Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Determinants

Question:

If $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$ then $\left|\begin{array}{ccc}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{array}\right|$ is equal to :

Options:

abc

-abc

0

1

Correct Answer:

abc

Explanation:

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$

$\Delta=\left|\begin{array}{ccc}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{array}\right|$

$\Delta=a b c\left|\begin{array}{ccc}\frac{1}{a}+1 & \frac{1}{a} & \frac{1}{a} \\ \frac{1}{b} & \frac{1}{b}+1 & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1\end{array}\right|$

(Multiplying by abc and dividing rows by a, b, c respectively)

$R_1 \rightarrow R_1+R_2+R_3$

$\Delta=a b c\left|\begin{array}{ccc}\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1 & \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1 & \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1 \\ \frac{1}{b} & \frac{1}{b}+1 & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1\end{array}\right|$

$\Delta=a b c\left|\begin{array}{ccc}1 & 1 & 1 \\ \frac{1}{b} & \frac{1}{b}+1 & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1\end{array}\right|$   (as $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$)

$C_2 → C_2 - C_1$

$C_3 → C_3 - C_1$

$\Delta=a b c\left|\begin{array}{ccc}1 & 0 & 0 \\ \frac{1}{b} & 1 & 0 \\ \frac{1}{c} & 0 & 1\end{array}\right|$

$\Delta=a b c$