SOLUTION: A = [5 0.25; 35 6] Find an invertible matrix X and a diagonal matrix D such that X^−1 * A * X=D.

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Question 1160967: A = [5 0.25; 35 6]
Find an invertible matrix X and a diagonal matrix D such that X^−1 * A * X=D.
Answer by CPhill(2285) About Me  (Show Source):
You can put this solution on YOUR website!
To find an invertible matrix $X$ and a diagonal matrix $D$ such that $X^{-1}AX = D$, we need to diagonalize the matrix $A$. The diagonal entries of $D$ will be the eigenvalues of $A$, and the columns of $X$ will be the corresponding eigenvectors.
Given:

$$A = \begin{bmatrix} 5 & 0.25 \\ 35 & 6 \end{bmatrix}$$
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### Step 1: Find the Eigenvalues of $A$
We solve the characteristic equation $\det(A - \lambda I) = 0$:
$$\det \begin{bmatrix} 5 - \lambda & 0.25 \\ 35 & 6 - \lambda \end{bmatrix} = 0$$
$$(5 - \lambda)(6 - \lambda) - (0.25)(35) = 0$$
$$30 - 5\lambda - 6\lambda + \lambda^2 - 8.75 = 0$$
$$\lambda^2 - 11\lambda + 21.25 = 0$$
Using the quadratic formula:

$$\lambda = \frac{11 \pm \sqrt{(-11)^2 - 4(1)(21.25)}}{2} = \frac{11 \pm \sqrt{121 - 85}}{2} = \frac{11 \pm \sqrt{36}}{2}$$
$$\lambda_1 = \frac{11 + 6}{2} = \frac{17}{2} = 8.5$$
$$\lambda_2 = \frac{11 - 6}{2} = \frac{5}{2} = 2.5$$
Thus, the diagonal matrix $D$ is:

$$D = \begin{bmatrix} 8.5 & 0 \\ 0 & 2.5 \end{bmatrix} \quad \text{or} \quad \begin{bmatrix} 17/2 & 0 \\ 0 & 5/2 \end{bmatrix}$$
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### Step 2: Find the Eigenvectors for Each Eigenvalue
#### 1. For $\lambda_1 = 8.5$ ($17/2$):
Solve $(A - 8.5I)\mathbf{v}_1 = \mathbf{0}$:
$$\begin{bmatrix} 5 - 8.5 & 0.25 \\ 35 & 6 - 8.5 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$
$$\begin{bmatrix} -3.5 & 0.25 \\ 35 & -2.5 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$
From the first row:

$$-3.5x_1 + 0.25x_2 = 0 \implies 0.25x_2 = 3.5x_1 \implies x_2 = 14x_1$$
Setting $x_1 = 1$, we get $x_2 = 14$. So, the first eigenvector is:

$$\mathbf{v}_1 = \begin{bmatrix} 1 \\ 14 \end{bmatrix}$$
#### 2. For $\lambda_2 = 2.5$ ($5/2$):
Solve $(A - 2.5I)\mathbf{v}_2 = \mathbf{0}$:
$$\begin{bmatrix} 5 - 2.5 & 0.25 \\ 35 & 6 - 2.5 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$
$$\begin{bmatrix} 2.5 & 0.25 \\ 35 & 3.5 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$
From the first row:

$$2.5x_1 + 0.25x_2 = 0 \implies 0.25x_2 = -2.5x_1 \implies x_2 = -10x_1$$
Setting $x_1 = 1$, we get $x_2 = -10$. So, the second eigenvector is:

$$\mathbf{v}_2 = \begin{bmatrix} 1 \\ -10 \end{bmatrix}$$
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### Final Answer
Construct matrix $X$ using the eigenvectors as columns:
$$X = \begin{bmatrix} 1 & 1 \\ 14 & -10 \end{bmatrix}$$
$$D = \begin{bmatrix} 8.5 & 0 \\ 0 & 2.5 \end{bmatrix} \quad \text{or} \quad \begin{bmatrix} 17/2 & 0 \\ 0 & 5/2 \end{bmatrix}$$