Difference between revisions of "Selftest: Matrix multiplication with a scalar"
From Robotics
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+ There is no <math>c</math> | + There is no <math>c</math> | ||
||<math>c=2</math> would be correct for all the green colored components in the result matrix: <math>\left[\begin{array}{ccc}{\color{Green}2}&{\color{Green}4}&{\color{Green}6}\\{\color{Green}0}&{\color{Green}4}&{\color{Green}2}\\{\color{Red}3}&{\color{Green}6}&{\color{Green}0}\end{array}\right]</math>. For the red component <math>c=1.5</math> would be right. So there is no general <math>c</math>, that holds for all the components. | ||<math>c=2</math> would be correct for all the green colored components in the result matrix: <math>\left[\begin{array}{ccc}{\color{Green}2}&{\color{Green}4}&{\color{Green}6}\\{\color{Green}0}&{\color{Green}4}&{\color{Green}2}\\{\color{Red}3}&{\color{Green}6}&{\color{Green}0}\end{array}\right]</math>. For the red component <math>c=1.5</math> would be right. So there is no general <math>c</math>, that holds for all the components. | ||
| + | |||
| + | <quiz display=simple> | ||
| + | {'''Is there any scalar constant <math>c</math>, so that the following equation holds?'''<br/> | ||
| + | :<math> | ||
| + | \left[\begin{array}{ccc}2&0&1\\0&2&3\\1&1&0\end{array}\right]\cdot c = \left[\begin{array}{ccc}4&0&2\\0&4&6\\2&2&0\end{array}\right] | ||
| + | </math><br/> | ||
| + | | typ="()" } | ||
| + | - <math>c=0.5</math> | ||
| + | - <math>c=1</math> | ||
| + | + <math>c=2</math> | ||
| + | - There is no <math>c</math> | ||
| + | ||If all the components of the left matrix are multiplied by <math>c=2</math>, it results in the right matrix. | ||
{'''Is the following equation correct?'''<br/> | {'''Is the following equation correct?'''<br/> | ||
Revision as of 16:21, 18 June 2014
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