Question

a + b & a + 2b & a + 3b \\ a + 2b & a + 3b & a + 4b \\ a + 4b & a + 5b & a + 6b \end{matrix} \right| =$$

• A. $a^{2} + b^{2} + c^{2} - 3abc$
• B. $3ab$
• C. $3a + 5b$
• D. 0

Answer 1

Answer: (D)

0

Answer 2

$\left| \begin{matrix} a + b & a + 2b & a + 3b \\ a + 2b & a + 3b & a + 4b \\ a + 4b & a + 5b & a + 6b \end{matrix} \right| = \left| \begin{matrix} a + b & a + 2b & a + 3b \\ b & b & b \\ 2b & 2b & 2b \end{matrix} \right|$ = 0

R_{2} \rightarrow R_{2} - R_{1} \\ R_{3} \rightarrow R_{3} - R_{2} \end{matrix} \right\}$$ **Trick:** Putting $a = 1 = b$. The determinant will be $\left| \begin{matrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 5 & 6 & 7 \end{matrix} \right| = 0$. Obviously answer is (4) **Note :** Students remember while taking the values of $a,b,c,.......$ that for there values, the options (1), (2), (3) and (4) should not be identical.