Q48. - Algebra 2

Question

Find the value of det A if   $A=\left[\begin{array}{ccc}0& 3& -2\\ -4& 0& 1\\ 3& 2& 0\end{array}\right]$.

 

$A.\text{ 0 B. 12 C. 25 D. 36}$

Step-by-Step Solution

Answer

The value of the determinant is$\left|A\right|=25$.

Step 1- Apply the concept of determinant.

The determinant is a scalar value that is a function of the entries of a square matrix. It allows characterizing some properties of the matrix and the linear map represented by the matrix.

Step 2- Step description.

Consider the matrix $A=\left(\begin{array}{ccc}0& 3& -2\\ -4& 0& 1\\ 3& 2& 0\end{array}\right)$.

 

 find the value of the determinant as follows:

 

 $\begin{array}{c}\left|A\right|=0\cdot \det \left(\begin{array}{cc}0& 1\\ 2& 0\end{array}\right)-3\cdot \det \left(\begin{array}{cc}-4& 1\\ 3& 0\end{array}\right)-2\cdot \det \left(\begin{array}{cc}-4& 0\\ 3& 2\end{array}\right)\\ =0\cdot \left(-2\right)-3\left(-3\right)-2\left(-8\right)\\ =25\end{array}$

Step 3- Step description.

Therefore, the value of the determinant is 25.