Question

If \( A = \begin{vmatrix} 8 & 27 & 125 \\ 2 & 3 & 5 \\ 1 & 1 & 1 \end{vmatrix} \), then the value of \( A^2 \) is equal to:

• A. $0$
• B. $36$
• C. $64$
• D. $2400$
• E. $3600$

Hint:

For determinants involving cubes or squares, row operations that target the '1s' in the bottom row usually yield the fastest path to the solution.

Answer 1

Answer: (E)

Concept: The determinant $|A|$ can be simplified using row or column operations. Note that the first row contains cubes: $2^3, 3^3, 5^3$.
• $A^2$ in this context refers to $|A|^2$.
• We can use properties like $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$ to simplify.

Step 1: Set up the determinant.
$|A| = \begin{vmatrix} 8 & 27 & 125
2 & 3 & 5
1 & 1 & 1 \end{vmatrix}$

Step 2: Apply Column Operations.
Apply $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$: \[ |A| = \begin{vmatrix} 8 & 19 & 117 2 & 1 & 3 1 & 0 & 0 \end{vmatrix} \]

Step 3: Expand along the third row.
Expanding along $R_3$: \[ |A| = 1 \begin{vmatrix} 19 & 117 1 & 3 \end{vmatrix} = 1(19 \times 3 - 117 \times 1) \] \[ |A| = 57 - 117 = -60 \]

Step 4: Calculate $A^2$.
\[ A^2 = (-60)^2 = 3600 \]