Question

If $x,y,z$ are in $A.P.$ with common difference $d$ and the rank of the matrix $\left|\begin{array}{ccc}4& 5& x\\ 5& 6& y\\ 6& k& z\end{array}\right|$ is $2$ then the value of $d$ and $k$ are

• A. $\frac{x}{4};$ arbitrary number
• B. arbitrary number, $7$
• C. $x,5$
• D. $\frac{x}{2},6$

Answer 1

The correct option is B arbitrary number, $7$

Let $A=\left|\begin{array}{ccc}4& 5& x\\ 5& 6& y\\ 6& k& z\end{array}\right|$

Since x,y,z are in AP with common difference d we get

$A=\left|\begin{array}{ccc}4& 5& x\\ 5& 6& x+d\\ 6& k& x+2d\end{array}\right|$

Apply $R_3\to R_3+R_1-2R_2$ to get

$A=\left|\begin{array}{ccc}4& 5& x\\ 5& 6& x+d\\ 0& k-7& 0\end{array}\right|$

Rank of A is given to be 2. So rank will be 2 when $k-7=0$

So $k=7$ and $d$ is any arbitrary number.