From Section 2.1 in Steven J. Leon & Lisette de Pillis's Linear Algebra with Applications (10th edition):
Problem 13. Let $A$ be a symmetric tridiagonal matrix (i.e. $A$ is symmetric and $a_{ij} = 0$ whenever $\lvert i-j \rvert > 1$). Let $B$ be the matrix formed from $A$ by deleting the first two rows and columns. Show that $$ \det (A) = a_{11}\det \left( M_{11} \right) - a_{12}^2 \det (B).$$
My Attempt:
Here $M_{11}$ is the matrix obtained from $A$ by deleting the first row and the first column of $A$.
Let $A$ be an $n \times n$ symmetric tridiagonal matrix. We use induction on $n$. Here we can assume without any loss of generality that $n = 3, 4, 5, \ldots$.
For $n = 3$, we have $$ A = \left[ \begin{matrix} a_{11} & a_{12} & 0 \\ a_{21} & a_{22} & a_{23} \\ 0 & a_{32} & a_{33} \end{matrix} \right], $$ where $$ a_{12} = a_{21} \qquad \mbox{ and } \qquad a_{23} = a_{32}. \tag{1} $$ Then $$ B = \left[ \begin{matrix} a_{33} \end{matrix} \right]. $$ So $$ \begin{align} \det (A) &= a_{11} \det \left( \left[ \begin{matrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{matrix} \right] \right) - a_{12} \det \left( \left[ \begin{matrix} a_{21} & a_{23} \\ 0 & a_{33} \end{matrix} \right] \right) \\ &= a_{11} \det \left( M_{11} \right) - a_{12}^2 a_{33} \\ &= a_{11}\det \left( M_{11} \right) - a_{12}^2 \det (B). \end{align} $$
Now suppose that the given result holds for all symmetric tridiagonal matrices of order $k \times k$, where $k = 3, \ldots, n-1$. Let $A$ be a symmetric tridiagonal matrix of order $n \times n$. Then $$ A = \left[ \begin{matrix} a_{11} & a_{12} & 0 & 0 & 0 & 0 & \ldots & 0 & 0 \\ a_{21} & a_{22} & a_{23} & 0 & 0 & 0 & \ldots & 0 & 0 \\ 0 & a_{32} & a_{33} & a_{34} & 0 & 0 & \ldots & 0 & 0 \\ 0 & 0 & a_{43} & a_{44} & a_{45} & 0 & \ldots & 0 & 0 \\ & & & \vdots & \\ 0 & 0 & 0 & 0 & 0 & 0 & \ldots & a_{n, n-1} & a_{nn} \end{matrix} \right], $$ where $$ a_{12} = a_{21}, a_{23} = a_{32}, a_{34} = a_{43}, \ldots, a_{n-1, n} = a_{n, n-1}. $$
How to proceed from here?
Or, is there some other more suitable approach?
The determinant of a matrix can be computed by using the Lagrange expansion along any one of its rows or columns.
By using the Lagrange expansion along the first row of the matrix $A$, you get that
$\det(A)=a_{11}\det(M_{11})-a_{12}\det(M_{12})\,.$
Moreover, by applying the Lagrange expansion to the first column of the matrix $M_{12}\,,\,$ you get that
$\det(M_{12})=a_{21}\det(B)\,.$
Since the matrix $A$ is symmetric, it follows that
$\det(M_{12})=a_{12}\det(B)\,.$
Consequently ,
$\det(A)=a_{11}\det(M_{11})-a_{12}^2\det(B)\,.$