Chapter 4, Problem 9STP

To determine

Tofind:the determinant of the matrix $\text{w}$ given below.

Answer to Problem 9STP

The determinant of $\text{w=1}\text{.}$

Explanation of Solution

Given:

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

Concept used:

To solve the determinant:

Step1: find the minor of the ${\text{a}}_{\text{11}}{\text{,a}}_{\text{12}}{\text{ and a}}_{\text{13}}$ .

Step2: multiply the minor with value of ${\text{a}}_{11},a_{12}{\text{ and a}}_{13}$ respectively.

Step3: if the sum of the position of the element is come up with even either multiply with one or remain whatever the value be. whereas if sum of the position of the element is come up with odd then the value should be multiplied by $\left(-1\right)$ .

Example:

  $a_{11}$ here the sum of the position $1+1=2$ is even so multiply the value by $\left(1\right)$ .

  $a_{12}$ here the sum of the position $1+2=3$ is odd so multiply the value by $\left(-1\right)$ .

Mathematically:

  $\left|\text{A}\right|\text{=}\left|{\text{a}}_{\text{11}}{\text{.minor of a}}_{\text{11}}\text{+}\left(\text{-1}\right){\text{a}}_{\text{12}}{\text{.minor of a}}_{\text{12}}{\text{+a}}_{\text{13}}{\text{.minor of a}}_{13}\right|$ .

Calculation:

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

Determinant of above matrices:

  $\left|\text{w}\right|\text{=}\left|\text{3}\left(\text{5-4}\right)\text{-1}\left(\text{2-0}\right)\text{+0}\right|\text{=}\left|\text{3-2}\right|\text{=1}\text{.}$

Hence the determinant of $\text{w=1}\text{.}$