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Assertion (A) : If every element of a th...

Assertion (A) : If every element of a third order determinant of value `Delta` is multiplied by 5, then the value of the new determinant is `125Delta`.
Reason (R ) : If k is a scalar and A is an `nxxn` matrix, then `|kA|=k^(n)|A|`

A

Both A and R are true and R is the correct explanation of A

B

Both A and R are true but R is NOT the correct explanation of A

C

A is true but R is false

D

A is false but R is true

Text Solution

AI Generated Solution

The correct Answer is:
To solve the assertion and reason provided in the question, let's break it down step by step. ### Step 1: Understand the assertion The assertion states that if every element of a third-order determinant (3x3 matrix) with a value of Δ is multiplied by 5, then the value of the new determinant becomes 125Δ. ### Step 2: Identify the order of the determinant The order of the determinant is 3, which means it is a 3x3 matrix. ### Step 3: Apply the property of determinants According to the property of determinants, if a scalar \( k \) is multiplied to every element of an \( n \times n \) matrix, the value of the determinant changes according to the formula: \[ |kA| = k^n |A| \] where \( n \) is the order of the matrix. ### Step 4: Substitute the values In this case, \( k = 5 \) and \( n = 3 \). Therefore, we can substitute these values into the formula: \[ |5A| = 5^3 |A| = 125 |A| \] Since the original determinant value is Δ, we can write: \[ |5A| = 125Δ \] ### Step 5: Conclusion Thus, the assertion is correct: if every element of a third-order determinant of value Δ is multiplied by 5, the value of the new determinant is indeed 125Δ. ### Step 6: Evaluate the reason The reason states that if \( k \) is a scalar and \( A \) is an \( n \times n \) matrix, then \( |kA| = k^n |A| \). This is a true statement based on the properties of determinants. ### Final Conclusion Both the assertion and the reason are correct. ### Summary of the Solution: 1. The assertion is true: multiplying every element of a 3x3 determinant by 5 results in a new determinant value of 125Δ. 2. The reason is also true: the property of determinants regarding scalar multiplication holds.
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