Let A = [aij] and B = [bij] be a square matrices

Solution:

Given:

$A$ and $B$ are square matrices of order 3

$b_{i 1}=2 a_{i 1}, b_{i 2}=3 a_{i 2}$ and $b_{i 3}=4 a_{i 3}, i=1,2,3$

$|A|=5$

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

Since, $b_{i 1}=2 a_{i 1}, b_{i 2}=3 a_{i 2}$ and $b_{i 3}=4 a_{i 3}$

Therefore,

$b_{11}=2 a_{11}, b_{21}=2 a_{21}, b_{31}=2 a_{31}$

$b_{12}=3 a_{12}, b_{22}=3 a_{22}, b_{32}=3 a_{32}$

$b_{13}=4 a_{13}, b_{23}=4 a_{23}, b_{33}=4 a_{33}$

$|B|=\left|\begin{array}{lll}b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33}\end{array}\right|$

$\Rightarrow|B|=\left|\begin{array}{lll}2 a_{11} & 3 a_{12} & 4 a_{13} \\ 2 a_{21} & 3 a_{22} & 4 a_{23} \\ 2 a_{31} & 3 a_{32} & 4 a_{33}\end{array}\right|$

Taking out 2,3 and 4 common from $C_{1}, C_{2}$ and $C_{3}$, respectively

$\Rightarrow|B|=2 \times 3 \times 4\left|\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right|$

$\Rightarrow|B|=24|A|$

$\Rightarrow|B|=24 \times 5 \quad(\because|A|=5)$

$\Rightarrow|B|=120$

Hence, $|B|=\underline{120}$.