Q. 48 - 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 

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

Step-by-Step Solution

Answer

The value of the determinant is $4$. 

Step 1. Given Information

We are given,

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

We have to solve the given determinant using properties of determinants.

Step 2. Solving the determinant

Using subtraction of determinant, we get 

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

Since the two columns of second determinant is same, so 

$$\begin{gathered} =\left|\begin{array}{ccc}x& y& z\\ u& v& w\\ 1& 2& 3\end{array}\right|-0 \\ =4 \end{gathered}$$