values of $t \in\mathbb R$ the matrix is not invertible For which values of $t \in\mathbb R$ the matrix is not invertible ?
$$
\begin{pmatrix}
\cos t &  - \sin t   \\
\sin t  &   \cos t 
\end{pmatrix}
$$ 
well computing the determinant we know that the determinant of this matrix is $1$,i.e, is invertible , i´m confused with this question, some help please.
 A: Since the determinant is $1$ regardless of the value of $t$, and since matrices with determinant $1$ are invertible, it follows that for all values of $t\in\mathbb R$, this matrix is invertible.  The set $\{t\in\mathbb R : \text{This is not invertible}\}$ is $\varnothing$.
A: The determinant, $$\Delta=\cos(t)[\cos(t)]-[-\sin(t)][\sin(t)]\equiv \overbrace{\cos^2(t)+\sin^2(t)\equiv1}^{\text{the famous Pythagorean trig. identity}} \neq0,$$ so there are no values of $t$ for which the determinant is 0, so the matrix is invertible for all $t$ (recall that a matrix is invertible iff its determinant is nonzero).
A: Geometrically, this matrix just represents a rotation by angle $t$ in the plane so it is clearly invertible.  In fact, its inverse is the same matrix with $t$ replaced by $-t$.
A: For all values $t \in \emptyset$.
Your matrix represents a rotation in 2D-space. The rectangular coordinate system with base vectors
$$
e_x = \left[
\begin{matrix}
1 \\
0
\end{matrix}
\right], 
e_y = \left[
\begin{matrix}
0 \\
1
\end{matrix}
\right] 
$$
is rotated by $R$ into
$$
e_x' = R e_x = \left[
\begin{matrix}
\cos t \\
\sin t
\end{matrix}
\right], 
e_y' = R e_y = \left[
\begin{matrix}
-\sin t \\
\cos t
\end{matrix}
\right] 
$$
That is why that matrix is probably called $R$ by the way.
And for every rotation of angle $t$, the inverse rotation of angle $-t$ is well-defined. That is why the solution set is empty.
A more algebraic way is to investigate the determinant of the matrix. If it is not zero, the matrix can be inverted. Here your determinant is $1$, independent from $t$.
It turns out that the rotations are the linear mappings with determinant $1$.
