Question

Let $P$ be a $3 \times 3$ matrix having characteristic roots $\lambda_1 = -\dfrac{2}{3}$, $\lambda_2 = 0$ and $\lambda_3 = 1$. Define $Q = 3P^3 - P^2 - P + I_3$ and $R = 3P^3 - 2P$. If $\alpha = \det(Q)$ and $\beta = \text{trace}(R)$, then $\alpha + \beta$ equals .......... (round off to two decimal places).

Hint:

For functions of matrices, the determinant and trace can be directly computed using eigenvalues instead of matrix expansion.

Answer 1

Correct Answer: 2

Step 1: Use eigenvalue properties.
If $\lambda_i$ are eigenvalues of $P$, then eigenvalues of $Q$ and $R$ are obtained by substituting into their defining expressions: \[ \text{Eigenvalues of } Q: 3\lambda_i^3 - \lambda_i^2 - \lambda_i + 1. \] \[ \text{Eigenvalues of } R: 3\lambda_i^3 - 2\lambda_i. \]

Step 2: Compute eigenvalues for each $\lambda_i$.
For $\lambda_1 = -\frac{2}{3}$: \[ Q_1 = 3\left(-\frac{2}{3}\right)^3 - \left(-\frac{2}{3}\right)^2 - \left(-\frac{2}{3}\right) + 1 = -\frac{8}{9} - \frac{4}{9} + \frac{2}{3} + 1 = \frac{7}{9}. \] For $\lambda_2 = 0$: $Q_2 = 1$. For $\lambda_3 = 1$: $Q_3 = 3 - 1 - 1 + 1 = 2$. Hence, \[ \alpha = \det(Q) = Q_1 Q_2 Q_3 = \frac{7}{9} \times 1 \times 2 = \frac{14}{9} \approx 1.56. \]

Step 3: Find trace of $R$.
For $\lambda_1 = -\frac{2}{3}$: $R_1 = 3\left(-\frac{2}{3}\right)^3 - 2\left(-\frac{2}{3}\right) = -\frac{8}{9} + \frac{4}{3} = \frac{4}{9}.$ For $\lambda_2 = 0$: $R_2 = 0$. For $\lambda_3 = 1$: $R_3 = 3(1)^3 - 2(1) = 1.$ \[ \beta = \text{trace}(R) = R_1 + R_2 + R_3 = \frac{4}{9} + 0 + 1 = \frac{13}{9} \approx 1.44. \] \[ \alpha + \beta = 1.56 + 1.44 = 3.00. \]

Final Answer: \[ \boxed{\alpha + \beta = 3.00.} \]