Question

a.find a basis for the span of the following set of vectors
|1 | |-4 | |1 | |5| |2| | -1| |11|
|3 | |-12| |-4| |8| |11| | -7| | 52 |
|-1| | 4 | |-5| |-17| |-21| |19| | -80|
|2| |-8| |4| |16| |17| | -3| | 82|

b. find the coordinate vector [x]B for the vector(see below) using the basis from a above
|-8|
|-58|
|151|
|-58|

Answer 1

Answer on Question #82533 – Math – Linear Algebra

Question

a. Find a basis for the span of the following set of vectors

|1 | |-4 | |1 | |5 | |2 | |-1 | |11|

|3 | |-12| |-4| |8| |11| |-7| |52|

|-1| |4| -5| -17| -21| 19| -80|

|2| -8| 4| 16| 17| -3| 82|

Solution

a

$$\left(\begin{array}{cccccc} 1 & -4 & 1 & 5 & 2 & -1 & 11 \\ 3 & -12 & -4 & 8 & 11 & -7 & 52 \\ -1 & 4 & -5 & -17 & -21 & 19 & -80 \\ 2 & -8 & 4 & 16 & 17 & -3 & 82 \end{array}\right) \rightarrow \begin{array}{c} r2 = r2 + (-3)r1 \\ r3 = r3 + r1 \\ r4 = r4 + (-2)r1 \end{array}$$ $$\rightarrow \left(\begin{array}{cccccc} 1 & -4 & 1 & 5 & 2 & -1 & 11 \\ 0 & 0 & -7 & -7 & 5 & -4 & 19 \\ 0 & 0 & -4 & -12 & -19 & 18 & -69 \\ 0 & 0 & 2 & 6 & 13 & -1 & 60 \end{array}\right) \rightarrow \begin{array}{c} r1 = r1 + \left(\frac{1}{7}\right) r2 \\ r3 = r3 + \left(\frac{-4}{7}\right) r2 \\ r4 = r4 + \left(\frac{2}{7}\right) r2 \end{array}$$ $$\rightarrow \left(\begin{array}{cccccc} & & & & \frac{19}{7} & -\frac{11}{7} & \frac{96}{7} \\ 1 & -4 & 0 & 4 & 5 & -4 & 19 \\ 0 & 0 & -7 & -7 & \frac{153}{7} & \frac{142}{7} & -\frac{559}{7} \\ 0 & 0 & 0 & 4 & \frac{101}{7} & -\frac{15}{7} & \frac{458}{7} \end{array}\right) \leftarrow \begin{array}{c} r1 = r1 + \left(\frac{1}{2}\right) r3 \\ r2 = r2 + \left(\frac{-7}{8}\right) r3 \\ r4 = r4 + \left(\frac{1}{2}\right) r3 \end{array}$$ \rightarrow \left(\begin{array}{cccccc} & & & & -\frac{115}{14} & \frac{60}{7} & -\frac{367}{14} \\ 1 & -4 & 0 & 0 & \frac{193}{8} & -\frac{87}{4} & \frac{711}{8} \\ 0 & 0 & -7 & 0 & \frac{193}{8} & -\frac{87}{4} & \frac{711}{8} \\ 0 & 0 & 0 & -8 & -\frac{153}{7} & \frac{142}{7} & -\frac{559}{7} \\ 0 & 0 & 0 & 0 & \frac{7}{2} & \frac{8}{2} \end{array}\right) \leftarrow \begin{array}{c} r1 = r1 + \left(\frac{115}{49}\right) r4 \\ r2 = r2 + \left(\frac{-193}{28}\right) r4 \\ r3 = r3 + \left(\frac{306}{49}\right) r4 \end{array} \right) $$\rightarrow \left(\begin{array}{cccccc} & & & & \frac{1340}{49} & \frac{1648}{49} \\ 1 & -4 & 0 & 0 & 0 & \frac{2153}{28} & -\frac{2433}{28} \\ 0 & 0 & -7 & 0 & 0 & \frac{2153}{28} & \frac{3890}{49} \\ 0 & 0 & 0 & 0 & \frac{2}{2} & \frac{3442}{49} & \frac{51}{2} \end{array}\right)$$

