Chapter 7, Problem 123CR

To determine

The determinant of the given matrix.

Answer to Problem 123CR

The value of the determinant of the given matrix is 24.

Explanation of Solution

Given: The given matrix is $A=\left[\begin{array}{ccc}-2& 5& 0\\ 2& -1& 0\\ -1& 1& -3\end{array}\right]$ .

Concept Used: The determinant value of any square matrix, say ${\left[a_{ij}\right]}_{n\times n}$ , is a number that is found by the summation of the product of all the elements of a single row/column with their corresponding cofactors, expanded along any row/column of choice. Thus, for any $n\times n$ matrix $A={\left[a_{ij}\right]}_{n\times n}$ , its determinant is given by-

  $\left|A\right|={\displaystyle \sum _{j=1}^n{a}_{mj}{C}_{mj}}$ , when expanded along the $m^{th}$ row.

Calculations: The given matrix is $A=\left[\begin{array}{ccc}-2& 5& 0\\ 2& -1& 0\\ -1& 1& -3\end{array}\right]$ . Thus,

  $\left|A\right|$

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

  $=0\left|\begin{array}{cc}2& -1\\ -1& 1\end{array}\right|-0\left|\begin{array}{cc}-2& 5\\ -1& 1\end{array}\right|+(-3)\left|\begin{array}{cc}-2& 5\\ 2& -1\end{array}\right|$ (Expanded along the third column)

  $=(-3)\times (2-5)$

  $=(-3)\times (-8)$

  $=24$

Thus, the determinant value of the given matrix is 24.