Question

If three-digit numbers $A28,3B9$ and $62C$, where $A,B$ and $C$ are integers between $0$ and $9$, are divisible by a fixed integer $k$, then the determinant $\left|\begin{array}{ccc}A& 3& 6\\ 8& 9& C\\ 2& B& 2\end{array}\right|$ is

• A. divisible by $k$
• B. divisible by $k^2$
• C. divisible by $2k$
• D. None of these

Answer 1

The correct option is A

divisible by $k$

Explanation for the correct option.

Find the divisibility of the determinant.

The numbers $A28,3B9$ and $62C$, are divisible by a fixed integer $k$. So let $A28=ka,3B9=kb,62C=kc$ where $a,b,c$ are integers.

Now in the given determinant $\left|\begin{array}{ccc}A& 3& 6\\ 8& 9& C\\ 2& B& 2\end{array}\right|$ perform the row operation: $R_2\to 100R_1+10R_3+R_2$

$\begin{array}{rcl}& & \left|\begin{array}{ccc}A& 3& 6\\ 8& 9& C\\ 2& B& 2\end{array}\right|\stackrel{R_2\to 100R_1+10R_3+R_2}{\to }\left|\begin{array}{ccc}A& 3& 6\\ 100·A+10·2+8& 100·3+10·B+9& 100·6+10·2+C\\ 2& B& 2\end{array}\right|\end{array}$

Now the determinant can be further simplified as:

$\begin{array}{rcl}\left|\begin{array}{ccc}A& 3& 6\\ 100·A+10·2+8& 100·3+10·B+9& 100·6+10·2+C\\ 2& B& 2\end{array}\right|& =& \left|\begin{array}{ccc}A& 3& 6\\ A28& 3B9& 62C\\ 2& B& 2\end{array}\right|\\ & =& \left|\begin{array}{ccc}A& 3& 6\\ ka& kb& kc\\ 2& B& 2\end{array}\right|\left[A28=ka,3B9=kb,62C=kc\right]\\ & =& k\left|\begin{array}{ccc}A& 3& 6\\ a& b& c\\ 2& B& 2\end{array}\right|\end{array}$

So it can be seen that the determinant $\left|\begin{array}{ccc}A& 3& 6\\ 8& 9& C\\ 2& B& 2\end{array}\right|$ is an integral multiple of $k$.

So the determinant $\left|\begin{array}{ccc}A& 3& 6\\ 8& 9& C\\ 2& B& 2\end{array}\right|$ is divisible by $k$.

Hence, the correct option is A.