Chapter 8.4, Problem 51E

To determine

Find the determinant of the given matrix.

Answer to Problem 51E

The determinant of the matrix $\left(\begin{array}{ccc}-1& 8& -3\\ 0& 3& -6\\ 0& 0& 3\end{array}\right)$ is -9.

Explanation of Solution

Given:

The matrix

  $\left(\begin{array}{ccc}-1& 8& -3\\ 0& 3& -6\\ 0& 0& 3\end{array}\right)$

Let consider the matrix

  $\left(\begin{array}{ccc}-1& 8& -3\\ 0& 3& -6\\ 0& 0& 3\end{array}\right)$

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

  $\left|A\right|=a_n{C}_n+a_{12}{C}_{i2}+\cdots +a_n{C}_{in}$

Or expanding along jth column is

  $\left|A\right|=a_1,C_{1j}+a_2,C_{2j}+\cdots +a_{nj}{C}_{nj}$

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

To expanding from column 1, to find $C_{11},C_{21}$ and $C_{31}$ .

  $\begin{array}{cc}{C}_{11}& ={(-1)}^{1+1}{M}_{11}\\ & =(3)(3)-(0)(-6)\\ & =9\end{array}$

Similarly,

  $\begin{array}{cc}{C}_{21}& ={(-1)}^{2+1}{M}_{21}\\ & =-24\\ C_{13}& ={(-1)}^{1+3}{M}_{13}\\ & =-39\end{array}$

Thus,

  $\begin{array}{cc}D& =a_{11}{C}_{11}+a_{21}{C}_{21}+a_{31}{C}_{31}\\ & =-1(9)+0(-24)+0(-39)\\ & =-9+0+0\\ & =-9\end{array}$