Question

If A,B,C are the angles of a triangle then \[\left[\begin{matrix}\cos^2A & CotA & 1 \\ \cos^2B & cotB & 1 \\ \cos^2C & cotC & 1\end{matrix}\right]\]

Find Its Determinant ?

Answer 1

what about it?

Answer 2

Oh..No....Sorry We Have to Find Its Determinant

Answer 3

cot A = cos(A)/sin(A) ?

Answer 4

I did Everything but they all Luk Werid.. :(

Answer 5

did you try
\(L_1\leftarrow L_1-L_3\)
\(L_2 \leftarrow L_2-L_3\)
?

Answer 6

the last column is a linear transform of the first column.
\[
\left[\begin{matrix}\cos^2A&\cot A &\cos^2A\\
\cos^2B&\cot B &\cos^2B\\
\cos^2C&\cot C &\cos^2C\\
\end{matrix}\right]+
\left[\begin{matrix} 0&0&\sin^2A\\
0&0&\sin^2B\\
0&0&\sin^2C\\
\end{matrix}\right]
\]

Answer 7

but det(A+B) is not something as easy as det(A)+det(B).
What I meant is:
\(\left[\begin{array}{ccc}
\cos^2A-\cos^2C & \cot A-\cot C & 0\\
\cos^2B-\cos^2C & \cot B-\cot C & 0\\
\cos^C & \cot C & 1
\end{array}\right]
\)
so the determinant is
\[
(\cos^2A-\cos^2C)(\cot B-\cot C) - (\cot A-\cot C)(\cos^2B-\cos^2C)\\
=\cos^2A\cot B - \cos^2 A\cot C - \cos^2C\cot B\\
\:\: -(\cot A\cos^2 - \cot A\cos^2C - \cot C\cos^2B)
\]