Let `A=|{:(,-2,7,sqrt3),(,0,0,-2),(,0,2,0):}| and A^(4)=lambda,I,"then"lambda "is"`

Options

A. `-16`
B. 16
C. 8
D. `-8`

Answer

To solve the problem, we need to find the value of \( \lambda \) such that \( A^4 = \lambda I \), where \( A \) is given as:

\[ A = \begin{pmatrix} -2 & 7 & \sqrt{3} \\ 0 & 0 & -2 \\ 0 & 2 & 0 \end{pmatrix} \]

### Step 1: Calculate \( A^2 \)

We start by calculating \( A^2 = A \cdot A \):

\[ A^2 = \begin{pmatrix} -2 & 7 & \sqrt{3} \\ 0 & 0 & -2 \\ 0 & 2 & 0 \end{pmatrix} \cdot \begin{pmatrix} -2 & 7 & \sqrt{3} \\ 0 & 0 & -2 \\ 0 & 2 & 0 \end{pmatrix} \]

Calculating the elements of \( A^2 \):

- First row:
- First element: \( (-2)(-2) + (7)(0) + (\sqrt{3})(0) = 4 \)
- Second element: \( (-2)(7) + (7)(0) + (\sqrt{3})(2) = -14 + 2\sqrt{3} \)
- Third element: \( (-2)(\sqrt{3}) + (7)(-2) + (\sqrt{3})(0) = -2\sqrt{3} - 14 \)

- Second row:
- First element: \( (0)(-2) + (0)(0) + (-2)(0) = 0 \)
- Second element: \( (0)(7) + (0)(0) + (-2)(2) = -4 \)
- Third element: \( (0)(\sqrt{3}) + (0)(-2) + (-2)(0) = 0 \)

- Third row:
- First element: \( (0)(-2) + (2)(0) + (0)(0) = 0 \)
- Second element: \( (0)(7) + (2)(0) + (0)(-2) = 0 \)
- Third element: \( (0)(\sqrt{3}) + (2)(-2) + (0)(0) = -4 \)

Thus, we have:

\[ A^2 = \begin{pmatrix} 4 & -14 + 2\sqrt{3} & -2\sqrt{3} - 14 \\ 0 & -4 & 0 \\ 0 & 0 & -4 \end{pmatrix} \]

### Step 2: Calculate \( A^4 \)

Now, we calculate \( A^4 = A^2 \cdot A^2 \):

\[ A^4 = \begin{pmatrix} 4 & -14 + 2\sqrt{3} & -2\sqrt{3} - 14 \\ 0 & -4 & 0 \\ 0 & 0 & -4 \end{pmatrix} \cdot \begin{pmatrix} 4 & -14 + 2\sqrt{3} & -2\sqrt{3} - 14 \\ 0 & -4 & 0 \\ 0 & 0 & -4 \end{pmatrix} \]

Calculating the elements of \( A^4 \):

- First row:
- First element: \( 4 \cdot 4 + (-14 + 2\sqrt{3}) \cdot 0 + (-2\sqrt{3} - 14) \cdot 0 = 16 \)
- Second element: \( 4 \cdot (-14 + 2\sqrt{3}) + (-14 + 2\sqrt{3}) \cdot (-4) + (-2\sqrt{3} - 14) \cdot 0 = -56 + 8\sqrt{3} + 56 - 8\sqrt{3} = 0 \)
- Third element: \( 4 \cdot (-2\sqrt{3} - 14) + (-14 + 2\sqrt{3}) \cdot 0 + (-2\sqrt{3} - 14)(-4) = -8\sqrt{3} - 56 + 8\sqrt{3} + 56 = 0 \)

- Second row:
- First element: \( 0 \cdot 4 + (-4) \cdot 0 + 0 \cdot 0 = 0 \)
- Second element: \( 0 \cdot (-14 + 2\sqrt{3}) + (-4)(-4) + 0 \cdot 0 = 16 \)
- Third element: \( 0 \cdot (-2\sqrt{3} - 14) + (-4) \cdot 0 + 0 \cdot (-4) = 0 \)

- Third row:
- First element: \( 0 \cdot 4 + 0 \cdot 0 + (-4) \cdot 0 = 0 \)
- Second element: \( 0 \cdot (-14 + 2\sqrt{3}) + 0 \cdot (-4) + (-4)(0) = 0 \)
- Third element: \( 0 \cdot (-2\sqrt{3} - 14) + 0 \cdot 0 + (-4)(-4) = 16 \)

Thus, we have:

\[ A^4 = \begin{pmatrix} 16 & 0 & 0 \\ 0 & 16 & 0 \\ 0 & 0 & 16 \end{pmatrix} = 16I \]

### Step 3: Conclusion

From the equation \( A^4 = \lambda I \), we can see that \( \lambda = 16 \).

### Final Answer

Thus, the value of \( \lambda \) is:

\[ \lambda = 16 \]