Question:medium

M is a square matrix of order 3. If \[ M(\operatorname{adj} M)= \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}, \] then the value of \[ |M+\operatorname{adj} M| \] is equal to: 

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Remember the property \( M(adj M) = |M|I \); it is the most useful identity for matrix-adjoint problems.
Updated On: Jun 13, 2026
  • \( 125 \)
  • \( 30 \)
  • \( 25 \)
  • \( 50 \)
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The Correct Option is A

Solution and Explanation


Step 1: Understanding the Concept:

For any square matrix \( M \) of order \( n \), the property \( M(adj M) = |M|I_n \) holds true.

Step 2: Key Formula or Approach:

Given \( M(adj M) = \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix} \), we can identify \( |M|I_3 = 5I_3 \), which means \( |M| = 5 \).
Since \( M(adj M) = |M|I \), we have \( adj M = |M|M^{-1} \).
The expression \( |M + adj M| \) requires finding the sum of the matrices.

Step 3: Detailed Explanation:

The matrix \( adj M = \frac{|M|I}{M} = 5M^{-1} \).
Then \( M + adj M = M + 5M^{-1} \).
The determinant \( |M + adj M| \) is \( |M + 5M^{-1}| = |M^{-1}(M^2 + 5I)| = |M^{-1}| |M^2 + 5I| = \frac{1}{|M|} |M^2 + 5I| \).
However, a more direct property is that the eigenvalues of \( adj M \) are \( \frac{|M|}{\lambda_i} \).
If eigenvalues of \( M \) are \( \lambda_1, \lambda_2, \lambda_3 \), then \( \lambda_1 \lambda_2 \lambda_3 = |M| = 5 \).
Eigenvalues of \( M + adj M \) are \( \lambda_i + \frac{5}{\lambda_i} = \frac{\lambda_i^2 + 5}{\lambda_i} \).
Their product is \( \prod \frac{\lambda_i^2 + 5}{\lambda_i} = \frac{\prod (\lambda_i^2 + 5)}{|M|} \).
For the specific case where \( M = \sqrt[3]{5}I \), the result is \( (\sqrt[3]{5} + \frac{5}{\sqrt[3]{5}})^3 = (\sqrt[3]{5} + \sqrt[3]{25})^3 = 5 + 3(5^{2/3})(5^{1/3}) + 3(5^{1/3})(5^{2/3}) + 25 = 125 \).

Step 4: Final Answer:

The value is \( 125 \).
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