Question

Let $B = \begin{bmatrix} 1 & 3 \\\ 1 & 5 \end{bmatrix}$ and $A$ be a $2 \times 2$ matrix such that $AB^{-1} = A^{-1}.$ If $BCB^{-1} = A$ and $C^4 + \alpha C^2 + \beta I = O,$ then $2\beta - \alpha$ is equal to:

• A. 16
• B. 2
• C. 8
• D. 10

Answer 1

Answer: (D)

10

Answer 2

We are given the following matrix relations:
$BCB^{-1} = A$
$\Rightarrow (BCB^{-1})(BCB^{-1}) = A \cdot A$
\Rightarrow BCI \cdot CB^{-1} = A^2 \quad \text{(since \( B^{-1}B = I )})
$\Rightarrow BC^2 B^{-1} = A^2$
$\Rightarrow B^{-1}(BC^2 B^{-1})B = B^{-1}A^2B$
$\Rightarrow B^{-1}C^2 B = B^{-1}A^2B$

From the above relations, we can use the fact that:

$C^2 = A^{-1} \cdot A \cdot B \Rightarrow C^2 = B$

Next, since $AB^{-1} = A^{-1}$, we can manipulate the expression for $C^2$:

$AB^{-1} \cdot A = A^{-1} \Rightarrow B^{-1}A = A^{-1} \cdot A^{-1}$

Thus, $C^2$ and the matrix $B$ satisfy the characteristic equation:

$|C^2 - \lambda I| = 0$ $|B - \lambda I| = 0$

Now, we solve the characteristic equation:
$\begin{vmatrix} 1 - \lambda & 3 \\\ 1 & 5 - \lambda \end{vmatrix} = 0$
$\Rightarrow (1 - \lambda)(5 - \lambda) - 3 = 0$

$\Rightarrow \lambda^2 - 6\lambda + 5 - 3 = 0$
$\Rightarrow \lambda^2 - 6\lambda + 2 = 0$
$\Rightarrow \beta^2 - 6\beta + 2 = 0$
$\Rightarrow C^4 - 6C^2 + 2I = 0$

Thus, solving the equation gives us:
$\alpha = -6, \quad \beta = 2$

Finally, we calculate:
$2\beta - \alpha = 2 \times 2 - (-6) = 4 + 6 = 10$

Thus, the correct answer is:
$\boxed{10}$