Chapter 10.6, Problem 27E

To determine

With the help of row and column operation the determinant of the given matrix is to be calculated:

  $\left|\begin{array}{cccc}0& 0& 4& 6\\ 2& 1& 1& 3\\ 2& 1& 2& 3\\ 3& 0& 1& 7\end{array}\right|$

Answer to Problem 27E

The determinant of given matrix $\left|\begin{array}{cccc}0& 0& 4& 6\\ 2& 1& 1& 3\\ 2& 1& 2& 3\\ 3& 0& 1& 7\end{array}\right|$ is $-18$ .

Explanation of Solution

Given:

  $\left|\begin{array}{cccc}0& 0& 4& 6\\ 2& 1& 1& 3\\ 2& 1& 2& 3\\ 3& 0& 1& 7\end{array}\right|$

Concept Used:

The determinant of every square matrix with the order $n$ is defined as $A=\left[a_{ij}\right]$ where,

  $a={\left(i,j\right)}^{th}$ of element $A$ .

Calculations:

The determinant of the given matrix is calculated as:

  $\begin{array}{l}\text{Expand}\text{by}\text{R3}\\ \left|\begin{array}{cccc}0& 0& 4& 6\\ 2& 1& 1& 3\\ 2& 1& 2& 3\\ 3& 0& 1& 7\end{array}\right|=R3\to R3-R2\left|\begin{array}{cccc}0& 0& 4& 6\\ 2& 1& 1& 3\\ 0& 0& 1& 0\\ 3& 0& 1& 7\end{array}\right|\\ ={\left(-1\right)}^{3+3}\left|\begin{array}{ccc}0& 0& 6\\ 2& 1& 3\\ 3& 0& 7\end{array}\right|\\ =\left|\begin{array}{ccc}0& 0& 6\\ 2& 1& 3\\ 3& 0& 7\end{array}\right|\\ =6\left(2\times 0-3\times 1\right)\\ =6\left(0-3\right)\\ =-18\end{array}$

Now, its determinant value is $D=-18$

Conclusion:

Hence, the determinant value of $\left|\begin{array}{cccc}0& 0& 4& 6\\ 2& 1& 1& 3\\ 2& 1& 2& 3\\ 3& 0& 1& 7\end{array}\right|$ is $D=-18$ .