Question

If A is a 3x3 Matrix such that \(|5 \cdot \text{adj}(A)| = 5, \text{ then } |A|\) is equal to

• A. \(\pm 1\)
• B. \(\pm \frac{1}{5}\)
• C. \(\pm \frac{1}{25}\)
• D. \(\pm 5\)

Answer 1

The Correct Option is B

The property states that for any square matrix A, \(|\text{adj}(A)| = |A|^{n-1}\), where n is the order of matrix A. 
In this case, A is a 3x3 matrix. 
So, \(|\text{adj}(A)| = |A|^{3-1} = |A|^2\)

From the given equation, \(|5 \cdot \text{adj}(A)| = 5\), we can substitute\( |adj A| \)with \(|A|^2 : |5A|^2 = 5\)
Taking the square root of both sides, we have: \(|5A| = \pm \sqrt{5}\) 
Now, we know that \(|cA| = c^n |A|,\) where c is a constant and n is the order of matrix A. 
In this case, n = 3, 
so we can write:
 \(5^n |A| = \pm \sqrt{5} \cdot 5^3 |A|\)
=\(\pm \sqrt{5} \cdot 125 |A|\)
= \(\pm \sqrt{5} \cdot |A|\)
= \(\pm \frac{\sqrt{5}}{125}.\)
Simplifying the expression, we have:
 \(|A| = \pm \sqrt{\frac{1}{25}} \quad \text{and} \quad |A| = \pm \frac{1}{5}\)
Therefore, \(|A|\) is equal to \(\pm \frac{1}{5}\) (option B).