Question

How do I find the determinant of a matrix Gaussian elimination?

Answer 1

Hint: Gaussian elimination is a process to solve the system of linear equations. We convert the given matrix to the upper triangular matrix. Then we will find the determinant of the upper triangular matrix.

Complete step-by-step solution:
The Gaussian elimination is normally used to solve the system of linear equations. We convert the system into the matrix form, reduce the augmented matrix into its upper triangular form or into its row echelon form.
Suppose that we are given with a $3\times 3$ matrix $\text{A}\text{.}$ We will reduce this matrix into an upper triangular matrix $\text{U}$ using elementary operations.
We can interchange two rows of the matrix $\text{A;}$ we can multiply any row of the matrix $\text{A}$ with a scalar and we can add a multiple of a row to another for reducing the matrix $\text{A}$ into an upper triangular matrix $\text{U}\text{.}$
After finding the upper triangular matrix $\text{U,}$ we calculate its determinant by multiplying the diagonal elements ${{u}_{11}},{{u}_{22}}$ and ${{u}_{33}}.$
Since the matrix $\text{A}$ is equivalent to the upper triangular matrix $\text{U,}$ the determinant of both the matrices are equal. Therefore, the determinant of the matrix $\text{A}$ is equal to the product of the diagonal elements of the upper triangular matrix $\text{U}\text{.}$ That is, $\det \text{A}={{u}_{11}}{{u}_{22}}{{u}_{33}}.$
This is true for any matrix.
Suppose $\text{A}=
  \left[ \begin{matrix}
   1 & 3 & 1 \\

   1 & 1 & -1 \\

   3 & 11 & 5 \\
\end{matrix} \right] .$
We are going to reduce $\text{A}$ into an upper triangular matrix as follows:
   $\text{A}=
  \left[ \begin{matrix}
   1 & 3 & 1 \\

   1 & 1 & -1 \\

   3 & 11 & 5 \\
\end{matrix} \right]
\sim
 \left[ \begin{matrix}
   1 & 3 & 1 \\

   0 & -2 & -2 \\

   0 & 2 & 2 \\
\end{matrix} \right]
; {{R}_{2}}\to {{R}_{2}}-{{R}_{1}}, {{R}_{3}}\to {{R}_{3}}-3{{R}_{1}} \\
  =
  \left[ \begin{matrix}
   1 & 3 & 1 \\

   0 & -2 & -2 \\

   0 & 0 & 0 \\
\end{matrix} \right]
; {{R}_{3}}\to {{R}_{2}}+{{R}_{3}} \\
     =
  \left[ \begin{matrix}
   1 & 3 & 1 \\

   0 & 1 & 1 \\

   0 & 0 & 0 \\
\end{matrix} \right]
 ; {{R}_{2}}\to -\dfrac{1}{2}{{R}_{2}} \\
$
Therefore, $\left| \text{A} \right|=1\cdot 1\cdot 0=0.$
$\left| \text{A} \right|=1\left( 5+11 \right)-3\left( 5+3 \right)+1\left( 11-3 \right)=16-3\cdot 8+8=24-24=0.$
Hence $\left| \text{A} \right|=\left| \text{U} \right|.$

Note: The determinant of a matrix is equal to the determinant of the corresponding upper triangular matrix. The determinant of an upper triangular matrix is the product of its diagonal elements.