Question

If $\left|\begin{array}{ccc}32& 24& 16\\ 8& 3& 5\\ 4& 5& 3\end{array}\right|=\lambda \left|\begin{array}{ccc}1& 3& 2\\ 2& 3& 5\\ 1& 5& 3\end{array}\right|$, then the value of $\lambda$ is

• A. $4$
• B. $8$
• C. $16$
• D. $24$
• E. $32$

Answer 1

Answer: (E)

$32$

We have $\left|\begin{array}{ccc}32& 24& 16\\ 8& 3& 5\\ 4& 5& 3\end{array}\right|=8\left|\begin{array}{ccc}1& 3& 2\\ 2& 3& 5\\ 1& 5& 3\end{array}\right|$
$=8\times 4\left|\begin{array}{ccc}1& 3& 2\\ 2& 3& 5\\ 1& 5& 3\end{array}\right|=32\left|\begin{array}{ccc}1& 3& 2\\ 2& 3& 5\\ 1& 5& 3\end{array}\right|$
But given $\left|\begin{array}{ccc}32& 24& 15\\ 8& 3& 5\\ 4& 5& 3\end{array}\right|=\lambda \left|\begin{array}{ccc}1& 3& 2\\ 2& 3& 5\\ 1& 5& 3\end{array}\right|$
On comparing, we get
$\lambda =32$.