Difference between revisions of "Selftest: Matrix multiplication with a scalar"

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<quiz display=simple>
 
<quiz display=simple>
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{'''Which ...?'''
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{
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[[File:Vektorrechnung_Aufgabe6.1.png|200px|left]]
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:<math>
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      <div style="float:left;">
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\left[\begin{array}{ccc}
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      <br style="clear:both;" />
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1&2&3\\
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0&2&1\\
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2&3&0
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\end{array}\right]
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\cdot c =
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\left[\begin{array}{ccc}
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2&4&6\\
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0&4&2\\
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3&6&0
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\end{array}\right]
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</math><br/>
 +
'''Is there any scalar constant <math>c</math>, so that the equation holds?'''
 
| typ="()" }
 
| typ="()" }
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+ <math>\begin{pmatrix} 0 \\ 6 \end{pmatrix}</math>
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- <math>c=0.5</math>
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- <math>\begin{pmatrix} 2 \\ 3 \end{pmatrix}</math>
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- <math>c=1</math>
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- <math>\begin{pmatrix} 5 \\ 0 \end{pmatrix}</math>
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- <math>c=2</math>
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- <math>\begin{pmatrix} -2 \\ 3 \end{pmatrix}</math>
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+ There is no <math>c</math>
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||The x-components of the two vectors cancel each other. Therefore the sum vector only has an y-component unequal zero. The length is the sum of the two y-components. Further information: see [[Simple arithmetic operations]]
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||<math>c=2</math> would be correct for all the components except the lower left one. Here <math>c=1.5</math> would be right.<br/>
 +
:<math>
 +
\left[\begin{array}{ccc}
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2&4&6\\
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0&4&2\\
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{\color{Red}3}&6&0
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\end{array}\right]
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</math>
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</div>
 
</div>
  

Revision as of 16:02, 18 June 2014

← Previous exercise: Minors and cofactors Exercises for chapter Matrices | Article: Multiplication with a scalar Next exercise: Transpose →

1. :<math> \left[\begin{array}{ccc

c=0.5
c=1
c=2
There is no c
→c=2 would be correct for all the components except the lower left one. Here c=1.5 would be right.

2. Which of the following vectors forms the sum of \vec{\mathbf{a}} and \vec{\mathbf{b}}?

Vektorrechnung Aufgabe6.2.png


\begin{pmatrix} -5 \\ 6 \end{pmatrix}
\begin{pmatrix} -2 \\ 0 \end{pmatrix}
\begin{pmatrix} 0 \\ -3 \end{pmatrix}
\begin{pmatrix} -6 \\ 5 \end{pmatrix}
→Vector \vec{\mathbf{a}} only has an x-component, vector \vec{\mathbf{b}} in contrast only has a y-component. The resulting vector consists of the x-component of \vec{\mathbf{a}} and the y-component of \vec{\mathbf{b}}. Further information: see Simple arithmetic operations

3. Which of the following vectors forms the sum of \vec{\mathbf{a}} and \vec{\mathbf{b}}?

Vektorrechnung Aufgabe6.3.png


\begin{pmatrix} 0 \\ 6 \end{pmatrix}
\begin{pmatrix} 2 \\ 3 \end{pmatrix}
\begin{pmatrix} 3 \\ 4 \end{pmatrix}
\begin{pmatrix} -4 \\ 3 \end{pmatrix}
→The x-component of vector \vec{\mathbf{a}} is directed opposite to the x-component of vector \vec{\mathbf{b}}. So the x-components are substracted. The y-components are added as usual. Further information: see Simple arithmetic operations

4. Which of the following vectors forms the substraction \vec{\mathbf{a}}-\vec{\mathbf{b}}?

Vektorrechnung Aufgabe7.1.png


\begin{pmatrix} 0 \\ -6 \end{pmatrix}
\begin{pmatrix} 7 \\ -3 \end{pmatrix}
\begin{pmatrix} 0 \\ 6 \end{pmatrix}
\begin{pmatrix} -4,5 \\ 2 \end{pmatrix}
→Because the x-components are equal, the resulting x-component is zero. The y-component of the resulting vector is the substraction of the y-components. Further information: see Simple arithmetic operations

5. Which of the following vectors forms the substraction \vec{\mathbf{a}}-\vec{\mathbf{b}}?

Vektorrechnung Aufgabe7.2.png


\begin{pmatrix} 3 \\ 0 \end{pmatrix}
\begin{pmatrix} -2 \\ -3 \end{pmatrix}
\begin{pmatrix} 4 \\ -2 \end{pmatrix}
\begin{pmatrix} -4,5 \\ 2 \end{pmatrix}
→The substraction of vectors can be traced back to vector addition because \vec{\mathbf{a}}-\vec{\mathbf{b}}=\vec{\mathbf{a}}+(-\vec{\mathbf{b}}). Further information: see Simple arithmetic operations
Vector subtraction

6. Which of the following vectors forms the substraction \vec{\mathbf{a}}-\vec{\mathbf{b}}?

Vektorrechnung Aufgabe7.3.png


\begin{pmatrix} 0 \\ -6 \end{pmatrix}
\begin{pmatrix} 3 \\ -3 \end{pmatrix}
\begin{pmatrix} 0 \\ 8 \end{pmatrix}
\begin{pmatrix} -2,5 \\ 0,5 \end{pmatrix}
→Because the x-components are equal, the resulting x-component is zero. The y-component of the resulting vector is the substraction of the y-components. Further information: see Simple arithmetic operations

7. Which statement is true?

For the calculation of the difference vector \vec{\mathbf{a}}-\vec{\mathbf{b}} first the vector -\vec{\mathbf{a}} is formed by inverting the direction of \vec{\mathbf{a}}.
For the calculation of the difference vector \vec{\mathbf{a}}-\vec{\mathbf{b}} first the vector -\vec{\mathbf{b}} is formed by inverting the direction of \vec{\mathbf{a}}.
For the calculation of the difference vector \vec{\mathbf{a}}-\vec{\mathbf{b}} first the vector -\vec{\mathbf{b}} is formed by inverting the direction of \vec{\mathbf{b}}.
→The substraction of vectors can be traced back to vector addition because \vec{\mathbf{a}}-\vec{\mathbf{b}}=\vec{\mathbf{a}}+(-\vec{\mathbf{b}}). Further information: see Simple arithmetic operations
Vector substraction

8. Fill-in-the-blank text:

Fill in the following words:

negative value, same direction, zero vector, factor

Multiplying a vector \vec{\mathbf{a}} by a real value p results in a vector \vec{\mathbf{a}}_p with and different magnitude. The magnitude changes with {p}. If the resulting vector \vec{\mathbf{a}}_p has an oppsite direction, p is a . The special case p=0 results in a .

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