The first, the third, the fourth, the fifth columns are linearly independent, hence a basis is:

$$B = \left\{\left(\begin{array}{c} 1 \\ 3 \\ -1 \\ 2 \end{array}\right), \left(\begin{array}{c} 1 \\ -4 \\ -5 \\ 4 \end{array}\right), \left(\begin{array}{c} 5 \\ 8 \\ -17 \\ 16 \end{array}\right), \left(\begin{array}{c} 2 \\ 11 \\ -21 \\ 17 \end{array}\right)\right\}$$

Answer a.: basis $\left\{\left( \begin{array}{l}1\\ 3\\ -1\\ 2 \end{array} \right),\left( \begin{array}{l}1\\ -4\\ -5\\ 4 \end{array} \right),\left( \begin{array}{l}5\\ 8\\ -17\\ 16 \end{array} \right),\left( \begin{array}{l}2\\ 11\\ -21\\ 17 \end{array} \right)\right\}.$

Question

b. find the coordinate vector $[x]B$ for the vector (see below) using the basis from a above

|-8|

|-58|

|151|

|-58|

Solution

b

$$x = P _ {B} \left[ x \right] _ {B} \rightarrow \left[ x \right] _ {B} = P _ {B} ^ {- 1} x$$ $$x = \left( \begin{array}{c} - 8 \\ - 5 8 \\ 1 5 1 \\ - 5 8 \end{array} \right), B = \left\{\left( \begin{array}{c} 1 \\ 3 \\ - 1 \\ 2 \end{array} \right), \left( \begin{array}{c} 1 \\ - 4 \\ - 5 \\ 4 \end{array} \right), \left( \begin{array}{c} 5 \\ 8 \\ - 1 7 \\ 1 6 \end{array} \right), \left( \begin{array}{c} 2 \\ 1 1 \\ - 2 1 \\ 1 7 \end{array} \right) \right\} \to P _ {B} = \left( \begin{array}{c c c c} 1 & 1 & 5 & 2 \\ 3 & - 4 & 8 & 1 1 \\ - 1 - 5 & - 1 7 & - 2 1 \\ 2 & 4 & 1 6 & 1 7 \end{array} \right)$$ $$\begin{array}{l} \det P _ {B} = \left| \begin{array}{c c c c} 1 & 1 & 5 & 2 \\ 3 & - 4 & 8 & 1 1 \\ - 1 & - 5 & - 1 7 & - 2 1 \\ 2 & 4 & 1 6 & 1 7 \end{array} \right| = \left\{ \begin{array}{l} r 2 = r 2 + (- 3) r 1 \\ r 3 = r 3 + r 1 \\ r 4 = r 4 + (- 2) r 1 \end{array} \right\} = \left| \begin{array}{c c c c} 1 & 1 & 5 & 2 \\ 0 & - 7 & - 7 & 5 \\ 0 & - 4 & - 1 2 & - 1 9 \\ 0 & 2 & 6 & 1 3 \end{array} \right| = \left\{ \begin{array}{l} r 3 = r 3 + \left(\frac {- 4}{7}\right) r 2 \\ r 4 = r 4 + \left(\frac {2}{7}\right) r 2 \end{array} \right. \\ = \left| \begin{array}{c c c c} 1 & 1 & 5 & 2 \\ 0 & - 7 & - 7 & 5 \\ 0 & 0 & - 8 & - 1 5 3 / 7 \\ 0 & 0 & 4 & 1 0 1 / 7 \end{array} \right| = \left\{r 4 = r 4 + \left(\frac {1}{2}\right) r 3 \right\} = \left| \begin{array}{c