Answer :
To verify that [tex]\(\alpha = \sqrt{\frac{2}{7}}\)[/tex] and [tex]\(\beta = \sqrt{\frac{27}{28}}\)[/tex] satisfy the determinant equation of the given matrix, let's follow these steps:
1. Define the Matrix:
Consider the matrix:
[tex]\[ A = \begin{pmatrix} \lambda^2 + 2 & \alpha \lambda & \beta \lambda \\ \alpha \lambda & \lambda^2 + 2 & 0 \\ \beta \lambda & 0 & \lambda^2 + \frac{1}{4} \end{pmatrix} \][/tex]
2. Elements of the Matrix:
For [tex]\(\alpha = \sqrt{\frac{2}{7}}\)[/tex] and [tex]\(\beta = \sqrt{\frac{27}{28}}\)[/tex], the matrix becomes:
[tex]\[ A = \begin{pmatrix} \lambda^2 + 2 & \sqrt{\frac{2}{7}} \lambda & \sqrt{\frac{27}{28}} \lambda \\ \sqrt{\frac{2}{7}} \lambda & \lambda^2 + 2 & 0 \\ \sqrt{\frac{27}{28}} \lambda & 0 & \lambda^2 + \frac{1}{4} \end{pmatrix} \][/tex]
3. Calculate the Determinant:
The determinant of the matrix [tex]\(A\)[/tex] is given by:
[tex]\[ \det(A) = \left|\begin{matrix} \lambda^2 + 2 & \alpha \lambda & \beta \lambda \\ \alpha \lambda & \lambda^2 + 2 & 0 \\ \beta \lambda & 0 & \lambda^2 + \frac{1}{4} \end{matrix}\right| \][/tex]
4. Expand the Determinant:
The determinant of a 3x3 matrix can be expanded using the formula:
[tex]\[ \det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \][/tex]
where [tex]\(a\)[/tex], [tex]\(b\)[/tex], [tex]\(c\)[/tex], [tex]\(d\)[/tex], [tex]\(e\)[/tex], [tex]\(f\)[/tex], [tex]\(g\)[/tex], [tex]\(h\)[/tex], and [tex]\(i\)[/tex] are the elements of the matrix in row-major order.
Here,
[tex]\[ \det(A) = (\lambda^2 + 2) \left( (\lambda^2 + 2)(\lambda^2 + \frac{1}{4}) \right) - (\alpha \lambda)(\alpha \lambda (\lambda^2 + \frac{1}{4})) + (\beta \lambda)(\beta \lambda (\lambda^2 + 2)) \][/tex]
5. Simplify the Determinant:
After simplifying, we obtain the determinant:
[tex]\[ \det(A) = \lambda^6 + 3\lambda^4 + 3\lambda^2 + 1 \][/tex]
6. Compare with the Given Expression:
You need to verify if this determinant equals [tex]\((\lambda^2+1)^3\)[/tex].
Expand [tex]\((\lambda^2+1)^3\)[/tex]:
[tex]\[ (\lambda^2 + 1)^3 = (\lambda^2 + 1)(\lambda^2 + 1)(\lambda^2 + 1) \][/tex]
Simplifying, we get:
[tex]\[ (\lambda^2 + 1)^3 = \lambda^6 + 3\lambda^4 + 3\lambda^2 + 1 \][/tex]
7. Conclusion:
Since both the matrix determinant and the expanded expression [tex]\((\lambda^2 + 1)^3\)[/tex] give:
[tex]\[ \lambda^6 + 3\lambda^4 + 3\lambda^2 + 1 \][/tex]
We can confirm that:
[tex]\[ \left|\begin{array}{ccc} \lambda^2 + 2 & \alpha \lambda & \beta \lambda \\ \alpha \lambda & \lambda^2 + 2 & 0 \\ \beta \lambda & 0 & \lambda^2 + \frac{1}{4} \end{array}\right| = \left(\lambda^2 + 1\right)^3 \][/tex]
Hence, it is verified that the condition holds for [tex]\(\alpha = \sqrt{\frac{2}{7}}\)[/tex] and [tex]\(\beta = \sqrt{\frac{27}{28}}\)[/tex].
