To find the constants $a$ and $b$ such that the matrix $A = \begin{pmatrix} a & 4 \\ 1 & b \end{pmatrix}$ has eigenvalues $\lambda_1 = 3$ and $\lambda_2 = -2$, we use the fact that the eigenvalues of a 2x2 matrix can be determined from its characteristic equation.

Characteristic Equation:

The characteristic equation of a matrix $A = \begin{pmatrix} a & 4 \\ 1 & b \end{pmatrix}$ is given by:

$\text{det}(A - \lambda I) = 0$

This gives us the following determinant:

$\text{det}\begin{pmatrix} a - \lambda & 4 \\ 1 & b - \lambda \end{pmatrix} = 0$

The determinant of this matrix is:

$(a - \lambda)(b - \lambda) - 4 = 0$

Expanding:

$(a - \lambda)(b - \lambda) = ab - a\lambda - b\lambda + \lambda^2$

Thus, the characteristic equation becomes:

$\lambda^2 - (a + b)\lambda + (ab - 4) = 0$

Since the eigenvalues are $\lambda_1 = 3$ and $\lambda_2 = -2$, we know that the characteristic equation can also be written as:

$(\lambda - 3)(\lambda + 2) = 0$

Expanding this:

$\lambda^2 - \lambda(3 - 2) - 6 = \lambda^2 - \lambda + 6 = 0$

Comparing the two characteristic equations:

1. The coefficient of $\lambda$ gives: $a + b = 1$
2. The constant term gives: $ab - 4 = -6$, or equivalently, $ab = -2$

Solving the system of equations:

We now have the following system of equations:

1. $a + b = 1$
2. $ab = -2$

Using substitution, solve for $a$ in terms of $b$:

$a = 1 - b$

Substitute this into the second equation:

$(1 - b)b = -2$

This simplifies to:

$b - b^2 = -2$

Rearranging:

$b^2 - b - 2 = 0$

Solve this quadratic equation:

$b = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-2)}}{2(1)}$

$b = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm \sqrt{9}}{2} = \frac{1 \pm 3}{2}$

So, the solutions for $b$ are:

$b = \frac{1 + 3}{2} = 2 \quad \text{or} \quad b = \frac{1 - 3}{2} = -1$

Finding $a$:

• If $b = 2$, then $a = 1 - 2 = -1$
• If $b = -1$, then $a = 1 - (-1) = 2$

Conclusion:

The constants $a$ and $b$ can either be:

• $a = -1$, $b = 2$
• $a = 2$, $b = -1$

Would you like further details or clarifications?

Here are 5 related questions to explore:

1. How do you find the eigenvectors of the matrix given the eigenvalues?
2. Can a matrix with distinct eigenvalues have more than two solutions for $a$ and $b$?
3. What is the physical significance of eigenvalues in a real-world application?
4. How does the determinant of a matrix relate to its eigenvalues?
5. What changes if both eigenvalues were the same (i.e., a repeated eigenvalue)?

Tip: For a 2x2 matrix, the trace (sum of the diagonal elements) is the sum of the eigenvalues, and the determinant is the product of the eigenvalues.