Q. 62 - Precalculus Enhanced with Graphing Utilities

Question

Interchange columns $1$ and $3$ of a $3$ by $3$ determinant. Show that the value of the new determinant is $-1$ times the value of the original determinant.

Step-by-Step Solution

Answer

It is proved that by interchanging the columns the value of the new determinant is $-1$ times the value of the original determinant.

Step 1. Given Information

We are given a determinant and we have to show that by interchanging the columns the value of the new determinant is $-1$times the value of the original determinant.

Step 2. Proving the statement

Let the determinant be,

$\left|\begin{array}{lll}a& b& c\\ d& e& f\\ g& h& i\end{array}\right|$

Expanding the determinant, we get

$$\begin{gathered} \left|\begin{array}{lll}a& b& c\\ d& e& f\\ g& h& i\end{array}\right|=(-1{)}^{1+1}\times a\times \left|\begin{array}{cc}e& f\\ h& i\end{array}\right|+(-1{)}^{1+2}\times b\times \left|\begin{array}{ll}d& f\\ g& i\end{array}\right|+(-1{)}^{1+2}\times c\times \left|\begin{array}{ll}d& e\\ g& h\end{array}\right| \\ =aei-afh-bdi+bfg+cdh-ceg \end{gathered}$$

Now, Interchanging the columns $1$ and $3$, we get

$\left|\begin{array}{ccc}c& b& a\\ f& e& d\\ i& h& g\end{array}\right|$

Step 3. Evaluating the determinant

Evaluating the changed determinant, we get

$$\begin{gathered} \left|\begin{array}{ccc}c& b& a\\ f& e& d\\ i& h& g\end{array}\right|=(-1{)}^{1+1}\times c\times \left|\begin{array}{ll}e& d\\ h& g\end{array}\right|+(-1{)}^{1+2}\times b\times \left|\begin{array}{ll}f& d\\ i& g\end{array}\right|+(-1{)}^{1+3}\times a\times \left|\begin{array}{ll}f& e\\ i& h\end{array}\right| \\ =g e c-h d c-g f b+i b d+h f a-a e i \\ =-(a e i-a f h-b d i+b f g+c d h-c e g) \\ =-\left|\begin{array}{lll}a& b& c\\ d& e& f\\ g& h& i\end{array}\right| \end{gathered}$$

Hence Proved.