Chapter 16.8, Problem 8WE

To determine

To calculate: The value of the determinant,

  $\left|\begin{array}{ccc}26& 29& 29\\ 25& 28& 26\\ 25& 30& 27\end{array}\right|$

Answer to Problem 8WE

The value of the determinant is $101$ .

Explanation of Solution

Given information:

The determinant:

  $\left|\begin{array}{ccc}26& 29& 29\\ 25& 28& 26\\ 25& 30& 27\end{array}\right|$

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

  $\left|\begin{array}{ccc}26& 29& 29\\ 25& 28& 26\\ 25& 30& 27\end{array}\right|$

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 three by $-1$ and add it to column two.

  $\left|\begin{array}{ccc}26& 29& 29\\ 25& 28& 26\\ 25& 30& 27\end{array}\right|=\left|\begin{array}{ccc}26& 0& 29\\ 25& 2& 26\\ 25& 3& 27\end{array}\right|$

Multiply the entries of rowtwo by $-1$ and add it to row three.

  $\begin{array}{c}\left|\begin{array}{ccc}26& 29& 29\\ 25& 28& 26\\ 25& 30& 27\end{array}\right|=\left|\begin{array}{ccc}26& 0& 29\\ 25& 2& 26\\ 25& 3& 27\end{array}\right|\\ =\left|\begin{array}{ccc}26& 0& 29\\ 25& 2& 26\\ 0& 1& 1\end{array}\right|\end{array}$

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

  $\begin{array}{c}\left|\begin{array}{ccc}26& 29& 29\\ 25& 28& 26\\ 25& 30& 27\end{array}\right|=\left|\begin{array}{ccc}26& 0& 29\\ 25& 2& 26\\ 25& 3& 27\end{array}\right|\\ =\left|\begin{array}{ccc}26& 0& 29\\ 25& 2& 26\\ 0& 1& 1\end{array}\right|\\ =\left|\begin{array}{ccc}26& 0& 29\\ -1& 2& -3\\ 0& 1& 1\end{array}\right|\end{array}$

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 third row,

  $\begin{array}{c}\left|\begin{array}{ccc}26& 29& 29\\ 25& 28& 26\\ 25& 30& 27\end{array}\right|=\left|\begin{array}{ccc}26& 0& 29\\ -1& 2& -3\\ 0& 1& 1\end{array}\right|\\ =0-1\left(-78+29\right)+1\left(52\right)\\ =52+49\\ =101\end{array}$

Thus, the value of the determinant is $101$ .