Chapter 16.8, Problem 5WE

To determine

To calculate: The value of the determinant,

  $\left|\begin{array}{ccc}2& -4& 6\\ 5& 0& -8\\ 3& -6& 9\end{array}\right|$

Answer to Problem 5WE

The value of the determinant is $0$ .

Explanation of Solution

Given information:

The determinant:

  $\left|\begin{array}{ccc}2& -4& 6\\ 5& 0& -8\\ 3& -6& 9\end{array}\right|$

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

  $\left|\begin{array}{ccc}2& -4& 6\\ 5& 0& -8\\ 3& -6& 9\end{array}\right|$

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 column three to column two,

  $\begin{array}{c}\left|\begin{array}{ccc}2& -4& 6\\ 5& 0& -8\\ 3& -6& 9\end{array}\right|=\left|\begin{array}{ccc}2& -4+6& 6\\ 5& 0+\left(-8\right)& -8\\ 3& -6+9& 9\end{array}\right|\\ =\left|\begin{array}{ccc}2& 2& 6\\ 5& -8& -8\\ 3& 3& 9\end{array}\right|\end{array}$

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

  $\begin{array}{c}\left|\begin{array}{ccc}2& -4& 6\\ 5& 0& -8\\ 3& -6& 9\end{array}\right|=\left|\begin{array}{ccc}2& -4+6& 6\\ 5& 0+\left(-8\right)& -8\\ 3& -6+9& 9\end{array}\right|\\ =\left|\begin{array}{ccc}2+2\left(-1\right)& 2& 6\\ 5+\left(-8\right)\left(-1\right)& -8& -8\\ 3+3\left(-1\right)& 3& 9\end{array}\right|\\ =\left|\begin{array}{ccc}0& 2& 6\\ 13& -8& -8\\ 0& 3& 9\end{array}\right|\end{array}$

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 column,

  $\begin{array}{c}\left|\begin{array}{ccc}2& -4& 6\\ 5& 0& -8\\ 3& -6& 9\end{array}\right|=\left|\begin{array}{ccc}0& 2& 6\\ 13& -8& -8\\ 0& 3& 9\end{array}\right|\\ =0-13\left|\begin{array}{cc}2& 6\\ 3& 9\end{array}\right|+0\\ =-13\left(18-18\right)\\ =0\end{array}$

Thus, the value of the determinant is $0$ .