Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Determinants

Question:

The value of $Δ=\begin{vmatrix}(a^x +a^{-x})^2&(a^x -a^{-x})^2&1\\(a^y +a^{-y})^2&(a^y -a^{-y})^2&1\\(a^z +a^{-z})^2&(a^z -a^{-z})^2&1\end{vmatrix}$, is

Options:

1

-1

0

none of these

Correct Answer:

0

Explanation:

Applying $C_1→C_1-C_2$, we get

$Δ=\begin{vmatrix}4&(a^x -a^{-x})^2&1\\4&(a^y -a^{-y})^2&1\\4&(a^z -a^{-z})^2&1\end{vmatrix}$ [Using: $(a+b)^2−(a - b)^ =4ab$] 

$⇒Δ=4\begin{vmatrix}1&(a^x -a^{-x})^2&1\\1&(a^y -a^{-y})^2&1\\1&(a^z -a^{-z})^2&1\end{vmatrix}$  [Taking out 4 common from $C_1$]

$⇒Δ=4×0=0$  [∵ $C_1$ and $C_3$ are identical]