Chapter 16.8, Problem 12WE

To determine

To calculate: The value of the determinant,

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

Answer to Problem 12WE

The value of the determinant is $-192$ .

Explanation of Solution

Given information:

The determinant:

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

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

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

The value of new determinant is same as the original one if all entries of either a row or a column are multiplied by a real number k , and the resultant products are added to corresponding entries of either another row or another column.

Multiply the entries of column two by $-1$ and add it to column one.

  $\left|\begin{array}{cccc}1& 1& 1& -1\\ 2& 2& -2& 2\\ -3& -3& 3& 3\\ -4& 4& 4& 4\end{array}\right|=\left|\begin{array}{cccc}0& 1& 1& -1\\ 0& 2& -2& 2\\ 0& -3& 3& 3\\ -8& 4& 4& 4\end{array}\right|$

Multiply the entries of column three by $-1$ and add it to column two.

  $\begin{array}{c}\left|\begin{array}{cccc}1& 1& 1& -1\\ 2& 2& -2& 2\\ -3& -3& 3& 3\\ -4& 4& 4& 4\end{array}\right|=\left|\begin{array}{cccc}0& 1& 1& -1\\ 0& 2& -2& 2\\ 0& -3& 3& 3\\ -8& 4& 4& 4\end{array}\right|\\ =\left|\begin{array}{cccc}0& 0& 1& -1\\ 0& 4& -2& 2\\ 0& -6& 3& 3\\ -8& 0& 4& 4\end{array}\right|\end{array}$

Multiply the entries of column four by 1 and add it to column three.

  $\begin{array}{c}\left|\begin{array}{cccc}1& 1& 1& -1\\ 2& 2& -2& 2\\ -3& -3& 3& 3\\ -4& 4& 4& 4\end{array}\right|=\left|\begin{array}{cccc}0& 1& 1& -1\\ 0& 2& -2& 2\\ 0& -3& 3& 3\\ -8& 4& 4& 4\end{array}\right|\\ =\left|\begin{array}{cccc}0& 0& 1& -1\\ 0& 4& -2& 2\\ 0& -6& 3& 3\\ -8& 0& 4& 4\end{array}\right|\\ =\left|\begin{array}{cccc}0& 0& 0& -1\\ 0& 4& 0& 2\\ 0& -6& 6& 3\\ -8& 0& 8& 4\end{array}\right|\end{array}$

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 fourth column,

  $\begin{array}{c}\left|\begin{array}{cccc}1& 1& 1& -1\\ 2& 2& -2& 2\\ -3& -3& 3& 3\\ -4& 4& 4& 4\end{array}\right|=1\left|\begin{array}{ccc}0& 4& 0\\ 0& -6& 6\\ -8& 0& 8\end{array}\right|\\ =1\left[-4\left(48\right)\right]\\ =-192\end{array}$

Thus, the value of the determinant is $-192$ .