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 - 2.3

Step 1: Find eigenvalues of $Q$

If $\lambda$ is an eigenvalue of $P$, then the eigenvalue of $Q$ is: $$\mu = 3\lambda^3 - \lambda^2 - \lambda + 1$$

For $\lambda_1 = -\frac{2}{3}$: $$\mu_1 = 3\left(-\frac{2}{3}\right)^3 - \left(-\frac{2}{3}\right)^2 - \left(-\frac{2}{3}\right) + 1$$ $$= 3\left(-\frac{8}{27}\right) - \frac{4}{9} + \frac{2}{3} + 1$$ $$= -\frac{8}{9} - \frac{4}{9} + \frac{6}{9} + \frac{9}{9}$$ $$= \frac{-8 - 4 + 6 + 9}{9} = \frac{3}{9} = \frac{1}{3}$$

For $\lambda_2 = 0$: $$\mu_2 = 3(0)^3 - (0)^2 - 0 + 1 = 1$$

For $\lambda_3 = 1$: $$\mu_3 = 3(1)^3 - (1)^2 - 1 + 1 = 3 - 1 - 1 + 1 = 2$$

Step 2: Calculate $\alpha = \det(Q)$

$$\alpha = \mu_1 \cdot \mu_2 \cdot \mu_3 = \frac{1}{3} \cdot 1 \cdot 2 = \frac{2}{3}$$

Step 3: Find eigenvalues of $R$

If $\lambda$ is an eigenvalue of $P$, then the eigenvalue of $R$ is: $$\nu = 3\lambda^3 - 2\lambda$$

For $\lambda_1 = -\frac{2}{3}$: $$\nu_1 = 3\left(-\frac{2}{3}\right)^3 - 2\left(-\frac{2}{3}\right) = -\frac{8}{9} + \frac{4}{3} = -\frac{8}{9} + \frac{12}{9} = \frac{4}{9}$$

For $\lambda_2 = 0$: $$\nu_2 = 3(0)^3 - 2(0) = 0$$

For $\lambda_3 = 1$: $$\nu_3 = 3(1)^3 - 2(1) = 3 - 2 = 1$$

Step 4: Calculate $\beta = \text{trace}(R)$

$$\beta = \nu_1 + \nu_2 + \nu_3 = \frac{4}{9} + 0 + 1 = \frac{4}{9} + \frac{9}{9} = \frac{13}{9}$$

Step 5: Calculate $\alpha + \beta$

$$\alpha + \beta = \frac{2}{3} + \frac{13}{9} = \frac{6}{9} + \frac{13}{9} = \frac{19}{9} \approx 2.11$$

Answer: $\frac{19}{9} \approx 2.11$