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# How to evaluate the determinant of the given matrix by reducing the matrix to row echelon form? First row $\left( {\begin{array}{*{20}{l}}0&3&1\end{array}} \right)$ , second row $\left( {\begin{array}{*{20}{l}}1&1&2\end{array}} \right)$ and third row $\left( {\begin{array}{*{20}{l}}3&2&4\end{array}} \right)$?

Last updated date: 21st Jun 2024
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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 $\left( {\begin{array}{*{20}{l}}0&3&1\\1&1&2\\3&2&4\end{array}} \right)$.
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
$\Rightarrow \left( {\begin{array}{*{20}{l}}0&3&1\\1&1&2\\3&2&4\end{array}} \right) = \left( {\begin{array}{*{20}{l}}1&1&2\\0&3&1\\3&2&4\end{array}} \right)$
Now, we will transform the first element of the third row as$1$ by using the operation ${R_3} \to {R_3} - 3{R_1}$. So, we get
$\Rightarrow \left( {\begin{array}{*{20}{l}}0&3&1\\1&1&2\\3&2&4\end{array}} \right) = \left( {\begin{array}{*{20}{l}}1&1&2\\0&3&1\\0&{ - 1}&{ - 2}\end{array}} \right)$
Now, we will transform the second element of the third row as $0$ by using the operation ${R_3} \to {R_3} + \dfrac{{{R_2}}}{3}$. So, we get
$\Rightarrow \left( {\begin{array}{*{20}{l}}0&3&1\\1&1&2\\3&2&4\end{array}} \right) = \left( {\begin{array}{*{20}{l}}1&1&2\\0&3&1\\0&0&{ - \dfrac{5}{3}}\end{array}} \right)$
We will now find the determinant of the above matrix which is in row-echelon form.
$\Rightarrow \left| {\begin{array}{*{20}{l}}1&1&2\\0&3&1\\0&0&{ - \dfrac{5}{3}}\end{array}} \right| = 1\left| {\begin{array}{*{20}{l}}3&1\\0&{ - \dfrac{5}{3}}\end{array}} \right| - 1\left| {\begin{array}{*{20}{l}}0&1\\0&{ - \dfrac{5}{3}}\end{array}} \right| + 2\left| {\begin{array}{*{20}{l}}0&3\\0&0\end{array}} \right|$
Simplifying the determinant, we get
$\Rightarrow \left| {\begin{array}{*{20}{l}}1&1&2\\0&3&1\\0&0&{ - \dfrac{5}{3}}\end{array}} \right| = 1\left( 3 \right)\left( { - \dfrac{5}{3}} \right) - 1\left( {0 - 0} \right) + 2\left( {0 - 0} \right)$
$\Rightarrow \left| {\begin{array}{*{20}{l}}1&1&2\\0&3&1\\0&0&{ - \dfrac{5}{3}}\end{array}} \right| = 1 \times \left( 3 \right) \times \left( { - \dfrac{5}{3}} \right)$
Multiplying the terms, we get
$\Rightarrow \left| {\begin{array}{*{20}{l}}1&1&2\\0&3&1\\0&0&{ - \dfrac{5}{3}}\end{array}} \right| = - 5$
Since a row has been interchanged, then the final determinant has to be multiplied by $\left( { - 1} \right)$ . Therefore, we get
$\Rightarrow \left| {\begin{array}{*{20}{l}}1&1&2\\0&3&1\\0&0&{ - \dfrac{5}{3}}\end{array}} \right| = \left( { - 1} \right) \times \left( { - 5} \right)$
$\Rightarrow \left| {\begin{array}{*{20}{l}}1&1&2\\0&3&1\\0&0&{ - \dfrac{5}{3}}\end{array}} \right| = 5$

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.