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lecture_3_slides
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| lecture_3_slides [2016/01/28 13:34] – rupert | lecture_3_slides [2016/02/01 17:29] (current) – rupert | ||
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| + | ~~REVEAL~~ | ||
| + | |||
| ==== Another look at the last example ==== | ==== Another look at the last example ==== | ||
| * $\begin{array}{ccccccrrr} x& | * $\begin{array}{ccccccrrr} x& | ||
| Line 13: | Line 15: | ||
| ==== ==== | ==== ==== | ||
| * $\begin{array}{ccccccrrr} x&&& | * $\begin{array}{ccccccrrr} x&&& | ||
| - | * Can now easily rearrange (1) to find $x$ in terms of $z$, and we rearrange (2) to find $y$ in terms of $z$. | + | * can easily rearrange (1) to find $x$ in terms of $z$ |
| + | * can easily | ||
| * Since $z$ can take any value, write $z=t$ where $t$ is a "free parameter" | * Since $z$ can take any value, write $z=t$ where $t$ is a "free parameter" | ||
| * (which means $t$ can be any real number, or $t\in \mathbb{R}$). | * (which means $t$ can be any real number, or $t\in \mathbb{R}$). | ||
| Line 30: | Line 33: | ||
| * take $t=1.5$ to see that $(-16+1.5\times 5, | * take $t=1.5$ to see that $(-16+1.5\times 5, | ||
| * etc. | * etc. | ||
| - | * This works for any value $t\in\mathbb{R}$, | + | * This works for any value $t\in\mathbb{R}$, |
| ==== Observations ==== | ==== Observations ==== | ||
| - The operations we applied to the original linear system don't change the set of solutions. This is because each operation is reversible. | - The operations we applied to the original linear system don't change the set of solutions. This is because each operation is reversible. | ||
| - | - Writing out the variables $x,y,z$ each time is unnecessary. If we erase the variables from the system $$\begin{array}{ccccccrrr} x& | + | - Writing out the variables $x,y,z$ each time is unnecessary: |
| + | |||
| + | * erase the variables from the system $$\begin{array}{ccccccrrr} x& | ||
| + | * write all the numbers in a grid, or a **matrix** | ||
| + | * we get $\begin{bmatrix} 1& | ||
| ==== ==== | ==== ==== | ||
| * System of linear equations: $\begin{array}{ccccccrrr} x& | * System of linear equations: $\begin{array}{ccccccrrr} x& | ||
| * $\begin{bmatrix} 1& | * $\begin{bmatrix} 1& | ||
| - | * Each row corresponds to one equation. | + | * Each **row** corresponds to one **equation**. |
| - | * The first column corresponds to the $x$ variable, the second to $y$, the third to $z$ | + | * Each **column** corresponds to one **variable** |
| - | * Numbers in the final column | + | * (except |
| - | * Instead of performing operations on equations, we can perform operations on the rows of this matrix. | + | * Instead of performing operations on equations, |
| ==== ==== | ==== ==== | ||
| Line 61: | Line 68: | ||
| This method always works: | This method always works: | ||
| * take any system of linear equations | * take any system of linear equations | ||
| - | * write down a corresponding matrix | + | * write down a corresponding matrix |
| - | * perform | + | * perform reversible operations on the rows of this matrix to get a " |
| * write down a new system of linear equations with the same solutions as the original system. | * write down a new system of linear equations with the same solutions as the original system. | ||
| - | * If we do things in a sensible way then the new system will be easy to solve, so we' | + | * Hopefully |
| - | * Let's give some terminology which will allow us to explain this clearly. | + | * and the solutions haven' |
| ===== The augmented matrix and elementary operations ===== | ===== The augmented matrix and elementary operations ===== | ||
| Line 86: | Line 94: | ||
| ==== Why do elementary operations leave the solutions of systems unchanged? ==== | ==== Why do elementary operations leave the solutions of systems unchanged? ==== | ||
| - | * we are doing the same thing to the left hand side and the right hand side of each equation, so any solution to the original system will also be a solution to the new system; and | + | * We do the same thing to the left hand side and the right hand side of each equation... |
| - | * these operations are reversible, using operations of the same type, so any solution to the new system will also be a solution to the original system. | + | * so any solution to the original system will also be a solution to the new system. |
| + | * These operations are all reversible | ||
| + | * so any solution to the new system will also be a solution to the original system. | ||
| ==== Elementary row operations on a matrix ==== | ==== Elementary row operations on a matrix ==== | ||
| {{page> | {{page> | ||
| + | |||
| + | ==== Example ==== | ||
| + | |||
| + | Use [[EROs]] to find the intersection of the planes | ||
| + | \begin{align*} 3x+4y+7z& | ||
| + | |||
| + | ==== Solution 1 ==== | ||
| + | |||
| + | \begin{align*} | ||
| + | \def\go# | ||
| + | \def\ar# | ||
| + | & | ||
| + | \ar{\text{reorder rows}}\go{1& | ||
| + | \ar{R3\to R3-3R1}\go{1& | ||
| + | \ar{R3\to R3-4R2}\go{1& | ||
| + | \ar{R3\to \tfrac16 R3}\go{1& | ||
| + | \end{align*} | ||
| + | |||
| + | ==== ==== | ||
| + | |||
| + | $\go{1& | ||
| + | |||
| + | * from the last row, we get $z=-3$ | ||
| + | * from the second row, we get $y-2z=5$ | ||
| + | * so $y-2(-3)=5$ | ||
| + | * so $y=-1$ | ||
| + | * from the first row, we get $x+3z=0$ | ||
| + | * so $x+3(-3)=0$ | ||
| + | * so $x=9$ | ||
| + | * Conclusion: $\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}9\\-1\\-3\end{bmatrix}$ is the only solution. | ||
| + | |||
| + | ==== Solution 2 ==== | ||
| + | |||
| + | We start in the same way, but by performing more [[EROs]] we make the algebra at the end simpler. | ||
| + | |||
| + | \begin{align*} | ||
| + | \def\go# | ||
| + | \def\ar# | ||
| + | & | ||
| + | \ar{\ldots \text{same EROs as above}\ldots}\go{1& | ||
| + | \ar{R2\to R2+2R3}\go{1& | ||
| + | \ar{R1\to R1-3R3}\go{1& | ||
| + | \end{align*} | ||
| + | |||
| + | ==== ==== | ||
| + | \[ \go{1& | ||
| + | |||
| + | * from the last row, we get $z=-3$ | ||
| + | * from the second row, we get $y=-1$ | ||
| + | * from the first row, we get $x=9$ | ||
| + | * So $\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}9\\-1\\-3\end{bmatrix}$ is the only solution. | ||
| + | |||
| + | ==== Discussion ==== | ||
| + | |||
| + | Both solutions use EROs to transform the [[augmented matrix]]. | ||
| + | |||
| + | * Solution 1: $\left[\begin{smallmatrix}1& | ||
| + | * " | ||
| + | * Called **row echelon form** | ||
| + | * Needed algebra to finish solution. | ||
| + | * Solution 2: $\left[\begin{smallmatrix}1& | ||
| + | * Staircase with zeros **above** 1s on steps (and below). | ||
| + | * Called **reduced row echelon form** | ||
| + | * No extra algebra needed to finish solution. | ||
| + | |||
lecture_3_slides.1453988053.txt.gz · Last modified: by rupert