Math Problem Statement

If A=⎡⎢⎣201213110⎤⎥⎦ then find A2−2A+5I.

Solution

To solve the expression A22A+5IA^2 - 2A + 5I, where A=(201213110)A = \begin{pmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & 1 & 0 \end{pmatrix} and II is the identity matrix, we'll follow these steps:

Step 1: Find A2A^2

First, we calculate the square of matrix AA:

A2=A×A=(201213110)×(201213110)A^2 = A \times A = \begin{pmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & 1 & 0 \end{pmatrix} \times \begin{pmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & 1 & 0 \end{pmatrix}

Let's perform the multiplication:

(2 \times 2) + (0 \times 2) + (1 \times 1) & (2 \times 0) + (0 \times 1) + (1 \times 1) & (2 \times 1) + (0 \times 3) + (1 \times 0) \\ (2 \times 2) + (1 \times 2) + (3 \times 1) & (2 \times 0) + (1 \times 1) + (3 \times 1) & (2 \times 1) + (1 \times 3) + (3 \times 0) \\ (1 \times 2) + (1 \times 2) + (0 \times 1) & (1 \times 0) + (1 \times 1) + (0 \times 1) & (1 \times 1) + (1 \times 3) + (0 \times 0) \end{pmatrix}$$ Simplifying each element: $$A^2 = \begin{pmatrix} 4 + 0 + 1 & 0 + 0 + 1 & 2 + 0 + 0 \\ 4 + 2 + 3 & 0 + 1 + 3 & 2 + 3 + 0 \\ 2 + 2 + 0 & 0 + 1 + 0 & 1 + 3 + 0 \end{pmatrix}$$ $$A^2 = \begin{pmatrix} 5 & 1 & 2 \\ 9 & 4 & 5 \\ 4 & 1 & 4 \end{pmatrix}$$ ### Step 2: Calculate $$2A$$ Next, we find $$2A$$: $$2A = 2 \times \begin{pmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & 1 & 0 \end{pmatrix} = \begin{pmatrix} 4 & 0 & 2 \\ 4 & 2 & 6 \\ 2 & 2 & 0 \end{pmatrix}$$ ### Step 3: Calculate $$5I$$ Now, calculate $$5I$$, where $$I$$ is the 3x3 identity matrix: $$I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$ $$5I = 5 \times \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix}$$ ### Step 4: Combine the results Finally, substitute the values into the expression $$A^2 - 2A + 5I$$: $$A^2 - 2A + 5I = \begin{pmatrix} 5 & 1 & 2 \\ 9 & 4 & 5 \\ 4 & 1 & 4 \end{pmatrix} - \begin{pmatrix} 4 & 0 & 2 \\ 4 & 2 & 6 \\ 2 & 2 & 0 \end{pmatrix} + \begin{pmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix}$$ Let's subtract and then add the matrices: $$A^2 - 2A = \begin{pmatrix} 5-4 & 1-0 & 2-2 \\ 9-4 & 4-2 & 5-6 \\ 4-2 & 1-2 & 4-0 \end{pmatrix} = \begin{pmatrix} 1 & 1 & 0 \\ 5 & 2 & -1 \\ 2 & -1 & 4 \end{pmatrix}$$ $$A^2 - 2A + 5I = \begin{pmatrix} 1 & 1 & 0 \\ 5 & 2 & -1 \\ 2 & -1 & 4 \end{pmatrix} + \begin{pmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix} = \begin{pmatrix} 6 & 1 & 0 \\ 5 & 7 & -1 \\ 2 & -1 & 9 \end{pmatrix}$$ Thus, the final result is: $$A^2 - 2A + 5I = \begin{pmatrix} 6 & 1 & 0 \\ 5 & 7 & -1 \\ 2 & -1 & 9 \end{pmatrix}$$ Would you like more details on any of the steps? Here are five related questions: 1. How do you multiply two matrices? 2. What is the identity matrix and its properties? 3. How do you calculate the determinant of a matrix? 4. What are eigenvalues and eigenvectors? 5. How do you find the inverse of a matrix? **Tip:** Remember, the order of multiplication matters in matrix operations; matrix multiplication is not commutative.

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Math Problem Analysis

Mathematical Concepts

Matrix Operations
Identity Matrix

Formulas

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Theorems

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Suitable Grade Level

Undergraduate Level