Chapter 8.4, Problem 58E

To determine

The determinant of the matrix by the expansion of row3.

Answer to Problem 58E

The determinant of the matrix by the expansion of row3 is 112.

Explanation of Solution

Given information:

The given matrix is shown below,

  $A=\left[\begin{array}{cccc}3& 6& -5& 4\\ -2& 0& 6& 0\\ 1& 1& 2& 2\\ 0& 3& -1& -1\end{array}\right]$

Formula used:

Matrix multiplication is used.

Calculation:

The second row has two zeroes as entries. Hence, it is easier to expand using that row.

By the definition of determinants, for row 2 expansions:

  $\left|A\right|=a_{21}{C}_{21}+a_{22}{C}_{22}+a_{23}{C}_{23}+a_{24}{C}_{24}$

Since, $a_{22}=0\text{and }{a}_{24}=0$ the second term and the fourth term in the expansion vanish.

Hence, there is no need to find $C_{22}{\text{and C}}_{24}$

To find the co-factor, use the relation:

  $C_{ij}={\left(-1\right)}^{i+j}{M}_{ij}$

To find the co-factor $C_{21}$ delete the second row and first column and find the determinant of the remaining matrix.

Thus,

  $C_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{ccc}6& -5& 4\\ 1& 2& 2\\ 3& -1& -1\end{array}\right|$

Expanding by co-factors in the first column yields:

  $\begin{array}{l}{C}_{21}=6{\left(-1\right)}^2\left|\begin{array}{cc}2& 2\\ -1& -1\end{array}\right|+1{\left(-1\right)}^2\left|\begin{array}{cc}-5& 4\\ -1& -1\end{array}\right|+3{\left(-1\right)}^2\left|\begin{array}{cc}-5& 4\\ 2& 2\end{array}\right|\\ =6\left[2\left(-1\right)-\left(-1\right)\left(2\right)\right]+1\left[-5\left(-1\right)-\left(-1\right)\left(4\right)\right]+3\left[-5\left(2\right)-2\left(4\right)\right]\\ =0+9-54\\ C_{21}=-45\end{array}$

Hence,

  $C_{31}=-45$

To find the co-factor $C_{23}$ , delete the second row and third column and find the determinant of the remaining matrix.

Thus,

  $C_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{ccc}3& 6& 4\\ 1& 1& 2\\ 0& 3& -1\end{array}\right|$

Since the one of the entries in the first column of the resultant matrix is zero, in order to simplify calculations it is easier to expand using column 1.

Expanding by cofactors in the first column yields:

  $\begin{array}{l}{C}_{23}=3{\left(-1\right)}^5\left|\begin{array}{cc}1& 2\\ 3& -1\end{array}\right|+1{\left(-1\right)}^5\left|\begin{array}{cc}6& 4\\ 3& -1\end{array}\right|+0{\left(-1\right)}^5\left|\begin{array}{cc}6& 4\\ 1& 2\end{array}\right|\\ =-3\left[1\left(-1\right)-3\left(2\right)\right]-1\left[6\left(-1\right)-3\left(4\right)\right]+0\\ =21+16\\ C_{23}=37\end{array}$

Hence,

  $C_{33}=37$

Therefore,

  $\begin{array}{l}\left|A\right|=a_{31}{C}_{31}+a_{32}{C}_{32}+a_{33}{C}_{33}+a_{34}{C}_{34}\\ =-2\left(-45\right)+0+6\left(37\right)+0\\ =90+222\\ \left|A\right|=112\end{array}$

Hence, determinant of the matrix $\left[\begin{array}{cccc}3& 6& -5& 4\\ -2& 0& 6& 0\\ 1& 1& 2& 2\\ 0& 3& -1& -1\end{array}\right]$ by the expansion of row 3 is 112.

Conclusion:

The determinant of the matrix by the expansion of row3 is 112.