Let α and β be real numbers. Consider a \(3 \times 3\) matrix A such that \(A^2 = 3A + \alpha I\). If \(A^4 = 21A + \beta I\), then

Question

Let \(λ ≠ 0\) be a real number. Let α, β be the roots of the equation \(14x^2 – 31x + 3λ = 0\) and α, γ be the roots of the equation \(35x^2 – 53x + 4λ = 0\). Then \(\frac{3\alpha}{\beta}\) and \(\frac{4\alpha}{\lambda}\) are the roots of the equation

Answer 1

To solve this problem, we need to determine the values of the constants \(\alpha\) and \(\beta\) that satisfy the given equations involving the matrix \(A\). We start with the matrix equations:

Given: \(A^2 = 3A + \alpha I\).
Given: \(A^4 = 21A + \beta I\).

First, let's express \(A^3\) in terms of \(A\) and \(\alpha\) using the given equation for \(A^2\):

From equation 1, multiply both sides by \(A\):

\(A^3 = A \cdot A^2 = A(3A + \alpha I) = 3A^2 + \alpha A\)

Substitute \(A^2 = 3A + \alpha I\) back for \(A^2\):

\(A^3 = 3(3A + \alpha I) + \alpha A = 9A + 3\alpha I + \alpha A = 9A + \alpha A + 3\alpha I = (9 + \alpha)A + 3\alpha I\)

Next, use this expression to find \(A^4\):

Multiply \(A^3\) by \(A\):

\(A^4 = A \cdot A^3 = A((9 + \alpha)A + 3\alpha I) = (9 + \alpha)A^2 + 3\alpha A\)

Substitute \(A^2 = 3A + \alpha I\) for \(A^2\):

\(A^4 = (9 + \alpha)(3A + \alpha I) + 3\alpha A = 3(9 + \alpha)A + (9 + \alpha)\alpha I + 3\alpha A\)

Simplify this expression:

\(A^4 = (27 + 3\alpha + 3\alpha)A + \alpha(9 + \alpha)I = (27 + 6\alpha)A + \alpha(9 + \alpha)I\)

Equating this to \(21A + \beta I\), we compare coefficients:

For \(A\): \(27 + 6\alpha = 21\)
For \(I\): \(\alpha(9 + \alpha) = \beta\)

Solving the first equation:

\(27 + 6\alpha = 21 \Rightarrow 6\alpha = 21 - 27 = -6 \Rightarrow \alpha = -1\)

Using \(\alpha = -1\), solve the equation for \(\beta\):

\(\beta = \alpha(9 + \alpha) = -1(9 - 1) = -1 \times 8 = -8\)

Therefore, the correct answer is:

\(\beta = -8\)

Answer 2