Question

By expanding along the first column. show that: $$ \text { det }\left[\begin{array}{ccccccc} 1 & 1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 1 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 1 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 1 & 1 \\ 1 & 0 & 0 & 0 & \cdots & 0 & 1 \end{array}\right]=1+(-1)^{n+1} $$ if the matrix is \(n \times n, n>2\)

Short answer

The determinant is \(1 + (-1)^{n+1}\).

Step-by-step solution

Understand the Matrix Structure

The given matrix is an upper triangular matrix with subdiagonals filled with 1s, except for the first and last rows. The first column contains 1 at the top and bottom, and the rest are 0s. We need to expand the determinant along the first column.

Apply Cofactor Expansion Along First Column

To find the determinant by expanding along the first column:\[det(A) = (-1)^{1+1} \cdot 1 \cdot det(A_{11}) + (-1)^{n+1} \cdot 1 \cdot det(A_{n1})\]where \(A_{11}\) is the minor of \(A\) by removing the first row and first column, and \(A_{n1}\) by removing the last row and first column.

Determine Det(A11)

The matrix \(A_{11}\) is a \((n-1)\times(n-1)\) submatrix that retains the upper triangular form with 1 on the subdiagonal, except at the last row which introduces a shift. Because it still holds an upper triangular form, the determinant of \(A_{11}\) is calculated from the product of its diagonal elements, which results in 1.

Determine Det(An1)

The submatrix \(A_{n1}\) retains the structure but removes the top left. It shifts the upper triangular structure downward. The determinant can be shown to be a matrix with one diagonal shift, maintaining its pattern and its determinant also resolves to 1.

Combine Results

Combine and conclude the determinant expression:\[det(A) = 1 \cdot det(A_{11}) + (-1)^{n+1} \cdot det(A_{n1})\]Thus,\[det(A) = 1 + (-1)^{n+1}\cdot 1 = 1 + (-1)^{n+1}\]

Conclusion

It is shown using the aligned expansion and matrix properties that the determinant of this matrix is indeed \(1 + (-1)^{n+1}\) for size \(n\times n\).

Key concepts

Cofactor Expansion

Cofactor expansion is a useful technique in linear algebra to compute the determinant of a matrix. It involves selecting a row or column and calculating the sum of each element's cofactor product.
This process breaks down a larger determinant problem into smaller, more manageable pieces.
In the context of the exercise, we expand along the first column of the matrix. The key steps in cofactor expansion include:

• Choose a row or column. Here, the first column of the matrix is selected because it has mostly zeros, simplifying the calculations.
• Calculate the cofactor for each non-zero element in that row or column. This includes the element values and the sign based on location.
• Multiply each element by its cofactor, and sum these products to find the determinant.

In our exercise, the matrix's first column's unique structure makes using cofactor expansion affectively manageable.

Matrix Structure

Understanding the structure of a matrix is critical for calculating the determinant, especially when using cofactor expansion. Each matrix has unique characteristics, like its dimensional arrangement and value positioning.
The given matrix in this exercise is viewed as an upper triangular matrix due to its non-zero elements forming a triangle on the upper side. This simplifies determinant calculations because:

• The matrix is square with size \( n \times n \) where each element on the main diagonal contributes to the determinant.
• Subdiagonals are filled with 1s except for shifts caused by specific rows (the first and the last).
• The first column has a structure that includes a 1 at the top and bottom, making cofactor expansion along its path a clever choice.

Being familiar with these structures allows predictions about the behavior during determinant calculations and aids in simplifying procedures.

Upper Triangular Matrix

An upper triangular matrix is one where all the elements below the main diagonal are zero. This structure simplifies finding determinants because it lets you multiply the diagonal elements directly.
Special patterns exist in this problem:

• Most elements below the main diagonal are zero, simplifying the determinant to be the product of the diagonal elements, normally.
• Sub matrices formed in the step by step solution also maintain a resemblance to the upper triangular structure.
• The diagonal in this matrix case comes into play importantly due to its expanded nature, allowing a representation of upper triangular identities that resolves to simple equations.

For matrices maintaining these structures, calculating the determinant simplifies into tracking their diagonal, heavily aiding in our problem's computational solution.