Chapter 16.8, Problem 3OE

To determine

To calculate: The value of the determinant,

  $\left|\begin{array}{ccc}0& 3& 1\\ 1& 2& 0\\ -1& 5& 2\end{array}\right|$

Answer to Problem 3OE

The value of the determinant is $1$ .

Explanation of Solution

Given information:

The value of the determinant:

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

Formula used:

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.

Calculation:

Consider the determinant

  $\left|\begin{array}{ccc}0& 3& 1\\ 1& 2& 0\\ -1& 5& 2\end{array}\right|$

It is provided that the value of the determinant

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

Observe that the first row is multiplied by 1 and then the resultant products are added to third row.

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.

  $\begin{array}{c}\left|\begin{array}{ccc}0& 3& 1\\ 1& 2& 0\\ -1& 2& 1\end{array}\right|=\left|\begin{array}{ccc}0\times 1& 3\times 1& 1\times 1\\ 1& 2& 0\\ -1& 2& 1\end{array}\right|\\ =\left|\begin{array}{ccc}0& 3& 1\\ 1& 2& 0\\ -1+\left(0\times 1\right)& 2+\left(3\times 1\right)& 1+\left(1\times 1\right)\end{array}\right|\\ =\left|\begin{array}{ccc}0& 3& 1\\ 1& 2& 0\\ -1& 5& 2\end{array}\right|\\ =1\end{array}$

Thus, the value of the determinant is $1$ .