Question

Let $A$ be a$3\times 3$matrix with $\text{det}\left(A\right)=4$. Let $R_i$ denote the $i^{th}$ row of $A$.

If a matrix $B$ is obtained by performing the operation $R_2\to 2R_2+5R_3$ on $2A$, then $\text{det}\left(B\right)$ is equal to:

• A. $64$
• B. $16$
• C. $80$
• D. $128$

Answer 1

The correct option is A

$64$

Explanation for correct option:

Finding the $\text{det}\left(B\right)$:

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

Multiplying by $2$ on both sides we get,

$2A=\left[\begin{array}{ccc}2R_{11}& 2R_{12}& 2R_{13}\\ 2R_{21}& 2R_{22}& 2R_{23}\\ 2R_{31}& 2R_{32}& 2R_{33}\end{array}\right]$

Given $B$ is obtained by performing the operation $R_2\to 2R_2+5R_3$ on $2A$,

$$\begin{gathered} B=\left[\begin{array}{ccc}2R_{11}& 2R_{12}& 2R_{13}\\ 2\left(2R_{21}\right)+5\left(2R_{31}\right)& 2\left(2R_{22}\right)+5\left(2R_{32}\right)& 2\left(2R_{23}\right)+5\left(2R_{33}\right)\\ 2R_{31}& 2R_{32}& 2R_{33}\end{array}\right] \\ =\left[\begin{array}{ccc}2R_{11}& 2R_{12}& 2R_{13}\\ 4R_{21}+10R_{31}& 4R_{22}+102R_{32}& 4R_{23}+10R_{33}\\ 2R_{31}& 2R_{32}& 2R_{33}\end{array}\right] \end{gathered}$$

Now, let's perform $R_2\to R_2-5R_3$

$$\begin{gathered} B=\left[\begin{array}{ccc}2R_{11}& 2R_{12}& 2R_{13}\\ 4R_{21}+10R_{31}-5\left(2R_{31}\right)& 4R_{22}+102R_{32}-5\left(2R_{32}\right)& 4R_{23}+10R_{33}-5\left(2R_{33}\right)\\ 2R_{31}& 2R_{32}& 2R_{33}\end{array}\right] \\ =\left[\begin{array}{ccc}2R_{11}& 2R_{12}& 2R_{13}\\ 4R_{21}& 4R_{22}& 4R_{23}\\ 2R_{31}& 2R_{32}& 2R_{33}\end{array}\right] \end{gathered}$$

Finding $\text{det}\left(B\right)$

$\begin{array}{rcl}\left|B\right|& =& 2\times 4\times 2\times \left|\begin{array}{ccc}{R}_{11}& R_{12}& R_{13}\\ R_{21}& R_{22}& R_{23}\\ R_{31}& R_{32}& R_{33}\end{array}\right|\\ & =& 16\times \left|A\right|\\ & =& 16\times 4\left[\because \left|A\right|=4\right]\\ \left|B\right|& =& 64\end{array}$

Hence, option (A) is the correct answer.