Chapter 4, Problem 25SGR

To determine

To find: the determinant of the given matrix.

Answer to Problem 25SGR

The determinant of matrix $\text{A}$ = $\text{-44}$ .

Explanation of Solution

Given:

  $\left|\begin{array}{ccc}2& 3& -1\\ 0& 2& 4\\ -2& 5& 6\end{array}\right|$ .

Concept used:

For finding the determinant of $2\times 2$ matrix.

It can be deduced by cross multiplying the numbers

To solve the determinant of $3\times 3$ .

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:

Here $\left|\begin{array}{ccc}2& 3& -1\\ 0& 2& 4\\ -2& 5& 6\end{array}\right|$ .

Using formula for solving the determinant.

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

Let the given determinant be $\text{A}$ .

  $\left|\begin{array}{ccc}2& 3& -1\\ 0& 2& 4\\ -2& 5& 6\end{array}\right|$ .

  $\left|\text{A}\right|$ = $\left|2\left(12-20\right)-3\left(0+8\right)-1\left(0+4\right)\right|=-16-24-4=-44$ .

Hence determinant of matrix $\text{A}$ = $\text{-44}$ .