## Question about #9 practice midterm |
Jackhe |
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In the problem we are given \begin{pmatrix} 1\\ 0\\ 2 \end{pmatrix}\begin{pmatrix} 2\\ 3\\ 2 \end{pmatrix}\begin{pmatrix} 2\\ 1\\ 0 \end{pmatrix} does this mean that \begin{pmatrix} 1 & 2 & 2\\ 0 & 3 & 1\\ 2 & 2 & 0 \end{pmatrix} is the identity matrix from \beta to \gamma ? and if so is the inverse equal to the identiry matrix from \gamma to \beta ? |

TLindberg: May 5, 2015, 12:24 a.m. |
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I'm pretty sure that's the idea. That's how I handled it when I did the question. |

aleksanderjanczewski: May 5, 2015, 12:40 a.m. |
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Yes, I did the same |

akaur: May 5, 2015, 11:34 a.m. |
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Yes, that's what I used as well |

James_Kercheval: May 5, 2015, 3:53 p.m. |
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I did the same thing. I found the inverse of this identity matrix and multiplied it the beta to beta matrix then the multiplied it by our friend's basis. The multiplicative of these three matrices is the answer. Note that the multiplication of these matrices is associative but not commutative. |

shellyhuo: May 5, 2015, 4:01 p.m. |
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Yes. |

danielkim: May 5, 2015, 4:28 p.m. |
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I got this for my answer. First row of the 3x3 matrix: (27/5 , 1/5, 12/5). Second row: (17/5, 11/5, 19/5). Third row: (16/5, -2/5, 17/5) |