Chapter 8.4, Problem 57E

To determine

The determinant of the matrix.

Answer to Problem 57E

The determinant value is D=412

Explanation of Solution

Given information:

The given matrix is shown below,

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

Formula used:

The expansion of the cofactor is $\left|A\right|=a_{i1}{C}_{i1}+a_{i2}{C}_{i2}+\mathrm{........}+a_{in}{C}_{in}$

Calculation:

If A is any square matrix, than the determinants of A in terms of cofactor expanding along i^th rowis $\left|A\right|=a_{i1}{C}_{i1}+a_{i2}{C}_{i2}+\mathrm{........}+a_{in}{C}_{in}$

Or expanding along i^th column is

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

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

To expanding from column 2, we need to find $C_{12},C_{22},C_{32},C_{42}\text{and}{C}_{52}$

  $\begin{array}{l}{C}_{12}={\left(-1\right)}^{1+2}{M}_{12}\\ =2\left[2+5-0\left(6+3\right)+4\left(30-6\right)\right]\\ =2\left[103\right]\\ C_{12}=206\end{array}$

Similarly we can find

  $\begin{array}{l}{C}_{22}={\left(-1\right)}^{2+2}{M}_{22}\\ =515\\ C_{32}={\left(-1\right)}^{3+2}{M}_{32}\\ =-443\\ C_{42}={\left(-1\right)}^{4+2}{M}_{42}\\ =-372\\ C_{52}={\left(-1\right)}^{5+2}{M}_{52}\\ =95\end{array}$

Thus we have

  $\begin{array}{l}D=a_{12}{C}_{12}+a_{22}{C}_{22}+a_{32}{C}_{32}+a_{42}{C}_{42}+a_{52}{C}_{52}\\ =2\left(206\right)+0\left(515\right)+0\left(-443\right)+0\left(-372\right)+0\left(95\right)\\ =412+0+0+0+0\\ D=412\end{array}$

Conclusion:

The determinant value is D=412