Difference between revisions of "031 Review Part 3, Problem 5"
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\end{bmatrix}^k</math> by diagonalizing the matrix. | \end{bmatrix}^k</math> by diagonalizing the matrix. | ||
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::<math>(\lambda+2)(\lambda+3)=0,</math> | ::<math>(\lambda+2)(\lambda+3)=0,</math> | ||
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| − | |we find that the eigenvalues of <math style="vertical-align: 0px">A</math> are <math style="vertical-align: 0px">-2</math> and <math style="vertical-align: 0px">3.</math> | + | |we find that the eigenvalues of <math style="vertical-align: 0px">A</math> are <math style="vertical-align: 0px">-2</math> and <math style="vertical-align: 0px">-3.</math> |
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| − | [[031_Review_Part_3|'''<u>Return to | + | [[031_Review_Part_3|'''<u>Return to Review Problems</u>''']] |
Latest revision as of 12:59, 15 October 2017
Find a formula for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{bmatrix} 1 & -6 \\ 2 & -6 \end{bmatrix}^k} by diagonalizing the matrix.
| Foundations: |
|---|
| Recall: |
| 1. To diagonalize a matrix, you need to know the eigenvalues of the matrix. |
| 2. Diagonalization Theorem |
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Solution:
| Step 1: |
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| To diagonalize this matrix, we need to find the eigenvalues and a basis for each eigenspace. |
| First, we find the eigenvalues of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A} by solving Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \text{det }(A-\lambda I)=0.} |
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\text{det }(A-\lambda I)} & = & \displaystyle{\text{det }\bigg(\begin{bmatrix} 1 & -6 \\ 2 & -6 \end{bmatrix}-\begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix}\bigg)}\\ &&\\ & = & \displaystyle{\text{det }\bigg(\begin{bmatrix} 1-\lambda & -6 \\ 2 & -6-\lambda \end{bmatrix}\bigg)}\\ &&\\ & = & \displaystyle{(1-\lambda)(-6\lambda)+12}\\ &&\\ & = & \displaystyle{\lambda^2+5\lambda+6}\\ &&\\ & = & \displaystyle{(\lambda+2)(\lambda+3).}\\ \end{array}} |
| Therefore, setting |
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| we find that the eigenvalues of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A} are Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -2} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -3.} |
| Step 2: |
|---|
| Now, we find a basis for each eigenspace by solving Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (A-\lambda I)\vec{x}=\vec{0}} for each eigenvalue Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda.} |
| For the eigenvalue Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda=-2,} we have |
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Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{A+2I} & = & \displaystyle{\begin{bmatrix} 1 & -6 \\ 2 & -6 \end{bmatrix}+\begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}}\\ &&\\ & = & \displaystyle{\begin{bmatrix} 3 & -6 \\ 2 & -4 \end{bmatrix}}\\ &&\\ & \sim & \displaystyle{\begin{bmatrix} 1 & -2 \\ 0 & 0 \end{bmatrix}.} \end{array}} |
| We see that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_2} is a free variable. So, a basis for the eigenspace corresponding to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -2} is |
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| Step 3: |
|---|
| For the eigenvalue Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda=-3,} we have |
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Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{A+3I} & = & \displaystyle{\begin{bmatrix} 1 & -6 \\ 2 & -6 \end{bmatrix}-\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}}\\ &&\\ & = & \displaystyle{\begin{bmatrix} 4 & -6 \\ 2 & -3 \end{bmatrix}}\\ &&\\ & \sim & \displaystyle{\begin{bmatrix} 1 & \frac{-3}{2} \\ 0 & 0 \end{bmatrix}.} \end{array}} |
| We see that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_2} is a free variable. So, a basis for the eigenspace corresponding to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -3} is |
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| Step 4: |
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| To diagonalize our matrix, we use the information from the steps above. |
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Using the Diagonalization Theorem, we have Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A=PDP^{-1}} where |
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| Step 5: |
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| Notice that |
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Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{A^k} & = & \displaystyle{(PDP^{-1})^k}\\ &&\\ & = & \displaystyle{PD^kP^{-1}}\\ &&\\ & = & \displaystyle{\begin{bmatrix} 2 & \frac{3}{2} \\ 1 & 1 \end{bmatrix}\bigg(\begin{bmatrix} -2 & 0 \\ 0 & -3 \end{bmatrix}\bigg)^k (-2)\begin{bmatrix} 1 & -\frac{3}{2} \\ -1 & 2 \end{bmatrix}}\\ &&\\ & = & \displaystyle{\begin{bmatrix} 2 & \frac{3}{2} \\ 1 & 1 \end{bmatrix}\begin{bmatrix} (-2)^k & 0 \\ 0 & (-3)^k \end{bmatrix} \begin{bmatrix} -2 & 3 \\ 2 & -4 \end{bmatrix}}\\ &&\\ & = & \displaystyle{\begin{bmatrix} 2 & \frac{3}{2} \\ 1 & 1 \end{bmatrix}\begin{bmatrix} (-2)^{k+1} & 3(-2)^k \\ 2(-3)^k & (-4)(-3)^k \end{bmatrix}} \\ &&\\ & = & \displaystyle{\begin{bmatrix} 2(-2)^{k+1}+3(-3)^k & 6(-2)^k+-6(-3)^k \\ (-2)^{k+1}+2(-3)^k & 3(-2)^k+(-4)(-3)^k \end{bmatrix}.} \\ \end{array}} |
| Final Answer: | |
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| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{bmatrix} 2(-2)^{k+1}+3(-3)^k & 6(-2)^k+-6(-3)^k \\ (-2)^{k+1}+2(-3)^k & 3(-2)^k+(-4)(-3)^k \end{bmatrix}} |