c c c} 1 & 1 & 5 & 2 \\ 0 & - 7 & - 7 & 5 \\ 0 & 0 & - 8 & - 1 5 3 / 7 \\ 0 & 0 & 4 & 7 / 2 \end{array} \right| \\ = \frac {1 \cdot (- 7) \cdot (- 8) \cdot 7}{2} = 1 9 6 \\ \end{array}$$ $$\begin{array}{l} P _ {1 1} = \left| \begin{array}{c c c} - 4 & 8 & 1 1 \\ - 5 & - 1 7 & - 2 1 \\ 4 & 1 6 & 1 7 \end{array} \right| \\ = (- 4) \cdot (- 1 7) \cdot 1 7 + 8 \cdot (- 2 1) \cdot 4 + 1 1 \cdot (- 5) \cdot 1 6 - 1 1 \cdot (- 1 7) \cdot 4 - (- 4) \\ \cdot (- 2 1) \cdot 1 6 - 8 \cdot (- 5) \cdot 1 7 = 1 1 5 6 - 6 7 2 - 8 8 0 + 7 4 8 - 1 3 4 4 + 6 8 0 = - 3 1 2 \\ \end{array}$$ $$P_{12} = \left| \begin{array}{ccc} 3 & 8 & 11 \\ -1 & -17 & -21 \\ 2 & 16 & 17 \end{array} \right| = 3 \cdot (-17) \cdot 17 + 8 \cdot (-21) \cdot 2 + 11 \cdot (-1) \cdot 16 - 11 \cdot (-17) \cdot 2 - 3 \cdot (-21) \cdot 16 - 8 \cdot (-1) \cdot 17 = -867 - 336 - 176 + 374 + 1008 + 136 = 139$$ $$P_{13} = \left| \begin{array}{ccc} 3 & -4 & 11 \\ -1 & -5 & -21 \\ 2 & 4 & 17 \end{array} \right| = 3 \cdot (-5) \cdot 17 + (-4) \cdot (-21) \cdot 2 + 11 \cdot (-1) \cdot 4 - 11 \cdot (-5) \cdot 2 - 3 \cdot (-21) \cdot 4 - (-4) \cdot (-1) \cdot 17 = -255 + 168 - 44 + 110 + 252 - 68 = 163$$ $$P_{14} = \left| \begin{array}{ccc} 3 & -4 & 8 \\ -1 & -5 & -17 \\ 2 & 4 & 16 \end{array} \right| = 3 \cdot (-5) \cdot 16 + (-4) \cdot (-17) \cdot 2 + 8 \cdot (-1) \cdot 4 - 8 \cdot (-5) \cdot 2 - 3 \cdot (-17) \cdot 4 - (-4) \cdot (-1) \cdot 16 = -240 + 136 - 32 + 80 + 204 - 64 = 84$$ $$P_{21} = \left| \begin{array}{ccc} 1 & 5 & 2 \\ -5 & -17 & -21 \\ 4 & 16 & 17 \end{array} \right| = 1 \cdot (-17) \cdot 17 + 5 \cdot (-21) \cdot 4 + 2 \cdot (-5) \cdot 16 - 2 \cdot (-17) \cdot 4 - 1 \cdot (-21) \cdot 16 - 5 \cdot (-5) \cdot 17 = -289 - 420 - 160 + 136 + 336 + 425 = 28$$ $$P_{22} = \left| \begin{array}{ccc} 1 & 5 & 2 \\ -1 & -17 & -21 \\ 2 & 16 & 17 \end{array} \right| = 1 \cdot (-17) \cdot 17 + 5 \cdot (-21) \cdot 2 + 2 \cdot (-1) \cdot 16 - 2 \cdot (-17) \cdot 2 - 1 \cdot (-21) \cdot 16 - 5 \cdot (-1) \cdot 17 = -289 - 210 - 32 + 68 + 336 + 85 = -42$$ $$P_{23} = \left| \begin{array}{ccc} 1 & 1 & 2 \\ -1 & -5 & -21 \\ 2 & 4 & 17 \end{array} \right| = 1 \cdot (-5) \cdot 17 + 1 \cdot (-21) \cdot 2 + 