Q. 63 - Precalculus Enhanced with Graphing Utilities

Question

Multiply each entry in row $2$ of a $3$ by $3$ determinant by the number $k$, $k\ne 0$. Show that the value of the new determinant is k times the value of the original determinant.

Step-by-Step Solution

Answer

It is proved that by multiplying by k, the new determinant is k times the value of the original determinant.

Step 1. Given Information

We are given a determinant and we have to show that by multiplying by k, the new determinant is k times the value of the original determinant.

Step 2. Proving the statement

Let the matrix be,

$\left|\begin{array}{lll}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right|$

The determinant of the matrix will be, 

$D=-a_{21}\left(a_{12}{a}_{33}-a_{13}{a}_{32}\right)+a_{22}\left(a_{11}{a}_{33}-a_{13}{a}_{31}\right)-a_{23}\left(a_{11}{a}_{32}-a_{12}{a}_{31}\right)$

Multiply each entry in the second row by k. The new matrix will be,

$\left|\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ k{a}_{21}& k{a}_{22}& k{a}_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right|$

Step 3. Proving the statement

The determinant of above matrix will be,

$$\begin{gathered} D_k=-k{a}_{21}\left(a_{12}{a}_{33}-a_{13}{a}_{32}\right)+k{a}_{22}\left(a_{11}{a}_{33}-a_{13}{a}_{31}\right)-k{a}_{23}\left(a_{11}{a}_{32}-a_{12}{a}_{31}\right) \\ {D}_k=k\left[-a_{21}\left(a_{12}{a}_{33}-a_{13}{a}_{32}\right)+a_{22}\left(a_{11}{a}_{33}-a_{13}{a}_{31}\right)-a_{23}\left(a_{11}{a}_{32}-a_{12}{a}_{31}\right)\right] \end{gathered}$$

So,

$\left|\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ k{a}_{21}& k{a}_{22}& k{a}_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right|=k\left|\begin{array}{lll}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right|$

Hence proved.