Consider the matrix  A= 
 (3 -b 
-2  -3)
. Find the value of b
 for which the matrix 
 has only one eigenvalue. Give only the numerical answer as a decimal.

The matrix \( A = \begin{pmatrix} 3 & -b \\ -2 & -3 \end{pmatrix} \) has only one eigenvalue when the discriminant of its characteristic equation is zero.

The characteristic equation is given by:

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

For matrix \( A \), the characteristic polynomial is:

\[
\text{det}\begin{pmatrix} 3 - \lambda & -b \\ -2 & -3 - \lambda \end{pmatrix} = 0
\]

Expanding the determinant:

\[
(3 - \lambda)(-3 - \lambda) - (-b)(-2) = 0
\]

Simplifying:

\[
(9 + 3\lambda - 3\lambda - \lambda^2) - 2b = 0
\]

This simplifies to:

\[
9 - \lambda^2 - 2b = 0
\]

Thus, the characteristic polynomial is:

\[
\lambda^2 - 9 + 2b = 0
\]

For the matrix to have only one eigenvalue, the discriminant of this quadratic equation must be zero. The discriminant of \( \lambda^2 + 0\lambda + (-9 + 2b) = 0 \) is:

\[
\Delta = 0^2 - 4(1)(-9 + 2b) = 0
\]

This gives:

\[
36 - 8b = 0
\]

Solving for \( b \):

\[
8b = 36
\]

\[
b = \frac{36}{8} = 4.5
\]

Thus, the value of \( b \) is \( \boxed{4.5} \).

Consider the matrix A = (3 -b, -2 -3). Find the value of b for which the matrix has only one eigenvalue. Give only the numerical answer as a decimal.

Learn how to find the value of b for the matrix A = (3 -b, -2 -3) such that it has only one eigenvalue. This problem involves linear algebra concepts like eigenvalues, quadratic equations, and discriminants. Suitable for undergraduate students studying linear algebra.

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