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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 \[\left( {\begin{array}{*{20}{l}}3&2&4\end{array}} \right)\]?

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Last updated date: 29th Feb 2024
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IVSAT 2024
Answer
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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.