Q. 65 - Precalculus Enhanced with Graphing Utilities

Question

Prove that, if row $2$ of a $3$ by $3$ determinant is multiplied by k, $k\ne 0$, and the result is added to the entries in row $1$, there is no change in the value of the determinant.

Step-by-Step Solution

Answer

It is proved that when a row $2$ of a determinant is multiplied by k, $k\ne 0$, and the result is added to the entries in row one, there is no change in the value of the determinant.

Step 1. Given Information

We are given a determinant.

We have to prove that there is no change in the value of the determinant if the second row is multiplied by some constant k and added to row one.

Step 2. Proving the statement

Let the determinant be,

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

Multiplying the entries in the second row by k and add these to the entries in the first row, we get

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

Step 3. Evaluating the determinant

Evaluating the determinant,

$$\begin{gathered} \left|\begin{array}{ccc}{a}_{11}+k{a}_{21}& a_{12}+k{a}_{22}& a_{13}+k{a}_{23}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right|=\left(a_{11}+k{a}_{21}\right)\left(a_{22}{a}_{33}-a_{23}{a}_{32}\right)-\left(a_{12}+k{a}_{22}\right)\left(a_{21}{a}_{33}-a_{31}{a}_{23}\right)+\left(a_{13}+k{a}_{23}\right)\left(a_{21}{a}_{32}-a_{31}{a}_{22}\right) \\ =a_{11}{a}_{22}{a}_{33}-a_{11}{a}_{23}{a}_{32}-a_{12}{a}_{21}{a}_{33}+a_{12}{a}_{31}{a}_{23}+a_{13}{a}_{21}{a}_{32}-a_{13}{a}_{31}{a}_{22} \\ =a_{11}\left(a_{22}{a}_{33}-a_{23}{a}_{32}\right)-a_{12}\left(a_{21}{a}_{33}-a_{31}{a}_{23}\right)+a_{13}\left(a_{21}{a}_{32}-a_{31}{a}_{22}\right) \\ =A \end{gathered}$$

Hence Proved.