Question

How to evaluate the determinant of the given matrix by reducing the matrix to row echelon form? First row $$\left(\begin{array}{lll}0& 3& 1\end{array}\right)$$ , second row $$\left(\begin{array}{lll}1& 1& 2\end{array}\right)$$ and third row $$\left(\begin{array}{lll}3& 2& 4\end{array}\right)$$?

Answer 1

Hint: Here, we have to find the determinant of the given matrix. We will 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 column by multiplying the determinant of a square matrix.

Complete Step by Step Solution:
We are given with a matrix $$\left(\begin{array}{lll}0& 3& 1\\ 1& 1& 2\\ 3& 2& 4\end{array}\right)$$.
Now, we will reduce the given matrix to row echelon form by using elementary row operations.
First, we will interchange the first row and the second row, so we get
$$\Rightarrow \left(\begin{array}{lll}0& 3& 1\\ 1& 1& 2\\ 3& 2& 4\end{array}\right)=\left(\begin{array}{lll}1& 1& 2\\ 0& 3& 1\\ 3& 2& 4\end{array}\right)$$
Now, we will transform the first element of the third row as$$1$$ by using the operation $$R_3\to R_3-3R_1$$. So, we get
$$\Rightarrow \left(\begin{array}{lll}0& 3& 1\\ 1& 1& 2\\ 3& 2& 4\end{array}\right)=\left(\begin{array}{lll}1& 1& 2\\ 0& 3& 1\\ 0& -1& -2\end{array}\right)$$
Now, we will transform the second element of the third row as $$0$$ by using the operation $$R_3\to R_3+{\displaystyle \frac{R_2}{3}}$$. So, we get
$$\Rightarrow \left(\begin{array}{lll}0& 3& 1\\ 1& 1& 2\\ 3& 2& 4\end{array}\right)=\left(\begin{array}{lll}1& 1& 2\\ 0& 3& 1\\ 0& 0& -{\displaystyle \frac{5}{3}}\end{array}\right)$$
We will now find the determinant of the above matrix which is in row-echelon form.
$$\Rightarrow \left|\begin{array}{lll}1& 1& 2\\ 0& 3& 1\\ 0& 0& -{\displaystyle \frac{5}{3}}\end{array}\right|=1\left|\begin{array}{ll}3& 1\\ 0& -{\displaystyle \frac{5}{3}}\end{array}\right|-1\left|\begin{array}{ll}0& 1\\ 0& -{\displaystyle \frac{5}{3}}\end{array}\right|+2\left|\begin{array}{ll}0& 3\\ 0& 0\end{array}\right|$$
Simplifying the determinant, we get
$$\Rightarrow \left|\begin{array}{lll}1& 1& 2\\ 0& 3& 1\\ 0& 0& -{\displaystyle \frac{5}{3}}\end{array}\right|=1\left(3\right)\left(-{\displaystyle \frac{5}{3}}\right)-1\left(0-0\right)+2\left(0-0\right)$$
$$\Rightarrow \left|\begin{array}{lll}1& 1& 2\\ 0& 3& 1\\ 0& 0& -{\displaystyle \frac{5}{3}}\end{array}\right|=1\times \left(3\right)\times \left(-{\displaystyle \frac{5}{3}}\right)$$
Multiplying the terms, we get
$$\Rightarrow \left|\begin{array}{lll}1& 1& 2\\ 0& 3& 1\\ 0& 0& -{\displaystyle \frac{5}{3}}\end{array}\right|=-5$$
Since a row has been interchanged, then the final determinant has to be multiplied by $$\left(-1\right)$$ . Therefore, we get
$$\Rightarrow \left|\begin{array}{lll}1& 1& 2\\ 0& 3& 1\\ 0& 0& -{\displaystyle \frac{5}{3}}\end{array}\right|=\left(-1\right)\times \left(-5\right)$$
$$\Rightarrow \left|\begin{array}{lll}1& 1& 2\\ 0& 3& 1\\ 0& 0& -{\displaystyle \frac{5}{3}}\end{array}\right|=5$$

Therefore the value of the determinant of the row echelon form of the given matrix is $$5$$.

Note:
We 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.