Question

Let A be the matrix of order 3 x 3 such that |A| = 1, $B = 2A^{-1}$ and $C = \frac{(adj A)}{\sqrt[3]{2}}$, then the value of $|AB^2 \cdot C.128^3|$ is

[Note: |A| represent determinant value of matrix A.]

• A. 8
• B. 4
• C. 64
• D. None of these

Answer 1

Answer: (C)

64

Answer 2

Given:

• A is a 3x3 matrix with |A| = 1
• $B = 2A^{-1}$
• $C = \frac{adj A}{\sqrt[3]{2}}$

We need to find the value of $|AB^2 \cdot C \cdot 128^3|$. We assume that the poorly formatted scalar is actually $\sqrt[3]{2}$

1. Calculate |B|:

   $|B| = |2A^{-1}| = 2^3 |A^{-1}| = 8 \cdot \frac{1}{|A|} = 8 \cdot 1 = 8$

2. Calculate |C|:

   $|C| = |\frac{adj A}{\sqrt[3]{2}}| = (\frac{1}{\sqrt[3]{2}})^3 |adj A| = \frac{1}{2} |A|^{3-1} = \frac{1}{2} |A|^2 = \frac{1}{2} \cdot 1^2 = \frac{1}{2}$

3. Calculate $|AB^2C|$:

   $|AB^2C| = |A| |B^2| |C| = |A| \cdot |B|^2 \cdot |C| = 1 \cdot 8^2 \cdot \frac{1}{2} = 1 \cdot 64 \cdot \frac{1}{2} = 32$

4. **Calculate $|AB^2 \cdot C \cdot 128^3|$ assuming it is $|AB^2 C \cdot \sqrt[3]{2}|$:

   $|AB^2 \cdot C \cdot 128^3| = |AB^2 C \cdot \sqrt[3]{2}| = |AB^2 C| \cdot (\sqrt[3]{2})^3 = 32 \cdot 2 = 64$

Therefore, $|AB^2 \cdot C.128^3| = 64$