Question

If $A = \begin{bmatrix} K & 4 \\\ 4 & K \end{bmatrix}$ and $|A^3| = 729$, then the value of $K^8$ is:

• A. $9^8$
• B. $5^8$
• C. $3^8$
• D. $(-3)^8$

Answer 1

Answer: (B)

$5^8$

Answer 2

The determinant of A is:
$|A| = \begin{vmatrix} K & 4 \\\ 4 & K \end{vmatrix} = K \cdot K - 4 \cdot 4 = K^2 - 16.$
Using the property of determinants:
$|A|^3 = (|A|)^3 = 729.$
Take the cube root:
$|A| = \sqrt[3]{729} = 9.$
Thus:
$K^2 - 16 = 9 \implies K^2 = 25.$
Therefore:
$K = \pm 5.$
The value of $K^8$ is:
$K^8 = (K^2)^4 = 25^4.$
Calculate:
$25^4 = (25^2)^2 = 625^2 = 390625.$
Final Answer:
$5^8$