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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.

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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 $\varnothing$.

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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).

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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 $-t$.

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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 $1$.

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