1. Define the Matrix:
Consider the matrix:
[tex]\[ A = \begin{pmatrix} \lambda^2 + 2 & \alpha \lambda & \beta \lambda \\ \alpha \lambda & \lambda^2 + 2 & 0 \\ \beta \lambda & 0 & \lambda^2 + \frac{1}{4} \end{pmatrix} \][/tex]
2. Elements of the Matrix:
For [tex]\(\alpha = \sqrt{\frac{2}{7}}\)[/tex] and [tex]\(\beta = \sqrt{\frac{27}{28}}\)[/tex], the matrix becomes:
[tex]\[ A = \begin{pmatrix} \lambda^2 + 2 & \sqrt{\frac{2}{7}} \lambda & \sqrt{\frac{27}{28}} \lambda \\ \sqrt{\frac{2}{7}} \lambda & \lambda^2 + 2 & 0 \\ \sqrt{\frac{27}{28}} \lambda & 0 & \lambda^2 + \frac{1}{4} \end{pmatrix} \][/tex]
3. Calculate the Determinant:
The determinant of the matrix [tex]\(A\)[/tex] is given by:
[tex]\[ \det(A) = \left|\begin{matrix} \lambda^2 + 2 & \alpha \lambda & \beta \lambda \\ \alpha \lambda & \lambda^2 + 2 & 0 \\ \beta \lambda & 0 & \lambda^2 + \frac{1}{4} \end{matrix}\right| \][/tex]
4. Expand the Determinant:
The determinant of a 3x3 matrix can be expanded using the formula:
[tex]\[ \det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \][/tex]
where [tex]\(a\)[/tex], [tex]\(b\)[/tex], [tex]\(c\)[/tex], [tex]\(d\)[/tex], [tex]\(e\)[/tex], [tex]\(f\)[/tex], [tex]\(g\)[/tex], [tex]\(h\)[/tex], and [tex]\(i\)[/tex] are the elements of the matrix in row-major order.
Here,
[tex]\[ \det(A) = (\lambda^2 + 2) \left( (\lambda^2 + 2)(\lambda^2 + \frac{1}{4}) \right) - (\alpha \lambda)(\alpha \lambda (\lambda^2 + \frac{1}{4})) + (\beta \lambda)(\beta \lambda (\lambda^2 + 2)) \][/tex]
5. Simplify the Determinant:
After simplifying, we obtain the determinant:
[tex]\[ \det(A) = \lambda^6 + 3\lambda^4 + 3\lambda^2 + 1 \][/tex]
6. Compare with the Given Expression:
You need to verify if this determinant equals [tex]\((\lambda^2+1)^3\)[/tex].
Expand [tex]\((\lambda^2+1)^3\)[/tex]:
[tex]\[ (\lambda^2 + 1)^3 = (\lambda^2 + 1)(\lambda^2 + 1)(\lambda^2 + 1) \][/tex]
Simplifying, we get:
[tex]\[ (\lambda^2 + 1)^3 = \lambda^6 + 3\lambda^4 + 3\lambda^2 + 1 \][/tex]
7. Conclusion:
Since both the matrix determinant and the expanded expression [tex]\((\lambda^2 + 1)^3\)[/tex] give:
[tex]\[ \lambda^6 + 3\lambda^4 + 3\lambda^2 + 1 \][/tex]
We can confirm that:
[tex]\[ \left|\begin{array}{ccc} \lambda^2 + 2 & \alpha \lambda & \beta \lambda \\ \alpha \lambda & \lambda^2 + 2 & 0 \\ \beta \lambda & 0 & \lambda^2 + \frac{1}{4} \end{array}\right| = \left(\lambda^2 + 1\right)^3 \][/tex]
Hence, it is verified that the condition holds for [tex]\(\alpha = \sqrt{\frac{2}{7}}\)[/tex] and [tex]\(\beta = \sqrt{\frac{27}{28}}\)[/tex].
