Question

Use Gaussian elimination to find the determinant of the matrices A in Exercises 1 through 10.

7. $\mathbf{\left[}\begin{array}{ccccc}\mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{1}\\ \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{1}& \mathbf{2}\\ \mathbf{0}& \mathbf{0}& \mathbf{1}& \mathbf{2}& \mathbf{3}\\ \mathbf{0}& \mathbf{1}& \mathbf{2}& \mathbf{3}& \mathbf{4}\\ \mathbf{1}& \mathbf{2}& \mathbf{3}& \mathbf{4}& \mathbf{5}\end{array}\mathbf{\right]}$

Short answer

Therefore, the determinant of given matrix is given by,

$detA=1$

Step-by-step solution

Definition

Gaussian elimination method is used to solve a system of linear equations.

Gaussian elimination provides a relatively efficient way of constructing the inverse to a matrix.

Interchanging two rows. Multiplying a row by a constant (any constant which is not zero).

Given

Given Matrix,

$A=\left[\begin{array}{ccccc}0& 0& 0& 0& 1\\ 0& 0& 0& 1& 2\\ 0& 0& 1& 2& 3\\ 0& 1& 2& 3& 4\\ 1& 2& 3& 4& 5\end{array}\right]$

To find determinant by using Gaussian Eliminations

We switch the first and fifth row, and then the second and fourth row. We get

We get,

$A=\left[\begin{array}{ccccc}1& 2& 3& 4& 5\\ 0& 1& 2& 3& 4\\ 0& 0& 1& 2& 3\\ 0& 0& 0& 1& 2\\ 0& 0& 0& 0& 1\end{array}\right]$

We had two row interchanges, so

$$\begin{gathered} detA={\left(-1\right)}^2·1·1·1·1·1 \\ detA=1 \end{gathered}$$