Chapter 10.6, Problem 18E

To determine

The minor and cofactor $M_{32},A_{32}$ of the given matrix is to be estimated:

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

Answer to Problem 18E

Minor $\left(M_{32}\right)$ and of Co-factor $\left(A_{32}\right)$ the matrix $A$ is $\frac{7}{2}$ and $-\frac{7}{2}$ respectively.

Explanation of Solution

Given:

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

Concept Used:

By deleting the row and column in which the element is present the minor of the square matrix is formed. The minor of the element with the appropriate sign is known as the cofactor.

The minor and cofactor are represented by $M_{ij}$ and $A_{ij}$ respectively.

The cofactor is calculated by using the following expression:

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

Calculations:

The given matrix is:

  $A=\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]=\left[\begin{array}{ccc}1& 0& \frac{1}{2}\\ -3& 5& 2\\ 0& 0& 4\end{array}\right]$

Minor $\left(M_{32}\right)$ of the matrix $A$ is:

  $\begin{array}{l}{M}_{32}=\left[\begin{array}{cc}1& \frac{1}{2}\\ -3& 2\end{array}\right]\\ M_{32}=\left(1\times 2\right)-\left(\frac{1}{2}\times -3\right)\\ M_{32}=2+\frac{3}{2}\\ M_{32}=\frac{7}{2}\end{array}$

Co-factor $\left(A_{32}\right)$ of the matrix $A$ is:

  $\begin{array}{l}{A}_{32}={\left(-1\right)}^{2+3}{M}_{32}\\ A_{32}={\left(-1\right)}^5\left(\frac{7}{2}\right)\\ A_{32}=-\frac{7}{2}\end{array}$

Conclusion:

Hence,

  $\begin{array}{l}{M}_{32}=\frac{7}{2}\\ A_{32}=-\frac{7}{2}\end{array}$