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Chapter 4: Problem 261

If $$ A=\left|\begin{array}{ll} k & 3 \\ 3 & k \end{array}\right| $$ and \(\left|\mathrm{A}^{3}\right|=343\), then find the value of \(\mathrm{k}=\ldots\) \(\begin{array}{llll}\text { (a) } \pm 1 & \text { (b) } \pm 2 & \text { (c) } \pm 3 & \text { (d) } \pm 4\end{array}\)

Short Answer

Expert verified
The value of k is \(k = \pm 4\).

Step by step solution

01

Find the determinant of A

To find the determinant of a 2x2 matrix, subtract the product of the diagonal elements from the product of the off-diagonal elements. In this case, the determinant is given by: \(Det(A) = k \times k - 3 \times 3\)
02

Raise the determinant to the power of 3

Since we are given that the determinant of A raised to the power of 3 is equal to 343, we can write the following equation: \((Det(A))^3 = 343\) Substitute the expression for the determinant of A from step 1: \((k^2 - 9)^3 = 343\)
03

Solve for k

To solve for k, take the cube root of both sides: \(k^2 - 9 = 7\) Now, we need to solve for k: \(k^2 = 16\) So, \(k = \pm 4\).
04

Choose the correct option

Since the possible values for k are ±4, the correct option is: (d) ±4

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