Chapter 16.9, Problem 2ST

To determine

To calculate: The value of the determinant,

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

Answer to Problem 2ST

The value of the determinant is $-2$ .

Explanation of Solution

Given information:

The determinant:

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

Formula used:

For the matrix of order $2\times 2$ ,

  $A=\left[\begin{array}{ll}p\hfill & q\hfill \\ r\hfill & s\hfill \end{array}\right]$

The determinant is given as $\left|A\right|=ps-qr$ .

Calculation:

Consider the determinant

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

Recall that for the matrix of order $2\times 2$ ,

  $A=\left[\begin{array}{ll}p\hfill & q\hfill \\ r\hfill & s\hfill \end{array}\right]$

The determinant is given as $\left|A\right|=ps-qr$ .

Expand by minors about third column,

  $\begin{array}{c}\left|\begin{array}{cccc}2& 1& 0& -2\\ -1& 0& 0& 4\\ 2& 3& 6& -1\\ 0& 1& 2& 2\end{array}\right|=0-0+6\left|\begin{array}{ccc}2& 1& -2\\ -1& 0& 4\\ 0& 1& 2\end{array}\right|-2\left|\begin{array}{ccc}2& 1& -2\\ -1& 0& 4\\ 2& 3& -1\end{array}\right|\\ =6\left[2\left(-4\right)+1\left(2+2\right)\right]-2\left[-1\left(1-8\right)-3\left(8-2\right)\right]\\ =6\left(-8+4\right)-2\left(7-18\right)\\ =-2\end{array}$

Thus, the value of the determinant is $-2$ .