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

The Correct Option is B

Given \( A = \begin{bmatrix} K & 4 \\ 4 & K \end{bmatrix} \), we need to find the value of \( K^8 \) knowing that \( |A^3| = 729 \). First, compute the determinant of \( A \): 

\(|A| = K^2 - 16\).

For \( A^3 \), the determinant is given by \(|A^3| = |A|^3 = 729\), so \((K^2 - 16)^3 = 729\). Recognizing \( 729 \) as \( 9^3 \), we equate:

\((K^2 - 16)^3 = 9^3\),

implies \( K^2 - 16 = 9 \).

Solve for \( K^2 \):

\(K^2 = 9 + 16 = 25\).

Find \( K^8 \) by raising to the fourth power:

\(K^8 = (K^2)^4 = 25^4 = (5^2)^4 = 5^8\).

Therefore, the value of \( K^8 \) is: \( 5^8 \).

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:
$5^8$