Matrices And Determinants Ques 61
Let the three digit numbers $A 28,3 B 9$ and $62 C$, where $A$, $B$ and $C$ are integers between $0$ and $9$ , be divisible by a fixed integer $k$. Show that the determinant
$\begin{vmatrix}A & 3 & 6 \\ 8 & 9 & C \\ 2 & B & 2\end{vmatrix}$ is divisible by $k$.
$(1990,4$ M)
Show Answer
Solution:
Formula:
- We know, $A 28=A \times 100+2 \times 10+8$
$ 3 B 9=3 \times 100+B \times 10+9 $
and $\quad 62 C=6 \times 100+2 \times 10+C$
Since, $A 28,3 B 9$ and $62 C$ are divisible by $k$, therefore there exist positive integers $m_{1}, m_{2}$ and $m_{3}$ such that,
$100 \times A+10 \times 2+8=m_{1} k, 100 \times 3+10 \times B+9=m_{2} k$
and $100 \times 6+10 \times 2+C=m_{3} k$ $\quad$ …….(i)
$ \therefore \quad \Delta=\begin{vmatrix} A & 3 & 6 \\ 8 & 9 & C \\ 2 & B & 2 \end{vmatrix} $
Applying $R_{2} \rightarrow 100 R_{1}+10 R_{3}+R_{2}$
$ \Rightarrow \quad \Delta= \begin{vmatrix} A & 3 \\ 100 A+2 \times 10+8 & 100 \times 3+10 \times B+9 \\ 2 & B \end{vmatrix} $
$ \begin{aligned} & \begin{vmatrix} 6 \\ 100 \times 6+10 \times 2+C \\ 2 \end{vmatrix} \\ & =\begin{vmatrix} A & 3 & 6 \\ A 28 & 3 B 9 & 62 C \\ 2 & B & 2 \end{vmatrix} \end{aligned} $
[from Eq. (i)]
$ =\begin{vmatrix} A & 3 & 6 \\ m_{1} k & m_{2} k & m_{3} k \\ 2 & B & 2 \end{vmatrix}=k\begin{vmatrix} A & 3 & 6 \\ m_{1} & m_{2} & m_{3} \\ 2 & B & 2 \end{vmatrix} $
$\therefore \quad \Delta=m k$
Hence, determinant is divisible by $k$.
BETA
