Question

How do I find the determinant of a matrix using the row echelon form?

Answer 1

To find the determinant of a matrix using the row echelon form, first we need to find the determinant of the given matrix. Then we will need to convert the given matrix into a row echelon form by using elementary row operations. We will then use the row echelon form of the matrix to find the determinant of the given matrix. The determinant of a matrix is a value obtained after crossing out a row and a column by multiplying the determinant of a square matrix.

Complete step by step answer:
To find the determinant of a matrix using the row echelon form:
Let us take an example to know;
Assume that we have given a matrix $\left( \begin{matrix} 0 & 3 & 1 \\\ 1 & 1 & 2 \\\ 3 & 2 & 4 \\\ \end{matrix} \right)$.
Now, we will reduce the given matrix to row echelon form by using the elementary row operations.
First, we will interchange the first row and the second row, we will get,

0 & 3 & 1 \\\ 1 & 1 & 2 \\\ 3 & 2 & 4 \\\ \end{matrix} \right)=\left( \begin{matrix} 1 & 1 & 2 \\\ 0 & 3 & 1 \\\ 3 & 2 & 4 \\\ \end{matrix} \right)$$ Now, We will transform the first element of the third row as 1 by using the row operation. Perform: $${{R}_{3}}\to {{R}_{3}}-3{{R}_{1}}$$ $$\Rightarrow \left( \begin{matrix} 0 & 3 & 1 \\\ 1 & 1 & 2 \\\ 3 & 2 & 4 \\\ \end{matrix} \right)=\left( \begin{matrix} 1 & 1 & 2 \\\ 0 & 3 & 1 \\\ 0 & -1 & -2 \\\ \end{matrix} \right)$$ Now, We will transform the second element of the third row as 0 by using the row operation. Perform: $${{R}_{3}}\to {{R}_{3}}-\dfrac{{{R}_{2}}}{3}$$ $$\Rightarrow \left( \begin{matrix} 0 & 3 & 1 \\\ 1 & 1 & 2 \\\ 3 & 2 & 4 \\\ \end{matrix} \right)=\left( \begin{matrix} 1 & 1 & 2 \\\ 0 & 3 & 1 \\\ 0 & 0 & -\dfrac{5}{3} \\\ \end{matrix} \right)$$ We will now find the determinant of the above matrix which is in row echelon form. $$\Rightarrow \left| \begin{matrix} 1 & 1 & 2 \\\ 0 & 3 & 1 \\\ 0 & 0 & -\dfrac{5}{3} \\\ \end{matrix} \right|=1\left| \begin{matrix} 3 & 1 \\\ 0 & -\dfrac{5}{3} \\\ \end{matrix} \right|-1\left| \begin{matrix} 0 & 1 \\\ 0 & -\dfrac{5}{3} \\\ \end{matrix} \right|+2\left| \begin{matrix} 0 & 3 \\\ 0 & 0 \\\ \end{matrix} \right|$$ Solving the above and simplifying the determinant, we will get $$\Rightarrow \left| \begin{matrix} 1 & 1 & 2 \\\ 0 & 3 & 1 \\\ 0 & 0 & -\dfrac{5}{3} \\\ \end{matrix} \right|=1\left( 3 \right)\left( -\dfrac{5}{3} \right)-1\left( 0-0 \right)+2\left( 0-0 \right)$$ $$\Rightarrow \left| \begin{matrix} 1 & 1 & 2 \\\ 0 & 3 & 1 \\\ 0 & 0 & -\dfrac{5}{3} \\\ \end{matrix} \right|=-5-0+0$$ Combining the terms, we will get $$\Rightarrow \left| \begin{matrix} 1 & 1 & 2 \\\ 0 & 3 & 1 \\\ 0 & 0 & -\dfrac{5}{3} \\\ \end{matrix} \right|=-5$$ Since a row has been interchanged, then the final determinant has to be multiplied by ‘-1’. Therefore, $$\Rightarrow \left| \begin{matrix} 1 & 1 & 2 \\\ 0 & 3 & 1 \\\ 0 & 0 & -\dfrac{5}{3} \\\ \end{matrix} \right|=-5\times \left( -1 \right)$$ $$\Rightarrow \left| \begin{matrix} 1 & 1 & 2 \\\ 0 & 3 & 1 \\\ 0 & 0 & -\dfrac{5}{3} \\\ \end{matrix} \right|=5$$ Therefore, The value of the determinant of the row echelon form of the given matrix is 5. Hence, In this way we will find the determinant of the matrix using the row echelon form. **Note:** Students need to know that for every square matrix, we can associate a number which is called as the determinant of the matrix. Row echelon form is any matrix that has the first non-zero element in the first row should be one and the elements below the main diagonal should be zero. Row echelon form of a matrix is also an upper triangular matrix. Whenever a row or a column is interchanged then the determinant has to be multiplied by a negative sign.