Math Problem Statement

Let AequalsPDP Superscript negative 1 and P and D as shown below. Compute Upper A Superscript 4. PequalsStart 2 By 2 Table 1st Row 1st Column 1 2nd Column negative 3 2nd Row 1st Column negative 2 2nd Column 7 EndTable ​, DequalsStart 2 By 2 Table 1st Row 1st Column 1 2nd Column 0 2nd Row 1st Column 0 2nd Column one fourth EndTable

Question content area bottom Part 1 Upper A Superscript 4equals    enter your response here ​(Simplify your​ answer.)

Solution

We are tasked with computing A4A^4 where A=PDP−1A = PDP^{-1}. Here's the step-by-step solution:

Step 1: Formula for Powers of AA

If A=PDP−1A = PDP^{-1}, then: A4=(PDP−1)4=PD4P−1.A^4 = (PDP^{-1})^4 = PD^4P^{-1}. This formula simplifies the computation by focusing on D4D^4, which is easier to compute as DD is diagonal.

Step 2: Compute D4D^4

The matrix DD is diagonal:

1 & 0 \\ 0 & \frac{1}{4} \end{bmatrix}.$$ The powers of a diagonal matrix are computed element-wise: $$D^4 = \begin{bmatrix} 1^4 & 0 \\ 0 & \left(\frac{1}{4}\right)^4 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{256} \end{bmatrix}.$$ ### Step 3: Compute $$A^4 = PD^4P^{-1}$$ We are given: $$P = \begin{bmatrix} 1 & -3 \\ -2 & 7 \end{bmatrix},$$ and we need $$P^{-1}$$. The formula for the inverse of a 2x2 matrix is: $$P^{-1} = \frac{1}{\text{det}(P)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix},$$ where the determinant of $$P$$ is: $$\text{det}(P) = (1)(7) - (-3)(-2) = 7 - 6 = 1.$$ Thus: $$P^{-1} = \begin{bmatrix} 7 & 3 \\ 2 & 1 \end{bmatrix}.$$ Now compute: $$A^4 = P D^4 P^{-1}.$$ ### Step 4: Perform the Matrix Multiplication We compute $$A^4$$ step-by-step. #### Compute $$P D^4$$: $$P D^4 = \begin{bmatrix} 1 & -3 \\ -2 & 7 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{256} \end{bmatrix} = \begin{bmatrix} 1 & -\frac{3}{256} \\ -2 & \frac{7}{256} \end{bmatrix}.$$ #### Compute $$(P D^4) P^{-1}$$: Multiply the result with $$P^{-1}$$: $$\begin{bmatrix} 1 & -\frac{3}{256} \\ -2 & \frac{7}{256} \end{bmatrix} \begin{bmatrix} 7 & 3 \\ 2 & 1 \end{bmatrix}.$$ Perform the multiplication: $$\begin{bmatrix} 1 & -\frac{3}{256} \\ -2 & \frac{7}{256} \end{bmatrix} \begin{bmatrix} 7 & 3 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} 7 + \left(-\frac{3}{256} \cdot 2\right) & 3 + \left(-\frac{3}{256} \cdot 1\right) \\ -14 + \left(\frac{7}{256} \cdot 2\right) & -6 + \left(\frac{7}{256} \cdot 1\right) \end{bmatrix}.$$ Simplify each term: $$\begin{bmatrix} 7 - \frac{6}{256} & 3 - \frac{3}{256} \\ -14 + \frac{14}{256} & -6 + \frac{7}{256} \end{bmatrix}.$$ Combine fractions: $$\begin{bmatrix} \frac{1792 - 6}{256} & \frac{768 - 3}{256} \\ \frac{-3584 + 14}{256} & \frac{-1536 + 7}{256} \end{bmatrix} = \begin{bmatrix} \frac{1786}{256} & \frac{765}{256} \\ \frac{-3570}{256} & \frac{-1529}{256} \end{bmatrix}.$$ ### Final Simplified Answer: $$A^4 = \begin{bmatrix} \frac{1786}{256} & \frac{765}{256} \\ \frac{-3570}{256} & \frac{-1529}{256} \end{bmatrix}.$$ If you want this in decimal form, let me know. --- Would you like further clarification or additional steps? ### Related Questions: 1. How is the formula $$A^n = PD^nP^{-1}$$ derived? 2. What happens if $$D$$ has repeated eigenvalues? 3. Can this method be generalized to higher-dimensional matrices? 4. What are the properties of diagonalizable matrices? 5. Why is diagonalization efficient for computing powers? ### Tip: Diagonalization is useful for efficiently computing matrix functions like powers or exponentials!

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Matrix Powers
Matrix Diagonalization

Formulas

A = PDP⁻¹
A⁴ = PD⁴P⁻¹
P⁻¹ = (1/det(P)) * [[d, -b], [-c, a]] for a 2x2 matrix [[a, b], [c, d]]
Matrix multiplication and powers of diagonal matrices: D⁴ = [[d₁⁴, 0], [0, d₂⁴]]

Theorems

Matrix Diagonalization
Properties of Powers of Matrices

Suitable Grade Level

Undergraduate Linear Algebra

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