Question:

If \(A=\left[\begin{array}{lll}a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c\end{array}\right]\) where \(a=7^x, b=7^{7^x}, c=7^{7^{7^x}}\) then \(\int|A| d x\), (Where \(|A|\) is the determinant of the matrix \(A\) ) is equal to

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The determinant of a diagonal matrix is the product of its diagonal entries, which is the derivative of a tower of exponentials.
Updated On: Nov 13, 2026
  • \(\frac{7^{7^{x}}}{(\mathrm{log}7)^{3}}+\mathrm{k}\), where k is constant of integration

  • \(\frac{7^{7^{7^{x}}}}{\mathrm{log}7}+\mathrm{k}\), where k is constant of integration

  • \(\frac{7^{7^{7^{x}}}}{(\mathrm{log}7)^{3}}+\mathrm{k}\), where k is constant of integration

  • \(7^{7^{7^{x}}}(\mathrm{log}7)^{3}+\mathrm{k}\), where k is constant of integration

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The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept
For a diagonal matrix, \(|A|=abc=7^{x}\cdot7^{7^{x}}\cdot7^{7^{7^{x}}}\).

Step 2: Substitution
Let \(t=7^{7^{7^{x}}}\). Differentiating step by step, \(\dfrac{dt}{dx}=t\,(\log7)^{3}\,7^{7^{x}}\,7^{x}\).

Step 3: Match the integrand
So \(|A|\,dx=7^{x}\cdot7^{7^{x}}\cdot t\,dx=\dfrac{dt}{(\log7)^{3}}\).

Step 4: Calculation
\(\int|A|\,dx=\dfrac{t}{(\log7)^{3}}+k=\dfrac{7^{7^{7^{x}}}}{(\log7)^{3}}+k\).

Final Answer:
The integral is \(\frac{7^{7^{7^{x}}}}{(\log7)^{3}}+k\). \[ \boxed{\text{(C) }\dfrac{7^{7^{7^{x}}}}{(\log7)^{3}}+k} \]
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