Chapter 8, Problem 116RE

a.

To determine

The minors of the matrix $\left[\begin{array}{ccc}8& 3& 4\\ 6& 5& -9\\ -4& 1& 2\end{array}\right]$ .

Answer to Problem 116RE

The minors of the given matrix are $M_{11}=19,M_{12}=-24,M_{13}=26,M_{21}=2,M_{22}=32,$

  $M_{23}=20,M_{31}=-47,M_{32}=-96,M_{33}=22.$

Explanation of Solution

Given information :

Given matrix $\left[\begin{array}{ccc}8& 3& 4\\ 6& 5& -9\\ -4& 1& 2\end{array}\right]$

The minor of an element $a_{11}$ of a matrix $\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]$ is $M_{11}=\left|\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right|\left(\text{delete}\text{first}\text{row}\text{and}\text{first}\text{column}\right)$

Similar way, minor of $8$ is $M_{11}=\left|\begin{array}{cc}5& -9\\ 1& 2\end{array}\right|=10+9=19$

  $\begin{array}{l}\text{minor of 3 is}{\text{M}}_{12}=\left|\begin{array}{cc}6& -9\\ -4& 2\end{array}\right|=12-36=-24\\ \text{minor of 4 is}{\text{M}}_{13}=\left|\begin{array}{cc}6& 5\\ -4& 1\end{array}\right|=6+20=26\end{array}$

  $\begin{array}{l}\text{minor of 6 is}{\text{M}}_{21}=\left|\begin{array}{cc}3& 4\\ 1& 2\end{array}\right|=6-4=2\\ \text{minor of 5 is}{\text{M}}_{22}=\left|\begin{array}{cc}8& 4\\ -4& 2\end{array}\right|=16+16=32\end{array}$

  $\begin{array}{l}\text{minor of -9 is}{\text{M}}_{23}=\left|\begin{array}{cc}8& 3\\ -4& 1\end{array}\right|=8+12=20\\ \text{minor of }-4\text{ is}{\text{M}}_{31}=\left|\begin{array}{cc}3& 4\\ 5& -9\end{array}\right|=-27-20=-47\\ \text{minor of 1 is}{\text{M}}_{32}=\left|\begin{array}{cc}8& 4\\ 6& -9\end{array}\right|=-72-24=-96\\ \text{minor of 2 is}{\text{M}}_{33}=\left|\begin{array}{cc}8& 3\\ 6& 5\end{array}\right|=40-18=22\end{array}$

Hence, minors of the given matrix are $M_{11}=19,M_{12}=-24,M_{13}=26,M_{21}=2,M_{22}=32,$

  $M_{23}=20,M_{31}=-47,M_{32}=-96,M_{33}=22.$

b.

To determine

The cofactors of the matrix $\left[\begin{array}{ccc}8& 3& 4\\ 6& 5& -9\\ -4& 1& 2\end{array}\right]$ .

Answer to Problem 116RE

The co-factors of the given matrix are $C_{11}=30,C_{12}=-12,C_{13}=-21,M_{21}=20,C_{22}=19,$

  $C_{23}=22,C_{31}=5,C_{32}=-2,C_{33}=19.$

Explanation of Solution

Given information :

Given matrix $\left[\begin{array}{ccc}8& 3& 4\\ 6& 5& -9\\ -4& 1& 2\end{array}\right]$

The cofactor of an element $a_{11}$ of a matrix $\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]$ is $M_{11}={\left(-1\right)}^{1+1}\left|\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right|\left(\text{delete}\text{first}\text{row}\text{and}\text{first}\text{column}\right)$

Similar way, cofactor of $8$ is $C_{11}={(-1)}^{1+1}{M}_{11}=\left|\begin{array}{cc}5& -9\\ 1& 2\end{array}\right|=10+9=19$

  $\begin{array}{l}\text{cofactor of 3 is}{C}_{12}={(}^{-}{\text{M}}_{12}=\left|\begin{array}{cc}6& -9\\ -4& 2\end{array}\right|=-\left(12-36\right)=24\\ \text{cofactor of 4 is}{C}_{13}={(}^{-}{\text{M}}_{13}=\left|\begin{array}{cc}6& 5\\ -4& 1\end{array}\right|=6+20=26\end{array}$

  $\begin{array}{l}\text{cofactor of 6 is}{\text{M}}_{21}=\left|\begin{array}{cc}3& 4\\ 1& 2\end{array}\right|=6-4=2\\ \text{cofactor of 5 is}{\text{M}}_{22}=\left|\begin{array}{cc}8& 4\\ -4& 2\end{array}\right|=16+16=32\end{array}$

  $\begin{array}{l}\text{cofactor of -9 is}{C}_{23}={(}^{-}{\text{M}}_{23}=-\left|\begin{array}{cc}8& 3\\ -4& 1\end{array}\right|=-\left(8+12\right)=-20\\ \text{cofactor of }-4\text{ is}{C}_{31}={(}^{-}{\text{M}}_{31}=\left|\begin{array}{cc}3& 4\\ 5& -9\end{array}\right|=-27-20=-47\\ \text{cofactor of 1 is}{C}_{33}={(}^{-}{\text{M}}_{32}=-\left|\begin{array}{cc}8& 4\\ 6& -9\end{array}\right|=-\left(-72-24\right)=96\\ \text{cofactor of 2 is}{C}_{33}={(}^{-}{\text{M}}_{33}=\left|\begin{array}{cc}8& 3\\ 6& 5\end{array}\right|=40-18=22\end{array}$

Hence, the cofactors of the given matrix are $C_{11}=30,C_{12}=-12,C_{13}=-21,M_{21}=20,C_{22}=19,$ $C_{23}=22,C_{31}=5,C_{32}=-2,C_{33}=19.$