Q. 45 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 43–50, use properties of determinants to find the value of each determinant if it is known that $\left|\begin{array}{ccc}x& y& z\\ u& v& w\\ 1& 2& 3\end{array}\right|=4$

$\left|\begin{array}{ccc}x& y& z\\ -3& -6& -9\\ u& v& w\end{array}\right|$

Step-by-Step Solution

Answer

The value of the determinant$\left|\begin{array}{ccc}x& y& z\\ -3& -6& -9\\ u& v& w\end{array}\right|$ is $12$

$\left|\begin{array}{ccc}x& y& z\\ -3& -6& -9\\ u& v& w\end{array}\right|=12$

Step 1. Given information

The given determinant is 

$\left|\begin{array}{ccc}x& y& z\\ -3& -6& -9\\ u& v& w\end{array}\right|$

Step 2. Value of determinant

Use the property of determinant and take $-3$ common form second row of determinants 

$$\begin{gathered} \left|\begin{array}{ccc}x& y& z\\ -3& -6& -9\\ u& v& w\end{array}\right|=\left|\begin{array}{ccc}x& y& z\\ -3\left(1\right)& -3\left(2\right)& -3\left(3\right)\\ u& v& w\end{array}\right| \\ =-3\left|\begin{array}{ccc}x& y& z\\ 1& 2& 3\\ u& v& w\end{array}\right| \end{gathered}$$

Use the property of determinant and interchange the second and third-row

$$\begin{gathered} \left|\begin{array}{ccc}x& y& z\\ -3& -6& -9\\ u& v& w\end{array}\right|=-3\left|\begin{array}{ccc}x& y& z\\ 1& 2& 3\\ u& v& w\end{array}\right| \\ =\left(-3\right)\left(-1\right)\left|\begin{array}{ccc}x& y& z\\ u& v& w\\ 1& 2& 3\end{array}\right| \\ =\left(-3\right)\left(-1\right)\left(4\right) \\ =12 \end{gathered}$$

So the value of the determinant is$\left|\begin{array}{ccc}x& y& z\\ -3& -6& -9\\ u& v& w\end{array}\right|=12$