Chapter 8.4, Problem 53E

To determine

Find the determinant of the given matrix.

Answer to Problem 53E

The determinant of the matrix $\left(\begin{array}{ccc}1& 4& -2\\ 3& 2& 0\\ -1& 4& 3\end{array}\right)$ is $-58$ .

Explanation of Solution

Given:

The matrix

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

Let consider the matrix

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

If A is any square matrix, than the determinants of A in terms of cofactor expanding along ith row is

  $\left|A\right|=a_n{C}_n+a_{12}{C}_{n2}+\cdots +a_{nn}{C}_{nn}$

Or expanding along jth column is

  $\left|A\right|=a_{1j}{C}_{1j}+a_{2j}{C}_{2j}+\cdots +a_{nj}{C}_{nj}$

The determinant of the above matrix by the method of expansion by cofactor,

To expanding from row 1, to find $C_{11},C_{12}$ and $C_{13}$ .

  $\begin{array}{cc}{C}_{11}& ={(-1)}^{1+1}{M}_{11}\\ & =(2)(3)-(4)(0)\\ & =6-0\\ & =6\end{array}$ Similarly,

  $\begin{array}{cc}{C}_{12}& ={(-1)}^{1+2}{M}_{12}\\ & =-9\\ C_{13}& ={(-1)}^{1+3}{M}_{13}\\ & =14\end{array}$

Thus,

  $\begin{array}{cc}D& =a_{11}{C}_{11}+a_{12}{C}_{12}+a_{13}{C}_{13}\\ & =1(6)+4(-9)-2(14)\\ & =6-36-28\\ & =[-58]\end{array}$