## 33A practice midterm #2 |
davidbendebury |
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First row reduce the given matrix. This gets you to: \left( \begin{array}{cccc} 1 & 0 & 0 & -7/6 \\ 0 & 1 & 0 & 17/3 \\ 0 & 0 & 1 & -1/2 \end{array} \right) Since this matrix is of rank 3, the image of this transformation has a dimension of 3. By the rank nullity theorem, the dimension of the kernel is 4 - 3 = 1 (we subtract the dimension of the image from the dimension of the transformation). For the final part, we have the equation: Ax=0 If we say that the row reduced matrix is A. We can say that x is equivalent to: \left( \begin{array}{c} x1 \\ x2 \\ x3 \\ x4 \end{array} \right) So we simply carry out the multiplication and set each equation equal to 0. We get: x1+x4*-7/6=0 x2+x4*17/3=0 x3-1/2*x4=0 We can express all four variables in terms of, for example, x3. \left( \begin{array}{c} 7/3*x3 \\ -34/3*x3 \\ x3 \\ 2*x4 \end{array} \right) |

davidbendebury: May 4, 2015, 7:56 p.m. |
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Looks like the formatting on this website is broken, but hopefully this is helpful to somebody. |

joseph_koh: May 4, 2015, 8:54 p.m. |
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Since x4 was common in each equation for each variable, why did you instead choose to use x3? I feel that x4 would have been the easiest in this situation. In terms of x4, I ended up with (7/6*x4, -17/3*x4, 1/2*x4, x4). |

aleksanderjanczewski: May 5, 2015, 1:04 a.m. |
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@davidbendebury: In case you decided to use x3 as a free variable you should have expressed x4 in terms of x3. So the final solutions should be {x1,x2,x3,x4}={7/3∗x3,−34/3∗x3,x3,2∗x3}, and then you may emphasize the free variable by introducing parameter, say, x3=t, which gives {7/3∗t,−34/3∗t,t,2∗t} |

davidbendebury: May 5, 2015, 9:59 p.m. |
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Sure, the variable you choose to use to express the final answer in is arbitrary. Any one of the four choices is correct. |