2 \cdot (-1) \cdot 4 - 2 \cdot (-5) \cdot 2 - 1 \cdot (-21) \cdot 4 - 1 \cdot (-1) \cdot 17 = -85 - 42 - 8 + 20 + 84 + 17 = -14$$ $$P_{24} = \left| \begin{array}{ccc} 1 & 1 & 5 \\ -1 & -5 & -17 \\ 2 & 4 & 16 \end{array} \right| = 1 \cdot (-5) \cdot 16 + 1 \cdot (-17) \cdot 2 + 5 \cdot (-1) \cdot 4 - 5 \cdot (-5) \cdot 2 - 1 \cdot (-17) \cdot 4 - 1 \cdot (-1) \cdot 16 = -80 - 34 - 20 + 50 + 68 + 16 = 0$$ $$P_{31} = \left| \begin{array}{ccc} 1 & 5 & 2 \\ -4 & 8 & 11 \\ 4 & 16 & 17 \end{array} \right| = 1 \cdot 8 \cdot 17 + 5 \cdot 11 \cdot 4 + 2 \cdot (-4) \cdot 16 - 2 \cdot 8 \cdot 4 - 1 \cdot 11 \cdot 16 - 5 \cdot (-4) \cdot 17 = 136 + 220 - 128 - 64 - 176 + 340 = 328$$ $$P_{32} = \left| \begin{array}{ccc} 1 & 5 & 2 \\ 3 & 8 & 11 \\ 2 & 16 & 17 \end{array} \right| = 1 \cdot 8 \cdot 17 + 5 \cdot 11 \cdot 2 + 2 \cdot 3 \cdot 16 - 2 \cdot 8 \cdot 2 - 1 \cdot 11 \cdot 16 - 5 \cdot 3 \cdot 17$$ $$= 136 + 110 + 96 - 32 - 176 - 255 = -121$$ $$P_{33} = \left| \begin{array}{ccc} 1 & 1 & 2 \\ 3 & -4 & 11 \\ 2 & 4 & 17 \end{array} \right| = 1 \cdot (-4) \cdot 17 + 1 \cdot 11 \cdot 2 + 2 \cdot 3 \cdot 4 - 2 \cdot (-4) \cdot 2 - 1 \cdot 11 \cdot 4 - 1 \cdot 3 \cdot 17$$ $$= -68 + 22 + 24 + 16 - 44 - 51 = -101$$ $$P_{34} = \left| \begin{array}{ccc} 1 & 1 & 5 \\ 3 & -4 & 8 \\ 2 & 4 & 16 \end{array} \right| = 1 \cdot (-4) \cdot 16 + 1 \cdot 8 \cdot 2 + 5 \cdot 3 \cdot 4 - 5 \cdot (-4) \cdot 2 - 1 \cdot 8 \cdot 4 - 1 \cdot 3 \cdot 16$$ $$= -64 + 16 + 60 + 40 - 32 - 48 = -28$$ $$P_{41} = \left| \begin{array}{ccc} 1 & 5 & 2 \\ -4 & 8 & 11 \\ -5 & -17 & -21 \end{array} \right|$$ $$= 1 \cdot 8 \cdot (-21) + 5 \cdot 11 \cdot (-5) + 2 \cdot (-4) \cdot (-17) - 2 \cdot 8 \cdot (-5) - 1 \cdot 11 \cdot (-17)$$ $$-5 \cdot (-4) \cdot (-21) = -168 - 275 + 136 + 80 + 187 - 420 = -460$$ $$P_{42} = \left| \begin{array}{ccc} 1 & 5 & 2 \\ 3 & 8 & 11 \\ -1 & -17 & -21 \end{array} \right|$$ $$= 1 \cdot 8 \cdot (-21) + 5 \cdot 11 \cdot (-1) + 2 \cdot 3 \cdot (-17) - 2 \cdot 8 \cdot (-1) - 1 \cdot 11 \cdot (-17) - 5$$ $$\cdot 3 \cdot (-21) = -168 - 55 - 102 + 16 + 187 + 315 = 193$$ $$P_{43} = \left| \begin{array}{ccc} 1 & 1 & 2 \\ 3 & -4 & 11 \\ -1 & -5 & -21 \end{array} \right|$$ $$= 1 \cdot (-4) \cdot (-21) + 1 \cdot 11 \cdot (-1) + 2 \cdot 3 \cdot (-5) - 2 \cdot (-4) \cdot (-1) - 1 \cdot 11 \cdot (-5)$$ $$-1 \cdot 3 \cdot (-21) = 84 - 11 - 30 - 8 + 55 + 63 = 153$$ $$P_{44} = \left| \begin{array}{ccc} 1 & 1 & 5 \\ 3 & -4 & 8 \\ -1 & -5 & -17 \end{array} \right|$$ $$= 1 \cdot (-4) \cdot (-17) + 1 \cdot 8 \cdot (-1) + 5 \cdot 3 \cdot (-5) - 5 \cdot (-4) \cdot (-1) - 1 \cdot 8 \cdot (-5)$$ $$-1 \cdot 3 \cdot (-17) = 68 - 8 - 75 - 20 + 40 + 51 = 56$$ $$P^{*} = \left( \begin{array}{rrr} P_{11} & -P_{12} & P_{13} & -P_{14} \\ -P_{21} & P_{22} & -P_{23} & P_{24} \\ P_{31} & -P_{32} & P_{33} & -P_{34} \\ -P_{41} & P_{42} & -P_{43} & P_{44} \end{array} \right) = \left( \begin{array}{rrr} -312 - 139 & 163 & -84 \\ -28 & -42 & 14 & 0 \\ 328 & 121 & -101 & 28 \\ 460 & 193 & -153 & 56 \end{array} \right)$$ $$P^{*T} = \left( \begin{array}{rrr} -312 - 28 & 328 & 460 \\ -139 - 42 & 121 & 193 \\ 163 & 14 & -101 - 153 \\ -84 & 0 & 28 & 56 \end{array} \right)$$ $$P _ {B} ^ {- 1} = \frac {P ^ {* T}}{d e t P _ {B}} = \left( \begin{array}{c c c c} - 7 8 / 4 9 & - 1 / 7 & 8 2 / 4 9 & 1 1 5 / 4 9 \\ - 1 3 9 / 1 9 6 & - 3 / 1 4 & 1 2 1 / 1 9 6 & 1 9 3 / 1 9 6 \\ 1 6 3 / 1 9 6 & 1 / 1 4 & - 1 0 1 / 1 9 6 & - 1 5 3 / 1 9 6 \\ - 3 / 7 & 0 & 1 / 7 & 2 / 7 \end{array} \right)$$ $$[ x ] _ {B} = P _ {B} ^ {- 1} x = - 8 \left( \begin{array}{c} - \frac {7 8}{4 9} \\ - \frac {1 3 9}{1 9 6} \\ \frac {1 6 3}{1 9 6} \\ - \frac {3}{7} \end{array} \right) - 5 8 \left( \begin{array}{c} - \frac {1}{7} \\ - \frac {3}{1 4} \\ \frac {1}{1 4} \\ 0 \end{array} \right) + 1 5 1 \left( \begin{array}{c} \frac {8 2}{4 9} \\ \frac {1 2 1}{1 9 6} \\ - \frac {1 0 1}{1 9 6} \\ \frac {1}{7} \end{array} \right) - 5 8 \left( \begin{array}{c} \frac {1 1 5}{4 9} \\ \frac {1 9 3}{1 9 6} \\ - \frac {1 5 3}{1 9 6} \\ \frac {2}{7} \end{array} \right) = \left( \begin{array}{c} \frac {6 7 4 2}{4 9} \\ \frac {1 0 6 2 5}{1 9 6} \\ - \frac {8 4 9 3}{1 9 6} \\ \frac {5 9}{7} \end{array} \right).$$

Answer: $[x]_B = \left( \begin{array}{c} \frac{6742}{49} \\ \frac{10625}{196} \\ -\frac{8493}{196} \\ \frac{59}{7} \end{array} \right).$

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