Questions: a1 -a2 0 ... 0 0 0 a2 -a3 ... 0 0 0 0 a3 ... 0 0 ... ... ... ... ... 0 0 0 0 ... an-1 -an 1 1 1 ... 1 1+an

Transcript text: $\left|\begin{array}{cccccc}a_{1} & -a_{2} & 0 & \ldots & 0 & 0 \\ 0 & a_{2} & -a_{3} & \ldots & 0 & 0 \\ 0 & 0 & a_{3} & \ldots & 0 & 0 \\ \ldots & \ldots & \ldots & \ldots & \ldots & 0 \\ 0 & 0 & 0 & \ldots & a_{n-1} & -a_{n} \\ 1 & 1 & 1 & \ldots & 1 & 1+a_{n}\end{array}\right|$

Solution

Solution Steps

To solve the determinant of the given matrix, we can use the properties of determinants and the structure of the matrix. The matrix is almost diagonal except for the last row. We can use cofactor expansion along the last row to simplify the calculation.

Step 1: Identify the Matrix Structure

The given matrix is a $ n \times n $ matrix with a specific pattern. Let's denote the matrix as $ A $:

\[ A = \begin{pmatrix} a_{1} & -a_{2} & 0 & \ldots & 0 & 0 \\ 0 & a_{2} & -a_{3} & \ldots & 0 & 0 \\ 0 & 0 & a_{3} & \ldots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \ldots & a_{n-1} & -a_{n} \\ 1 & 1 & 1 & \ldots & 1 & 1+a_{n} \end{pmatrix} \]

Step 2: Perform Row Operations

To simplify the determinant calculation, we can perform row operations. Specifically, we will subtract the first row from the last row:

\[ R_n \rightarrow R_n - R_1 \]

This operation modifies the last row:

\[ \begin{pmatrix} 1 & 1 & 1 & \ldots & 1 & 1+a_{n} \end{pmatrix} \rightarrow \begin{pmatrix} 1-1 & 1+1 & 1+0 & \ldots & 1+0 & 1+a_{n}-0 \end{pmatrix} = \begin{pmatrix} 0 & 2 & 1 & \ldots & 1 & 1+a_{n} \end{pmatrix} \]

Step 3: Simplify the Matrix

The matrix now looks like this:

\[ A' = \begin{pmatrix} a_{1} & -a_{2} & 0 & \ldots & 0 & 0 \\ 0 & a_{2} & -a_{3} & \ldots & 0 & 0 \\ 0 & 0 & a_{3} & \ldots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \ldots & a_{n-1} & -a_{n} \\ 0 & 2 & 1 & \ldots & 1 & 1+a_{n} \end{pmatrix} \]

Step 4: Calculate the Determinant

We can now expand the determinant along the first column, which has a single non-zero element $ a_1 $:

\[ \det(A') = a_1 \cdot \det\begin{pmatrix} a_{2} & -a_{3} & \ldots & 0 & 0 \\ 0 & a_{3} & \ldots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \ldots & a_{n-1} & -a_{n} \\ 2 & 1 & \ldots & 1 & 1+a_{n} \end{pmatrix} \]

Step 5: Recursive Determinant Calculation

Notice that the submatrix is of the same form as the original matrix but of size $ (n-1) \times (n-1) $. We can continue this process recursively until we reach a $ 2 \times 2 $ matrix:

\[ \det\begin{pmatrix} a_{n-1} & -a_{n} \\ 1 & 1+a_{n} \end{pmatrix} = a_{n-1}(1+a_{n}) - (-a_{n}) = a_{n-1} + a_{n-1}a_{n} + a_{n} \]