# Prove the following

Question:

Let $\mathrm{A}$ be a $3 \times 3$ matrix with $\operatorname{det}(\mathrm{A})=4$. Let $\mathrm{R}_{\mathrm{i}}$ denote the $\mathrm{i}^{\text {th }}$ row of $\mathrm{A}$. If a matrix $\mathrm{B}$ is obtained by performing the operation $\mathrm{R}_{2} \rightarrow 2 \mathrm{R}_{2}+5 \mathrm{R}_{3}$ on $2 \mathrm{~A}$, then $\operatorname{det}(\mathrm{B})$ is equal to:

1. (1) 64

2. (2) 16

3. (3) 80

4. (4) 128

Correct Option: 1

Solution:

$A=\left[\begin{array}{lll}\mathrm{R}_{11} & \mathrm{R}_{12} & \mathrm{R}_{13} \\ \mathrm{R}_{21} & \mathrm{R}_{22} & \mathrm{R}_{23} \\ \mathrm{R}_{31} & \mathrm{R}_{32} & \mathrm{R}_{33}\end{array}\right]$

$2 \mathrm{~A}=\left[\begin{array}{lll}2 \mathrm{R}_{11} & 2 \mathrm{R}_{12} & -2 \mathrm{R}_{13} \\ 2 \mathrm{R}_{21} & 2 \mathrm{R}_{22} & 2 \mathrm{R}_{23} \\ 2 \mathrm{R}_{31} & 2 \mathrm{R}_{32} & 2 \mathrm{R}_{33}\end{array}\right]$

$\mathrm{R}_{2} \rightarrow 2 \mathrm{R}_{2}+5 \mathrm{R}_{3}$

$B=\left[\begin{array}{ccc}2 \mathrm{R}_{11} & 2 \mathrm{R}_{12} & 2 \mathrm{R}_{13} \\ 4 \mathrm{R}_{21}+10 \mathrm{R}_{31} & 4 \mathrm{R}_{22}+10 \mathrm{R}_{32} & 4 \mathrm{R}_{23}+10 \mathrm{R}_{33} \\ 2 \mathrm{R}_{31} & 2 \mathrm{R}_{32} & 2 \mathrm{R}_{33}\end{array}\right]$

$\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-5 \mathrm{R}_{3}$

$\mathrm{B}=\left[\begin{array}{lll}2 \mathrm{R}_{11} & 2 \mathrm{R}_{12} & 2 \mathrm{R}_{13} \\ 4 \mathrm{R}_{21} & 4 \mathrm{R}_{22} & 4 \mathrm{R}_{23} \\ 2 \mathrm{R}_{31} & 2 \mathrm{R}_{32} & 2 \mathrm{R}_{33}\end{array}\right]$

$|B|=\left|\begin{array}{lll}-2 \mathrm{R}_{11} & 2 \mathrm{R}_{12} & 2 \mathrm{R}_{13} \\ 4 \mathrm{R}_{21} & 4 \mathrm{R}_{22} & 4 \mathrm{R}_{23} \\ 2 \mathrm{R}_{31} & 2 \mathrm{R}_{32} & 2 \mathrm{R}_{33}\end{array}\right|$

$|\mathrm{B}|=2 \times 2 \times 4\left|\begin{array}{lll}\mathrm{R}_{11} & \mathrm{R}_{12} & \mathrm{R}_{13} \\ \mathrm{R}_{21} & \mathrm{R}_{22} & \mathrm{R}_{23} \\ \mathrm{R}_{31} & \mathrm{R}_{32} & \mathrm{R}_{33}\end{array}\right|$

$=16 \times 4$

$=64$