Q. 64 - Precalculus Enhanced with Graphing Utilities

Question

Prove that a $3$ by $3$ determinant in which the entries in column $1$ equal those in column $3$ has the value $0$.

Step-by-Step Solution

Answer

It is proved that the determinate in which two columns has the same set of values is zero.

Step 1. Given Information

We are given that a matrix same values in column $1$ and $3$.

We have to prove that the determinant of matrix with the same values in two columns is zero.

Step 2. Proving the statement

Let the determinant be,

$A=\left|\begin{array}{lll}a& b& a\\ d& e& d\\ g& h& g\end{array}\right|$

Performing row operation $C_3\to C_3-C_1$, we get

$$\begin{gathered} =\left|\begin{array}{lll}a& b& a-a\\ d& e& d-d\\ g& h& g-g\end{array}\right| \\ =\left|\begin{array}{lll}a& b& 0\\ d& e& 0\\ g& h& 0\end{array}\right| \\ =0 \end{gathered}$$

Hence proved.