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).

Answer 2

We are given: \[ |5 \cdot \text{adj}(A)| = 5 \] We use the property: \[ |\text{adj}(A)| = |A|^{n-1} \] for an \( n \times n \) matrix. Since \( A \) is a 3x3 matrix, we have: \[ |\text{adj}(A)| = |A|^2 \] Also, for any scalar \( k \) and matrix \( B \) of order \( n \): \[ |kB| = k^n |B| \] So: \[ |5 \cdot \text{adj}(A)| = 5^3 \cdot |\text{adj}(A)| = 125 \cdot |A|^2 \] Given: \[ 125 \cdot |A|^2 = 5 \Rightarrow |A|^2 = \frac{5}{125} = \frac{1}{25} \Rightarrow |A| = \pm \frac{1}{5} \] Correct answer: \( \pm \frac{1}{5} \) 

Answer 3

We are given that A is a 3x3 matrix such that |5 * adj(A)| = 5, and we need to find |A|.

We know that adj(A) = |A| * A^-1, and for an n x n matrix A,

|k * A| = kⁿ |A|

Also, |adj(A)| = |A|^n-1, where n is the order of the matrix.

In this case, n = 3.

Using the properties, we have:

|5 * adj(A)| = 5³ * |adj(A)| = 125 * |adj(A)|

We are given that |5 * adj(A)| = 5

Therefore, 125 * |adj(A)| = 5

|adj(A)| = 5 / 125 = 1 / 25

We also know that |adj(A)| = |A|^n-1 = |A|^3-1 = |A|²

So, |A|² = 1 / 25

Taking the square root of both sides:

|A| = \(\pm \sqrt{\frac{1}{25}}\)

|A| = \(\pm \frac{1}{5}\)

Answer:

\(\pm \frac{1}{5}\)