Let $y_1$, $y_2$, $y_3$ be the eigenvalues of

$\mathrm{M\; =}$

$\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML">\; <mrow>\; <mo><br></mo>\; </mrow>\; </math>}$

Where $t$ ∈ $\left[[-\mathrm{\pi ,},\pi \right]$ ]  and

$y_1+y_2+y_3=1+\sqrt{2}$

Find the value of $t$.

The correct option is:
$$\frac{\pi }{4},-\frac{\pi }{4}$$

Step-by-Step Explanation:

We are given the 3 × 3 matrix:
$$M=\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos t& \sin t\\ 0& \sin t& \cos t\end{array}\right]$$

Observe that the matrix $M$ is block diagonal. It consists of a 1 × 1 block:
$$M_1=\left[1\right]$$
and a 2 × 2 block:
$$M_2=\left[\begin{array}{cc}\cos t& \sin t\\ \sin t& \cos t\end{array}\right]$$

The eigenvalues of a block diagonal matrix are the union of the eigenvalues of its diagonal blocks.
Therefore, the first eigenvalue is:
$$y_1=1$$

The remaining two eigenvalues, $y_2$ and $y_3$, are the eigenvalues of the block $M_2$.
To find the eigenvalues of $M_2$, we solve the characteristic equation:
$$det(M_2-\lambda I)=0$$
Substituting the components of $M_2$:
$$det\left[\begin{array}{cc}\cos t-\lambda & \sin t\\ \sin t& \cos t-\lambda \end{array}\right]=0$$
Evaluating the determinant:
$${(\cos t-\lambda )}^2-{\sin }^{2}t=0$$
Using the difference of squares factorization:
$$(\cos t-\lambda -\sin t)(\cos t-\lambda +\sin t)=0$$
This gives the two solutions:
$$y_2=\cos t+\sin t$$
$$y_3=\cos t-\sin t$$

Now, let's sum all three eigenvalues of the matrix $M$:
$$y_1+y_2+y_3=1+(\cos t+\sin t)+(\cos t-\sin t)$$
Simplifying this expression:
$$y_1+y_2+y_3=1+2\cos t$$

We are given that:
$$y_1+y_2+y_3=1+\sqrt{2}$$
Equating the two expressions for the sum of eigenvalues:
$$1+2\cos t=1+\sqrt{2}$$
Subtracting 1 from both sides:
$$2\cos t=\sqrt{2}$$
Dividing by 2:
$$\cos t=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$$

We are given that $t\in [-\pi ,\pi ]$. Within this interval, the cosine function takes the value $\frac{1}{\sqrt{2}}$ at:
$$t=\frac{\pi }{4}\text{and}t=-\frac{\pi }{4}$$