Chapter 8.4, Problem 55E

To determine

Find the determinant of the given matrix.

Answer to Problem 55E

The determinant of the given matrix is 0.

Explanation of Solution

Given:

The matrix

  $\left(\begin{array}{cccc}5& 3& 0& 6\\ 4& 6& 4& 12\\ 0& 2& -3& 4\\ 0& 1& -2& 2\end{array}\right)$

Let consider the matrix

  $\left(\begin{array}{cccc}5& 3& 0& 6\\ 4& 6& 4& 12\\ 0& 2& -3& 4\\ 0& 1& -2& 2\end{array}\right)$

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

  $\left|A\right|=a_{i1}{C}_{i1}+a_{i2}{C}_{i2}+\cdots +a_{in}{C}_{in}$

Or expanding along $j$ th 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 column 1, to find $C_{11},C_{21},C_{31}$ and $C_{41}$ .

  $\begin{array}{cc}{C}_{11}& ={(-1)}^{1\cdot 1}{M}_{11}\\ & =6(-6+8)-4(4-4)+12(-4+3)\\ & =12-0-12\\ & =0\end{array}$

Similarly,

  $\begin{array}{cc}{C}_{23}& ={(-1)}^{2\cdot 3}{M}_{23}\\ & =0\\ C_{33}& ={(-1)}^{343}{M}_{33}\\ & =0\\ C_{43}& ={(-1)}^{4+3}{M}_{43}\\ & =0\end{array}$

Thus,

  $\begin{array}{cc}D& =a_{11}{C}_{11}+a_{21}{C}_{21}+a_{31}{C}_{31}+a_{41}{C}_{41}\\ & =5(0)+4(0)+0(0)+0(0)\\ & =0\end{array}$