Math Problem Statement

Suppose A = [[-1, 2], [-4, 7]] and A^2 - αA + I = 0 for some α ∈ ℝ. Find the value of α.

Solution

We are tasked with solving for α\alpha in the given matrix equation:

A2−αA+I=0,A^2 - \alpha A + I = 0, where A=[−12−47]A = \begin{bmatrix} -1 & 2 \\ -4 & 7 \end{bmatrix}, and II is the 2×22 \times 2 identity matrix. Let’s work through this step by step.

Step 1: Compute A2A^2

We calculate A2=A⋅AA^2 = A \cdot A: A=[−12−47].A = \begin{bmatrix} -1 & 2 \\ -4 & 7 \end{bmatrix}. A2=[−12−47]⋅[−12−47].A^2 = \begin{bmatrix} -1 & 2 \\ -4 & 7 \end{bmatrix} \cdot \begin{bmatrix} -1 & 2 \\ -4 & 7 \end{bmatrix}.

Using matrix multiplication:

(-1)(-1) + (2)(-4) & (-1)(2) + (2)(7) \\ (-4)(-1) + (7)(-4) & (-4)(2) + (7)(7) \end{bmatrix} = \begin{bmatrix} 1 - 8 & -2 + 14 \\ 4 - 28 & -8 + 49 \end{bmatrix} = \begin{bmatrix} -7 & 12 \\ -24 & 41 \end{bmatrix}.$$ --- ### Step 2: Substitute into the equation The given equation is: $$A^2 - \alpha A + I = 0.$$ Substituting $$A^2$$, $$A$$, and $$I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$, we have: $$\begin{bmatrix} -7 & 12 \\ -24 & 41 \end{bmatrix} - \alpha \begin{bmatrix} -1 & 2 \\ -4 & 7 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}.$$ Simplify: $$\begin{bmatrix} -7 & 12 \\ -24 & 41 \end{bmatrix} - \alpha \begin{bmatrix} -1 & 2 \\ -4 & 7 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = 0.$$ --- ### Step 3: Expand and equate entries Combining terms: $$\begin{bmatrix} -7 + 1 & 12 + 0 \\ -24 + 0 & 41 + 1 \end{bmatrix} - \alpha \begin{bmatrix} -1 & 2 \\ -4 & 7 \end{bmatrix} = \begin{bmatrix} -6 & 12 \\ -24 & 42 \end{bmatrix} - \alpha \begin{bmatrix} -1 & 2 \\ -4 & 7 \end{bmatrix}.$$ Now: $$\begin{bmatrix} -6 + \alpha & 12 - 2\alpha \\ -24 + 4\alpha & 42 - 7\alpha \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}.$$ This gives four scalar equations: 1. $$-6 + \alpha = 0$$, 2. $$12 - 2\alpha = 0$$, 3. $$-24 + 4\alpha = 0$$, 4. $$42 - 7\alpha = 0$$. --- ### Step 4: Solve for $$\alpha$$ From the first equation: $$\alpha = 6.$$ Verify with other equations: - Second equation: $$12 - 2(6) = 0 \quad \text{(true)}$$, - Third equation: $$-24 + 4(6) = 0 \quad \text{(true)}$$, - Fourth equation: $$42 - 7(6) = 0 \quad \text{(true)}$$. --- ### Final Answer: $$\boxed{\alpha = 6}$$ Let me know if you want further details! --- ### 5 Related Questions: 1. How is the Cayley-Hamilton theorem related to this problem? 2. What are eigenvalues and eigenvectors of matrix $$A$$? 3. How can this problem be solved using determinants? 4. How does matrix addition/subtraction work element-wise? 5. What is the general form of matrix equations involving powers of $$A$$? ### Tip: Always verify your solution by substituting it back into the original equation to ensure correctness.

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Math Problem Analysis

Mathematical Concepts

Matrix Multiplication
Matrix Equations
Algebraic Substitution

Formulas

Matrix Multiplication Rule
Matrix Identity Property

Theorems

Properties of Matrices
Cayley-Hamilton Theorem

Suitable Grade Level

Undergraduate (Linear Algebra Level)

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