Chapter 16.8, Problem 6WE

To determine

To calculate: The value of the determinant,

  $\left|\begin{array}{ccc}-1& 4& 12\\ 1& -2& -12\\ 8& 16& 6\end{array}\right|$

Answer to Problem 6WE

The value of the determinant is $-204$ .

Explanation of Solution

Given information:

The determinant:

  $\left|\begin{array}{ccc}-1& 4& 12\\ 1& -2& -12\\ 8& 16& 6\end{array}\right|$

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

  $\left|\begin{array}{ccc}-1& 4& 12\\ 1& -2& -12\\ 8& 16& 6\end{array}\right|$

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.

Add the corresponding entries of row two to row one,

  $\begin{array}{c}\left|\begin{array}{ccc}-1& 4& 12\\ 1& -2& -12\\ 8& 16& 6\end{array}\right|=\left|\begin{array}{ccc}-1+1& 4-2& 12-12\\ 1& -2& -12\\ 8& 16& 6\end{array}\right|\\ =\left|\begin{array}{ccc}0& 2& 0\\ 1& -2& -12\\ 8& 16& 6\end{array}\right|\end{array}$

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 first row,

  $\begin{array}{c}\left|\begin{array}{ccc}-1& 4& 12\\ 1& -2& -12\\ 8& 16& 6\end{array}\right|=\left|\begin{array}{ccc}0& 2& 0\\ 1& -2& -12\\ 8& 16& 6\end{array}\right|\\ =0-2\left|\begin{array}{cc}1& -12\\ 8& 6\end{array}\right|+0\\ =-2\left(6+96\right)\\ =-204\end{array}$

Thus, the value of the determinant is $-204$ .