Chapter 16.8, Problem 13WE

To determine

To prove: If the corresponding elements of either two entire rows or a columns are same then the value of determinant is 0.

Explanation of Solution

Given information:

The determinant of a 3 by 3 matrix.

The value of new second order 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.

If either an entire row or an entire column is 0 then the value of determinant is 0.

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 $\left|A\right|=ps-qr$ .

Proof:

Consider a 3 by 3 matrix say, $\left(\begin{array}{ccc}a& b& c\\ d& e& f\\ a& b& c\end{array}\right)$ .

Multiply all entries of row 2 by a real number say k and add the corresponding product to row 1 to compute the determinant.

  $\begin{array}{c}\left|\begin{array}{ccc}a+dk& b+ek& c+fk\\ d& e& f\\ a& b& c\end{array}\right|=\left(a+dk\right)\left|\begin{array}{cc}e& f\\ b& c\end{array}\right|-\left(b+ek\right)\left|\begin{array}{cc}d& f\\ a& c\end{array}\right|+\left(c+fk\right)\left|\begin{array}{cc}d& e\\ a& b\end{array}\right|\\ =\left(a+dk\right)\left(ec-bf\right)-\left(b+ek\right)\left(dc-af\right)+\left(c+fk\right)\left(db-ae\right)\\ =aec-abf-bdc+baf+cdb-cae+k\left(ecd-bfd-dce+eaf+dbf-aef\right)\\ =0+k\left(0\right)\\ =0\end{array}$

The value of determinant is 0.

Hence, it is proved that if the corresponding elements of either two entire rows or a columns are same then the value of determinant is 0.