Question

The value of the determinant $\begin{vmatrix}(10^{5}+10^{-5})^{2}&(10^{5}-10^{-5})^{2}& 1\\ (100^{5}+100^{-5})^{2}&(100^{5}-100^{-5})^{2}& 1\\ (6^{100}+6^{-100})^{2}&(6^{100}-6^{-100})^{2}& 1\end{vmatrix}$ is equal to

• A. 100
• B. 200
• C. 0
• D. 6000
• E. 60600

Hint:

Math Tip: Whenever you see frighteningly large exponents inside a determinant, look for algebraic identities (like difference of squares) or row/column operations that will collapse the complex terms into constants.

Answer 1

The Correct Option is C

Concept:
Determinants - Column Operations and Algebraic Identities.

• $(a+b)^2 - (a-b)^2 = 4ab$
• If two columns (or rows) of a determinant are proportional, the value of the determinant is $0$.

Step 1: Apply the algebraic identity to the determinant columns.
Let's perform a column operation: replace Column 1 ($C_1$) with $C_1 - C_2$. $$ C_1 \rightarrow C_1 - C_2 $$
Step 2: Evaluate the new elements in Column 1.
Using the identity $(x+y)^2 - (x-y)^2 = 4xy$:

• Row 1: $(10^5+10^{-5})^2 - (10^5-10^{-5})^2 = 4(10^5)(10^{-5}) = 4(1) = 4$
• Row 2: $(100^5+100^{-5})^2 - (100^5-100^{-5})^2 = 4(100^5)(100^{-5}) = 4(1) = 4$
• Row 3: $(6^{100}+6^{-100})^2 - (6^{100}-6^{-100})^2 = 4(6^{100})(6^{-100}) = 4(1) = 4$

Step 3: Rewrite the simplified determinant.
After the column operation, the determinant becomes: $$ \begin{vmatrix} 4 & (10^{5}-10^{-5})^{2} & 1 \\ 4 & (100^{5}-100^{-5})^{2} & 1 \\ 4 & (6^{100}-6^{-100})^{2} & 1 \end{vmatrix} $$
Step 4: Identify proportional columns.
Observe that Column 1 ($C_1$) is exactly $4$ times Column 3 ($C_3$). $$ C_1 = 4 C_3 $$ Since $C_1$ and $C_3$ are proportional, the value of the determinant is zero.