide solve it \[\left[\begin{matrix}1 & a & -b \\ a & 1&-1 \\ 0&1&-1\end{matrix}\right]\]
they are using that one as a determinate, is my solution equally right or should I spend some time learn there way?
equally right. it's a property of the determinant that transposing it will not change the value of the determinant