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Math  /  Algebra

QuestionLet M=[1010−5−5]M=\left[\begin{array}{cc} 10 & 10 \\ -5 & -5 \end{array}\right]
Find formulas for the entries of MnM^{n}, where nn is a positive integer. Submit answer Next item

Studdy Solution

STEP 1

1. M M is a 2×2 2 \times 2 matrix.
2. We are tasked with finding a general formula for the entries of Mn M^n .
3. Matrix exponentiation involves repeated multiplication of the matrix by itself.
4. The matrix M M can potentially be simplified using eigenvalues and eigenvectors or diagonalization.

STEP 2

1. Analyze the structure of the matrix M M .
2. Determine if M M can be diagonalized.
3. Use diagonalization to find Mn M^n .
4. Derive formulas for the entries of Mn M^n .

STEP 3

Analyze the structure of the matrix M M :
M=[1010−5−5] M = \begin{bmatrix} 10 & 10 \\ -5 & -5 \end{bmatrix}
Notice that the rows of M M are linearly dependent, which implies that M M is not invertible and has a determinant of zero.

STEP 4

Determine if M M can be diagonalized:
1. Calculate the characteristic polynomial of M M :
det(M−λI)=∣10−λ10−5−5−λ∣ \text{det}(M - \lambda I) = \begin{vmatrix} 10 - \lambda & 10 \\ -5 & -5 - \lambda \end{vmatrix}
=(10−λ)(−5−λ)−(10)(−5) = (10 - \lambda)(-5 - \lambda) - (10)(-5)
=λ2−5λ=λ(λ−5) = \lambda^2 - 5\lambda = \lambda(\lambda - 5)
2. The eigenvalues are λ1=0 \lambda_1 = 0 and λ2=5 \lambda_2 = 5 .
3. Find the eigenvectors for each eigenvalue.

STEP 5

Find eigenvectors for each eigenvalue:
1. For λ1=0 \lambda_1 = 0 :
Mv=0v⇒[1010−5−5][xy]=[00] Mv = 0v \Rightarrow \begin{bmatrix} 10 & 10 \\ -5 & -5 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}
10x+10y=0⇒x=−y 10x + 10y = 0 \Rightarrow x = -y
Eigenvector: v1=[1−1] v_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}
2. For λ2=5 \lambda_2 = 5 :
Mv=5v⇒[1010−5−5][xy]=5[xy] Mv = 5v \Rightarrow \begin{bmatrix} 10 & 10 \\ -5 & -5 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = 5 \begin{bmatrix} x \\ y \end{bmatrix}
10x+10y=5x⇒5x+10y=0⇒x=−2y 10x + 10y = 5x \Rightarrow 5x + 10y = 0 \Rightarrow x = -2y
Eigenvector: v2=[2−1] v_2 = \begin{bmatrix} 2 \\ -1 \end{bmatrix}

STEP 6

Use diagonalization to find Mn M^n :
1. Form the matrix P P using the eigenvectors:
P=[12−1−1] P = \begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix}
2. Form the diagonal matrix D D using the eigenvalues:
D=[0005] D = \begin{bmatrix} 0 & 0 \\ 0 & 5 \end{bmatrix}
3. M=PDP−1 M = PDP^{-1}
4. Calculate Mn=PDnP−1 M^n = PD^nP^{-1}
Dn=[0005n] D^n = \begin{bmatrix} 0 & 0 \\ 0 & 5^n \end{bmatrix}
5. Calculate P−1 P^{-1} and multiply to find Mn M^n .

STEP 7

Calculate P−1 P^{-1} and find Mn M^n :
1. Calculate P−1 P^{-1} :
P−1=1(−1)(1)−(−1)(2)[−1−211]=[12−1−1] P^{-1} = \frac{1}{(-1)(1) - (-1)(2)} \begin{bmatrix} -1 & -2 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix}
2. Multiply to find Mn M^n :
Mn=PDnP−1=[12−1−1][0005n][12−1−1] M^n = P D^n P^{-1} = \begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 0 & 5^n \end{bmatrix} \begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix}
3. Calculate the product:
Mn=[1⋅0+2⋅01⋅0+2⋅5n−1⋅0−1⋅0−1⋅0−1⋅5n][12−1−1] M^n = \begin{bmatrix} 1 \cdot 0 + 2 \cdot 0 & 1 \cdot 0 + 2 \cdot 5^n \\ -1 \cdot 0 - 1 \cdot 0 & -1 \cdot 0 - 1 \cdot 5^n \end{bmatrix} \begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix}
=[02⋅5n0−5n][12−1−1] = \begin{bmatrix} 0 & 2 \cdot 5^n \\ 0 & -5^n \end{bmatrix} \begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix}
=[0⋅1+2⋅5n⋅(−1)0⋅2+2⋅5n⋅(−1)0⋅1−5n⋅(−1)0⋅2−5n⋅(−1)] = \begin{bmatrix} 0 \cdot 1 + 2 \cdot 5^n \cdot (-1) & 0 \cdot 2 + 2 \cdot 5^n \cdot (-1) \\ 0 \cdot 1 - 5^n \cdot (-1) & 0 \cdot 2 - 5^n \cdot (-1) \end{bmatrix}
=[−2⋅5n−2⋅5n5n5n] = \begin{bmatrix} -2 \cdot 5^n & -2 \cdot 5^n \\ 5^n & 5^n \end{bmatrix}

STEP 8

Derive formulas for the entries of Mn M^n :
The entries of Mn M^n are given by:
Mn=[−2⋅5n−2⋅5n5n5n] M^n = \begin{bmatrix} -2 \cdot 5^n & -2 \cdot 5^n \\ 5^n & 5^n \end{bmatrix}
Thus, the formulas for the entries of Mn M^n are:
- Top left: −2⋅5n -2 \cdot 5^n - Top right: −2⋅5n -2 \cdot 5^n - Bottom left: 5n 5^n - Bottom right: 5n 5^n

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