Chapter 11, Problem 27RE

To determine

To find: The value of determinant.

Answer to Problem 27RE

The value of determinant is $108$ .

Explanation of Solution

Given information:

The determinant is $\left|\begin{array}{ccc}1& 4& 0\\ -1& 2& 6\\ 4& 1& 3\end{array}\right|$ .

Calculation:

Consider the determinant.

  $\left|\begin{array}{ccc}1& 4& 0\\ -1& 2& 6\\ 4& 1& 3\end{array}\right|$

The formula for solving a $3\times 3$ determinant is given by,

  $\left|\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right|=a_{12}\left|\begin{array}{cc}{a}_{22}& a_{23}\\ a_{32}& a_{33}\end{array}\right|-a_{12}\left|\begin{array}{cc}{a}_{21}& a_{23}\\ a_{31}& a_{33}\end{array}\right|-a_{13}\left|\begin{array}{cc}{a}_{21}& a_{22}\\ a_{31}& a_{32}\end{array}\right|$

The formula for solving the determinant is,

  $\begin{array}{c}D=\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|\\ =ad-bc\end{array}$

Solve the determinant using above formula.

  $\begin{array}{c}\left|\begin{array}{ccc}1& 4& 0\\ -1& 2& 6\\ 4& 1& 3\end{array}\right|=1\left|\begin{array}{cc}2& 6\\ 1& 3\end{array}\right|-4\left|\begin{array}{cc}-1& 6\\ 4& 3\end{array}\right|+0\left|\begin{array}{cc}-1& 2\\ 4& 1\end{array}\right|\\ =1\left(6-6\right)-a\left[-3-24\right]+0\left(-1-8\right)\\ =108\end{array}$

Therefore, the value of determinant is $108$ .